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Okay it's, it's ten past and we're probably going to be sparsely attended due
to the weather so let's . So here we are, first winter storm, you guys already got
your midterms back, so what do we have like twenty people out of 100 we're about
to, take off in a very different direction, talking about set theory and
propositional logic. Set theory. How many of you guys had an introduction to sets in
like elementary school or middle school, whatever it's called these days? A few
people. So I grew up with what they called the New Math, which was a lot of set
theory in, you know, third and forth grade or something like that. I'm not quite sure
why it was called the New Math, but I find this stuff really helps to organize ones
thinking about a variety of things that are, that are very important for counting
and for probability, and its very easy to get set theoretic notions inside out. And
so we're going to work through a bunch of this stuff, set it up with a little bit
formalism and a little bit of notation. But trying to illustrate things with
examples so that it all makes sense, to the extent that this is a review you can
you know, nod off or encourage me to go faster or whatever you think makes sense.
I don't mean to torture you with stuff that you've already, that you already have
see and understand well, so we'll just see how it goes. We're going to talk about
sets and relations between sets. The fundamental thing we start with is sort of
a universe of objects, and we're going to call that S. And do, do you. Venn diagrams
familiar for, to everybody? Yeah? Okay do Venn diagrams. So Venn diagrams are part
of the new math right. This is part of, part of the same thing. Okay so the, the
set of things that, that were, that are under consideration, the things under
discussion, the universal set is going to be called S. And when we start talking
about probability and random events, S is going to be called the outcome spaces. The
set of all things that could happen. It, as the result of a particular experiment.
the experim ent might be tossing a coin once. What could happen? Well, the coin
could land heads, the coin could land tails. And then if you really want to be
pointy headed the coin could land on it's edge or the coin might not land at all.
Right? The those are pretty much all the possibilities. Okay, so S is the set of
things we're, we're talking about. And the, kind of the most fundamental notion
we have is to represent that. Something is an element of this set. So a set is a
collection of things without regard for their order. so this, this symbol, this
funny E is pronounced, is an element of or is a member of, or is in. So this is
pronounced as x is in A. x is a member of the Set A. x is an element of A. so here
is the idea that we have some set A, it's a subset of the universal set S, it's you
know, it's a subswet of all the things are under consideration and here's the point
x, x is an element of A. x is an A. what is mean for two sets to be equal, it means
that they have the same numbers. Or, again order doesn't matter, so if two sets have
the same numbers, they are the same set and in some philosophic you know, sense,
there is only one set with a given collection of members. Every other set
that has the collection, same collection of numbers isn't a different set it's that
same set. So in particular a set, we can, we can write a set by listing it's
elements so here's a particular set, its elements are little a little b and little
c , that is one in the same set as the set that we make by listing the elements in
this order because it contains exactly the same things. So one, one way of writing a
set is to put it's elements in curly braces. Another way, just listing its
elements. Another way of specifying a set is to say, I'm considering all things that
satisfy particular property, so all things such that some property is true. So I
might list, I might have a set that is the even numbers less than or equal to ten.
And I can either write that as, you know, zero, two, four, six, eight, ten or I
could write that as x, such as x is greater than zero, x is even and x is less
than ten, right? It's the same set. Just a different way of describing it. One case
you list the elements, the other case you list some properties that the elements
satisfy. Okay. the compliment of a set is the collection of things that aren't in
the set. So if this is the set A, the, the yellow region, then the blue region
outside is A compliment. It's the stuff that isn't in A. So it's all the elements
of the universal set under consideration that are not elements of this particular
set we're talking about. . There is a special symbol for this set that doesn't
have any elements at all. It's the empty set. Either write it with just two curly
braces and nothing in between them. Or this, this thing that's like a Pi. , the,
the empty set is the complement of the universal set. What, what are all those
things that are not in S? Well, there isn't anything that isn't in S because
that's everything that's under consideration. So what isn't in S is
empty. There, there's. It has no elements. . so we'd write that S compliment as the
empty set, the compliment of the empty set is S, right? Everything that, that, that
you know, what, what's the compliment of nothing? Everything. Okay if a very
important relationship between two sets is the subset relationship. A is a subset of
B if every element of A is also an element of B. And you can write that two ways.
This is writing it, A is a subset of B and this is writing the other way around, B is
a superset of A or B contains A. So both of these mean that every element of A is a
element of B. It's a containment relationship. Containment is different
from the little E is an element of. Right? Is an element of, for A to be an element
of B, B would have to be a collection of things that includes A as one of its
members. This relationship with the subset means that everything that is an element
of a is also an element of B. It doesn't mean that a itself is an element of B. .
Okay the way the Venn diagram per set containment s ubset relation looks is like
this, A is a subset of B because every element of A is also an element of B. The
empty set is a subset of every set, okay? Because all of its elements are in every
other set. It doesn't have any elements. It's not very hard for its elements to be
in every other set, right? . Okay. The subset relationship is transitive. So,
everybody remember what transitive means? The relation is transitive if you know, if
A bears a certain, bears, bears the relation to B, and B bears the relation to
C, then A bears the relation to C. So set containment is like that. If A is a subset
of B and B is a subset of C, then A is a subset of C. So, other things that are
like that are things like inequality. If A is less than you know, if, if the number X
is less than Y and Y is less than Z, then X is less than Z. In, in, inequality
satisfy the transitive relationship. okay so I mean just to put this into words,
this example. Every, every raven is a bird. Every bird is an animal. So every
raven is an animal. So a set of ravens is a subset of a set of birds. A set of birds
is a subset of a set of animals. So a set of ravens is a subset of a set of animals.
Now if you take complements, that actually reverses the subset relationship. So, if
every A is a B, then every not B is a not A. Okay if, if A is a subset of B, the B
compliment is a subset of A compliment. So if every raven is a bird, then every
non-bird is a non-raven. You can't be a raven without being a bird, so if you
aren't a bird, you're not a raven. Does that make sense? It, it flips it around.
Alright this is the negation of the subset relationship, it's the subset with the
slash through it. A is not a subset of B, means that, there is some element of a
that is not an element of B. And that is also reversed under set under compliments.
So if A is not a subset of B, then B compliment is not a subset of A
compliment. Alright. Somebody come up with an example other than the one that's
written down here. No, you talk. So you want to start with not e very something is
a something. Sorry. Not every something is a something else. Not every food is a
hamburger. Okay. So. . . Alright. So not every, not every non hamburger is a non
food right? So if not all food is hamburger and so, not all non hamburger
isn't food. There are some, there are some foods that are not hamburgers. So there
are non, there are some not hamburgers that are foods. Right? That's, that's the
rhythm of it. . Okay, I mentioned this before. But you know, A is not the same
thing as the set that contains A. Alright? These are two, there's, there's this great
joke. What-, what's the difference between zero and nothing? Like that's zero. That's
nothing. Right, I mean it, a set is not the same as the elements as it contains,
so A is an element of the set that contains a as an element, but A is not a
subset of a set that contains a as an element. Okay. The set that contains A as
a subset of the set that contains A. this thing is a set with one element A.
Alright. Intersection and union are the next two topics. I'm sure you've seen
these before. The intersection of two sets is the set of things they have in common.
You know, all the members that they have in common. So in particular, if we had
this set zero, one, two, three. And this set one, three, five. What elements do
they have in common? They share the elements one and three. The intersection
is the set containing one and three. In a Venn diagram, it's the overlap between the
two sets. So if yellow is the set A and blue is the set b, the intersection is the
green, A intersect B. If you take the intersection of the empty set and any
other set, what you get the empty set. What members do they have in common. None
right, because that's all that the empty set has in the first place. If you take
the intersection of S and any other set you get back that set because all of its
elements are already elements of S. When you take the intersection with S you don't
lose anything. If we're talking about the set of all foods, and A is the set of all
hamburg ers, then depending on what you think about people's cooking, you know, if
you take intersection of food and hamburger, you still get back hamburgers,
right? Not necessarily edible hamburger, but hamburger. Okay, Oops. I'm miss,
missing a. Oh. No. I'm not missing . Okay. intersection is associative. So, forget
about these auto parenthesis. They're just blocking of a positive phrase here. But A
intersect B intersect C is the same as A intersect B intersect C. And so there's no
ambiguity in just writing A intersect B intersect C. When we start mixing unions
in our sections, that's not true. Oh its true for intersections, its also going to
be true for unions, but its not true for mixtures up under rules for how you
distribute intersection over union and union over intersection. Alright. Why this
have to be true? Well this is, these are all the things, all the elements that are
shared by B and C and then these are those that are also shared by A. So these are
things that are shared by A, B and C. These are things showed by A and B but
they are more insisting that they are also be shared by C. So its the same set of
things. if things are shared by A and B they're shared by A and B. They're shared
by B and A. There's no, the, You, you can reverse the order. It doesn't change. The,
the elements that you get doesn't change the elements they have in common. Okay,
unions. the union of two sets. Is the set of all things that are in either or both
of the two sets. So, in Venn diagram it's this. If this is A, and this is B. Then
their union is anything. It is an either A, or B. Or both, A and B. if I take the
union of the empty set and some other set, it doesn't change it because I haven't had
any, I haven't added any elements since the, the empty set doesn't have any
elements in the first place. If I take the union of S and any other set, I just get
back s because S already has everything in it. I'm not adding anything when I take
the union. Of a with it. as I mentioned before unions are associative and
commutative so A union, B union, C is the same as A union, B union, C etcetera.
These are all the things that are in any of these. Any there either in A or B or C
or in any combination of those. Excuse me. okay. Something I forgot to say before
about subsets is that if A is a subset of B and B is a subset of A, then A and B are
equal. Okay? The set is a subset of another and vice versa then they must have
exactly the same elements. Every element of one is an element of the other. if A is
a subset of B, then what happens when you take the union? Well, every element of A
is already an element of B, so when I take the union I haven't added anything to B
that wasn't already there, I just get back B, this make sense to everybody? We're
going kind of quickly but I'm assuming this is mostly review, so far so good?
Slow me down, good time to ask questions. Okay I mentioned before, alright the. What
happens if I take the complement of an intersection? The answer is, I get the
union of the complement. So let's just try to think this through. If we wanted to be
really clever, we could click the footnote and get an example. All right. So let's
give an example of that in words. What's a set A? Pick something. People who, people
who take step classes. Okay. And B could be people who go on to get business
degrees for example? Okay. So, A intersect B are those people who take step classes
and want to get business degrees? Okay. What's the compliment of that? How do you
end up not being in that set? So you can either, you can, you can be in the
compliment of the set, either because you don't take a stack class or because you
don't go on to get a business degree. But that's exactly this, these are people who
don't take stat classes, these are people who don't get business degrees. Okay, so,
either of those. And it, and, and, there's the, there could be overlap between these
two things right. There could be people who, who don't take stat classes and don't
go on to get business degrees or there could be people who do take stat classes
but don 't get business degrees. I don't know if there are any people who do get
business degrees but don't take stat classes. But that's a separate issue.
Alright, so, the compliment of an intersection is the union of the
compliments. And, what about this? People who either take stat classes or get
business degrees. We want to take the compliment of that. So that's got to be
people who neither take stat classes nor get business degrees. It's an intersection
of things. You have to both not take a stat class and not get a business degree.
So that's this. It's the intersection of these two things. Didn't take a stat class
and didn't get a business degree. Though the intersection can be pronounced, and,
right? It's, it's a conjunction of conditions. Is there a question, or just a
head scratch? Okay. . Union can be pronounced or, either of these has to
hold. And it's not a, it's not a, an exclusive or. It's an inclusive or. Either
this or that or both. Okay. Mentioned before, that. While, intersection, is
associative, and union is associative, they're not sort of associative across
each other. So what happens if I take the intersection of A, with the union of B and
C? So we need C. What is C? Goes, does graduate work? Alright. So we've got, took
a stat class and, either gets a business degree or does graduate work. Okay. So, we
want our, this is equivalent to, you have to either have taken a Stat class and
gotten a business degree or taken a Stat class and gone on to do graduate work. Or
both. Okay, , so these are people who take stat classes and get business degrees.
These are people who take stat classes and go on to do graduate work. This, this
makes sense, so we've written this a intersect a union as a union of two
intersections. Okay. Similarly, those people who take a stat class or both get a
business degree and go on and do graduate work. And this is saying that that's,
that's people who take a stat class, or get a business degree and either take a
stat class or, go on to do graduate work. And they can do, they can do both. They
can do both, yeah. Sorry? The, or always means this or this or both. It's things
that is in either or both of these things. The union always means that. This means it
has to be in both. Alright, so this just sort of make sense I mean let's suppose
that... I mean you want to prove this stuff mathematically what you do is you
say okay suppose that there is a element x in this set. Show that it must be in this
set. Conversely suppose there's an element Y of this set show that it must be in this
set. That, that's the way you'd prove that, that two sets are equal. That every
element of one is an element of the other. Every element of the second is an element
of the first. That's doing this, this sort of thing I said if A is A subset of B and
B is A subset of A, then a and b are equal. That, that's the way you would
prove this. So if we have an element that, here, what do we know about it? What we
know that either it is an a or it's in both of these. Well if it's an A, then
it's an A, so it's on both sides of this intersection and so it must be over here.
Right? Because both sides of this intersection include A so this thing,
that's so, anything that's an A is on this side. Well, what if it isn't an A? What if
it's a B intersect C? That's in both B and C then what happens on the right? It's got
to be over here, because its in B. And it's got to be over here, because it's in
C. So it's got to be in both, so it's got to be in the intersection of these two
things. But that's the, the rhythm of how you prove these things. Alright, now we're
going to talk about some labels for relationships. Disjoint or mutually
exclusive, so those are synonyms. Two sets are mutually exclusive if they don't have
any elements in common. Another way of saying that is if their intersection is
the empty set. Your intersection is the empty set they don't have any elements in
common. So that's another way of writing that two sets are disjoint, the
intersection is empty. . Here's a Venn diagram of what that looks like. They
don't overlap. They don't have any elements in common. We talk about a
collection of sets being disjoined if every pair of sets in that collection is
disjoined. So one way of writing an intersection is with this, this cap if you
like but you'd also if you write two sets down next to each other it's an equivalent
way of expression this. This mean AI intersect AJ. And so, this collection a1,
a2, a3 etc, is disjoint if every. Pair AI and AJ, the intersection of AI and j is
empty unless I and J are equal. So the intersection of any set with itself is
just keeping back that original set which might or might not be empty, but when you
take an intersection of any pair, any, any distinct pair of these, you get the empty
set. So for instance oh I didn't, I've changed, fixed the typos this, this
morning but did not upload the file yet, there should be commas here. So here is a
list of four sets, so this is A1, A2, A3 and A4. This collection of sets is
disjoint because if I look... Does any other, this set has the elements one, two
and three. Are those elements in any of these other sets? No. So if I take the
intersection of this one with any of the others, I get the empty set. Well this one
has the elements zero and four. Are those elements shared by any other set? So if I
take the intersection of A2 and any of the others, I get the empty set. Similarly,
the intersection of A3 and any of the others is empty, the intersection of A4
and any of the others is empty, so this is a disjoint collection of sets. All right.
The empty set and any other set are mutually exclusive, are disjoint because
when you take their intersection, you get back the empty set. Now. When we were
talking about counting, one of our strategies for counting was to take a big
collection of things, and sub divide it into non overlapping groups in such a way
that every element of the original set was in exactly one of those groups, okay?
We're going to get at a more mathematical set of terms to describe that situation.
It's going to end up being useful for fin ding probabilities of complicated events
as well. So the first notion is that of a collection being exhaustive. or, and then
from that we're going to talk about partitions. So, a collection of sets
exhausts another set if every element of, of that second set is in at least one of
the other sets. So, for example, the set of even numbers and the set of odd numbers
exhausts the set of integers. Okay, together every integer is in at least one
of those other sets. In fact, it's exactly in one of those sets. alright To be
exhausted, but things don't need to be disjoint. You could have, you could have,
For example, I think I, I think I give this example there. Yeah, so these sets.
1,, two, three, one, four, three, five and -one, -two, -three, -four, exhaust that
set. Exhaust two, three, four, 5,/ Those sets are not disjoint. one, two, three has
an intersection with this. It also has an intersection with this non-empty
intersection with that. Right? So the element. One appears in more than one of
those sets. The element two appears in more than one of those sets, but every
element of this set appears in at least one of those collections, one member of
those, that collection, yes. so is S equal to all of those different numbers or
that's equal to ? S isn't any of these. S might be the set of integers or something
like that in this case. So what we're saying here is that this is A. And this
could be A1, A2, A2, A3, A4. And so every element of A is in at least one of the
other sets. A1, two, three or four. Okay? So we've c-, we've exhausted, you know,
the, the, the, together, if you go through all of the elements of A1, two, three
four, etcetera you have found all of the elements of the set A, and then some. And
you might have found some of them more than once. But it exhausts it. . Partition
is, closer to what we were talking about for that strategy for counting. The idea
of a partition is that every element of the set a occurs in exactly one of the
other things. What's the question? Yeah. I was wondering about the negative, the
negative. Is that the fourth set? Yeah. How does that ex haust the. Okay, what's
involved in exhausting this set is this collection, so I should probably, I'm
going to let me just write this down. So we have A1. , I1, is equal to one, two,
3.1, A2 is equal one, four. A33, is equal to five. I'm sorry, three, five. Could you
see that through here? Or is this, is the lectern in the way? A4 is equal to -one,
-two, -three, -four. And A is equal to two, three, four, five. That brings
specific reason that, that that's what A is? Well, its just what I've. This reason
right there? Yeah, but that, that's what. Yeah, I'm. So the assertion is that these
four sets exhaust this set. What that means is that every element of this set is
in at least one of those. And that set's in. The, yeah. But all of, the implicit in
this is that there is some universe S which might, for the sake of argument, be
the positive and negative integers, okay. So all the integers, . So. Is the first
element in one of, in at least one of these sets. Yes, it's in this one, okay?
What about the second element, three. That's in this one and in this one. four,
that's in this one, five, that's in this one. So every element of this is in at
least one of those. This one isn't helping. Oh. Okay, but, but this
collection still exhausts the set A. So you can have other ones too, They just,
they don't help. Yeah. Is there a way to show a Venn Diagram of that or is that
not. Sure. what's it's basically saying is that A is a subset of A1, union A2, union
A3, union A4. So Venn diagram for this might look something like the following.
here's A. Here's A1. Here's A2. A2 has an interception with that, so we should draw
it that way. A3 intersects the first but not the second. So we could draw A3 like
this. Oops. I've got to draw it so that it includes the rest of A3 and then A4 is
over here somewhere. Okay, so a4 doesn't share any elements with the set A. And
that in fact doesn't ensure any elements with the sets A1, two or three either,
right? but A is contained in the union of A1 through A4. All of those numbers have
to be. Everything at A has to be at least one of the other g uys. Okay. so partition
is saying that every element of the set A is in exactly one of the others. And,
there's nothing else. That they don't together include anything that is in set
A. So basically what that's saying is that you're carving up the set A. So, again, if
here's S, and here's some set A. Then, to partition A is basically to just divide it
into some pieces. So that this might be A1, A2, A3, A4, A5. So that these pieces
are, are disjoint. They don't have any elements in common. And their union, is
all of A. So a different way to write that would be A is equal to the union over J,
of AJ, so that's like A1, union A2, union A3, etcetera. And AI, AJ is empty for I
not equal to J. Yeah. Well it's hard to draw this, but I was intending for this
not to be overlapped with but be as separate piece. So, I mean, maybe I should
draw it this way. Is that easier? The question is, how do I know A5 is not part
of A4. I, there isn't a good way to represent it graphically. But the idea
was, that they don't share any elements in common. Uh-huh Okay. So, What's. Okay so,
the difference between a collection exhausting some other set and the
collection being a partition of some other set, has to do with whether those sets can
overlap. And whether, they contain anything that isn't in that set that
you're, that, the, the, the second set, the set so the set of integers can be, it,
it is exhausted by the set of integers that are divisible by that are either
prime, or are divisible by two, three, four, five, six, seven, eight, nine, you
know etc, right? They're either, they're either prime, or they're divisible by two
or they're divisible by three or they're divisible and so on... But many numbers
are getting occur in more than one of those sets. Right, that, that the numbers
that are divisible by four are also divisible by two. Right, so that's, that
exhausts the set of integers, but it's not a partition of a set of integers, because
you're getting some numbers many, many, many, many times once in each of, in each
of, you know , in a bunch of these subsets. Okay, a partition is saying that
everything occurs in exactly one. Of those subsets. And, there's nothing else. You
haven't included anything that isn't in the set A, when you take the union. Let's
look at these properties, so. This is just to summarise what we just did. if A is a
subset of B and B is a subset of C, then A is a subset of C. Taking compliments
reverses said inclusion, so if A is a subset of B then, B compliment is a subset
of A compliment. If every A is a B, then every non B is a non A, . if I take the
intersection of two things, what I get is going to be a subset of either of them.
Right, when I take the intersection. At, at best, I'm not losing anything. Because
every element of a is already an element of b. But if not every element of a is an
element of b then I've thrown something away by taking the intersection. I've lost
something. And so what I end up with can't be any bigger than what I started with.
That's what this is saying. . in fact, I only end up with what I started with if A
was already a subset of B. So if every element of a was already A subset of B,
then when I take the intersection, I haven't lost anything. Any elements of A.
All right. Similarly, unions are including more stuff. So, when I include the stuff
that's in B as well as the stuff that's in A, I still have all the stuff that's in A.
I can only have gained things, so A is a subset of A union B. these rules for
taking complements of unions and their sections, the complement. You know, if
something isn't in both A and B, either it isn't in A, or it isn't in B. If something
isn't in either A or B, then it isn't in A and it isn't in B. . Okay. A union B is
the same as B union A. It's the stuff that's in both, sorry, it's the stuff
that's in either. then we have these, associative rules. If it's in A and, B and
C, then it's in A and B, and C. If it's in A or, B or C, then it's in A or B, or C.
If it's in A and, B or C, then either it's in A and B, or it's in A and C, or both.
If it's an A or B and C, it's either in, it's both in A or B and it's in A or C.
Okay. So, we're going to do some examples. Why don't we take a two minute break, and
then we'll, we'll do some examples. This is getting closer to the homework and
stuff, so. And is it subset of a? Sure. Yes. Yep. The, these, these symbols are
arbitrary, so the, this is also true. If it's in both A and B, it's in B. If it's
in both A and B, it's in A. Yep. . Yeah. . . Sure. . It was just that like UFOs. .
Yeah. . It, It said and there have been no reported sightings. And therefore . No.
It, it said that But. Yeah. Basically nobody has proved they exist, therefore,
they don't exist. We can look at the specific answers to that one, but I think
that. Yeah. Okay, lets, Let's start again. All right. So this is an add collection of
things. Dog, Fred, albatross, income tax and cancer is the set A. And the set B is
cancer, income tax, albatross, Fred and dog. What's the difference between those
two sets? Is dog in both of them? Mm-hm. Is Fred in both of them? Yeah. Albatross?
Yeah. Income tax? Mm-hm. Cancer? No. Left anything out? No. Those are the same set.
Right. A and B are one and the same set. Okay? The, the art, the elements are
listed in a different order, but order doesn't matter because we're talking about
collections, sets. Okay, so, which of these things is true? Is, is, is A not
equal to B? No. No, okay, that's false. Is A-union-A equal to A? That's always true.
The union of set in itself is always the original set. You're taking all the things
that are either in A or in A or both. Right? That's just the things that are in
A. So that's, that's got to be true. Is A compliment or sub set of B compliment. A
compliment is in fact, equal to B compliment so, it's certainly a subset. Is
a equal to a? Yeah, we like that. Is A, whoops, A union B equal to A? Yup cuz
that's really the same as A union A. Is A, a subset of B? Yes. Yup, every set is a
subset of itself. A intersect B, is that equal to B? Yes it is. Yup, cause A
intersect I'm sorry that's A union or A inter, A intersect. Okay A intersect A is
A. It's always true that the intersection of a set in itself is the original set. A
compliment is equal to B compliment. . Is that ? I don't know. My vision isn't good
enough. . . J, J is not a signif- the one above is equal. Okay. J is not a subset so
that's false, right? Because in fact A is a subset of B. Is A complement a subset of
A complement? Every set is a subset of itself. Okay this one is false because it
is in fact a subset. A is always a subset of itself. A equals B. Yup, B is a subset
of A. Yup. Okay. So this should be right, yes? We helped. Alright. Why you not L
again? L? So, I don't know if you can see that but there's a slash through the
subset sign. It's saying A compliment is not a subset of B compliment. Can you see
that? Okay, yeah it's *** the screen. okay. Which of these are always true? A
union B is a subset of D. Right, okay. When I, when I take B and then I include,
in addition to B, all the things, things that are in A. What I end up with might
not all B in B cuz there might be some elements of A that are not in B. In which
case, A union B has stuff that isn't in B. Alright? So that's not necessarily true.
Alright, if A union B equals C, and A intersect C is empty, then A is empty.
Okay, everything that's in C is either in A or B or both. Right? So what part of C
came from A? Let's put it this way. If, if, if A, if C is equal to A union B, then
what's the relationship of A to C? It's a subset right? So, if I take the
intersection of a subset with something that, that it is a subset of what do I get
back? Same thing. So A intersects C is A. Right, so this is saying, this is saying
that A is empty, therefore A is empty. That makes sense? No? Not yet. Okay Wait .
Could you zoom in . The problem is that when I zoom in it, it, it runs off the,
the edge of the you don't get to see the whole answer. Oh. it's a, a limitation of
the browser. if I hit control. And then it's like, kind of like. Hm. I'm not sure
how to do it. You want to show, want to show, come show me on my? Yeah. Yeah. Let
me squeeze this guy a little bit. okay, great. Awesome. Terrific. So you just hold
it down and then move it up or down. Yeah. Excellent. Woo-hoo, thank you, new tricks.
Okay. Alright, so we're looking at B the item B. And what's going on? . Well, I've
got that C is equal to A union B. Okay? So it follows from that, that A is a subset
of C. Right? Everything that is in, that is in A is in C. Also, everything that is
in B is in C. Okay? So, if A is a subset of C, then what do I know? I know that A
intersect C is = to A. And so picture for this, A, B and C is all of this. C is A .
C is A union B. Yeah, okay so I know that A is itself a subset of that, so if I take
A intersect that, I just get back A. Yep. Is it a union B. C is the union of A and
B. So C is all those things that are either in A, or B, or both. And all of
this is. Cuz like, what if it's, what if C is just A And not B? Well, if this is
saying it isn't, this is saying that, that eh, everything that it, what does C
consist of? It consists of all those things that are either in A or in B or
both. That, this is, this is, this is defining C to be that. Okay. So eh, you
know, if, if, if A were the numbers zero, one, two and B were the numbers two,
three, four, then C would be zero, one, two, three, four. It's all the things that
are in either or both. Okay. But now when I take the union of those things. So I, I
know that every element of A is in here, because this is a union. It's, it's got
every element of A in it. So if I, if I take the intersection of A, remember, if,
if I have, if I have a situation where Q is a subset of R. If, then Q intersect R.
Write it that way, is Q. Right? If, if every element of Q is an element of R.
Then when I take the intersection. I haven't taken anything away from Q. I just
get Q back. Okay. So here. I know that every element of A, is an element of C,
because C is defined to contain everything that's in A, and other stuff. So, if I
take the intersection of A and C, I got to get bac k A, but this says, that when I
take the intersection of A and C, I get back nothing. Okay. So that means nothing
is A. So A must be empty. So far, so good? Yes, the question was did that mean C is
equal to B and the answer is yes. And, it could be that B is empty also. Could be
that C is empty. But that, that's certainly true. Okay, if A is a sub set of
B, then A union B is equal to D.Is that always true? Saying, everything that's in
A is in B. So everything that's in A or B, already includes the stuff that's in A.
So, we have them change it. Yes? How can we . Okay. So subsets can be empty. So I,
I can, I, the question was how does it make sense for the intersection with the
subset to be empty and it, it, it, basically, the scenario you know, here
would be, this is. A is the empty set, , B equals that then C is equal to AU, B is
equal to two, three, four. This is still true, this is equal to A and that is equal
to in fact the empty set. Drawn more elegantly by somebody with better
penmanship than I have. I got Cs in handwriting as a kid. It never got better.
I'm sorry for you guys. I try. Okay. Does that? So the empty set is a subset of
every set. Okay, you're willing to take it, Uh-huh. All right. Okay, so we agreed
on this one. If A is a subset of B then A union B equals A. Is that true? No. Okay,
A is a subset of B then everything that's an A is already included here, but there
could be stuff that's in B but not in A. So that's not true. If A intersect B is
empty. Another way of saying that is, is if A and B are mutually exclusive, or if A
and B are just joint. And if B and C are mutually exclusive, then A and C are
mutually exclusive. Okay, here, we can, we can come up with a pretty simple
counter-example of that, to that. Right? Let's lift. . . A is the Set one. B is the
Set zero. C is the Set one. Right? These have no elements in common, these have no
elements in common, these are in fact the same set. That's not true. A is a subset
of A intersect B. No. It's the other way around. Right? A intersect B is a subs et
of A. When you, when you take A intersect B, you can lose stuff that was in A,
because it happen, it's only the things that are in both A and B. Okay? So, A is
potentially larger, has potentially more elements. Okay. A is subset of A union B.
Right? Yeah. That's, that's going to be true, because the right-hand side is,
includes everything that's in A, and possibly more, if B has stuff that isn't
in A. a intersect b is a subset of a. Yes, okay, cuz when we take intersect b, we
have only things that are in a, but perhaps not all of them. Because we only
get those things in a that are also in b. So, what we end up left over with is
certainly n a. . If A union B equals A, then B is a subset of A. If when we take
everything that's in A and add to it, include with it, everything that's in B.
And by doing that we haven't gotten anything more then we had in the first
place. Then it must be the case that every element of B was already an element of A.
Right, so this ones true. Okay. If A is a subset of B, and B is a subset of C, then
A is a subset of C. If every element of A is an element of B. and everything in b is
in c, then everything in a is in c. This is the transitivity property, so this is
true as well. somewhere over here, this better give me a green check off-screen.
Control. Whoo-hoo. Alright. Okay. Now it's going to get really fun. , fun. Okay.
Oops. Control. Zoom. Okay s is spherical objects. B is ping-pong balls. Whoops. C
is basketballs. No. A is a set of. A is balls. B is ping pong balls. C is
basketballs. Okay. And we're pretending that basketballs are perfect spheres for
the sake of argument. If. Alright, is the set so, here, I mean, just. it's probably
pretty obvious. This means B complement. That, that's a subset mark. But this is
supposed to be. If I could make it a superscript, I would. But, you can't do
that in HTML, in, in the body of a, of a multiple selection thing. So this is B
complement. So this is saying A, complement is a subset of B complement, .
So let's look at the first one. Is B is a subset of C compliment? So, is, are, by
set of all ping pong balls, are subset of the things that are not basketballs. There
are spherical objects that are. They're spherical objects that are not basketballs
or ping pong balls or balls. Right? And this is saying this is basically saying:
are there any basketballs that are ping pong balls? No. Okay. Every ping pong ball
is a non-basketball. Yeah. Right? That's another way of saying this. Yes? Okay. So
that's true. Okay. So if that's true the next one's got to be false. Right? That's
easy enough. Okay. So, C is, say, is asking, If I look at the things that are
either balls or ping pong balls, are they all, ping pong balls? No. No, okay. if I
look at things that are either, that are both balls and ping pong balls, are they
balls? Yes. Yes. Okay. Because, we've, for the sake of arguing, you know, for the
missing sample, every ping pong ball is a ball. Alright, are ping pong balls a sub
set of the set of balls? Ev-, every ping pong ball is a ball. That's this one. Is
every ping pong ball a basket ball? No. No. Are there ping pong balls that are not
balls? That's what G's asking, right? people getting this translation of symbols
to words, cuz that's really what this is about, right? The logic isn't hard, it's
the translation that's hard. Yes. Yes, so D what in fact is true is that A intersect
B is B. Oh. I am sorry so D is wrong yeah so did I. Do, d, d. Eight or. . This is
false. It should be. Why is ? That's false, I didn't, I think I read it as
union rather than intersection. Yeah. It, that, that would be true if. So, what is A
intersect B? It's the set of all balls that are ping pong balls. That is, it's
the set of all ping pong balls. There are balls that are not ping, ping balls. So A
intersect B is not a set up of balls, its a set up of ping pong balls, its a smaller
set. Right. So, yeah. Thank you. . But, but, wait. In D there's the circle and
then. Circles his name. What shares there . Right. The intersect means it has to be
both. Right. Okay. So for something to be both a ball and the ping pong ball, it has
to be a ping pong ball. It can't be a basketball, for example. Mm-hm. So A
intersect B doesn't include basketballs. A includes basketballs. A intersect B does
not include basketballs. Does. But any besides B cannot be . Intersect B is, is
ping pong balls. It's exactly the set of ping pong balls. But the idea is not to.
They, they, they both share the, the same objects in, in, inside it's circle. But
not all of them. Right, oh. The relationship we have is here's, here's the
set of all spherical objects. Okay? Here's the set of all balls. This is A. Here's
the set of all ping pong balls. Here's the set of all basket balls. Every basketball
is a ball, every basketball is a sphere, every ball is a sphere, every ping pong
ball is a ball, every ping pong ball is a sphere, but. Not everything that is a ping
pong ball. . No, every ping pong ball is a sphere, but not every, is a ball, but not
every ball is a ping pong ball. Right. Right, that's, that's what's going on in
D. Does that make sense? Okay. But, A. The in section D it's a. It's part of it. The,
the one. D? We're look. Yeah. I'm talking about D. D. Okay. It wouldn't be true if
it was an element of A, right? A. no, it's, it's a subset. You mean a subset? So
that mean. Yeah, a subset. Yeah A, A intersect B is a subset of A. Yeah. Yeah.
Yeah. Yep. that's true. So basically what's going on is going is when you take
A intersect B you have lost something from A. Right? You only, you, you, all you have
left are the ping pong balls. You have excluded the other balls. You don't get
back A. You get back only part of A. That part that is, that, that, that is, the
elements of A that are ping pong balls . Okay. B is not a subset of C. Right? F is
false. B is a subset of a so g is false. B intersects C is empty. Right? You, you
don't play ping pong with basketballs. Yeah. Okay. Okay, so that's H. Is every
ball a ping-pong ball? No. No. Okay, so I is false. now we've got, are those balls
that are ping-pong balls, p ing-pong balls and nothing else. Yes, okay so J is true.
Okay A complement not a sub set of B complement. Okay, what is this saying?
This is saying. A compliment is the set of non balls. B compliment is the set of non
ping pong balls. . Oh, wait. We got an inside out. Yes, that's exactly why this
is interesting. okay, which is a bigger set. A compliment or B compliment? B
compliment. B compliment is bigger. So the compliment, if you take a smaller set and
take its compliment, if, if B is a subset of A. Then a compliment is a subset of B
compliment, right? That taking compliments reverses the subset inclusion. Okay. So,
in this case, B is a subset of A, right? Ping pong balls are only some of the balls
that there are, okay. So. B compliment contains A compliment. A compliment is a
subset of B compliment. Every non ball is a non ping pong ball. Sorry. If you're
not, if you're not a ball, you're not a ping pong ball. Yup, okay. So, in fact,
this should be subset, not, not subset. So this is false, because it has the slash
through it. Oh. It's false. Yup. Okay. Is C a subset of A? Yes. Basketballs are
balls. If you take the balls and add to them the Ping-Pong balls, you end up with
the balls. Right. So this is true. is every spherical object either a ping pong
ball or a basketball? No. No. Okay. So N is false. Is every basketball a non ping
pong ball? Yes, okay. So looking at the picture, what is this saying? So, we're,
we're looking at O, be complement. Is everything out here that includes C? Let.
Can you go over add one more time? I'm confused. Sure. N? M. M. A union B is A.
Okay, so we're looking at M. A is set of all balls. B is the set of all ping pong
balls. So A is this, B is this. Every element of B is already in A. So when I
take their union, I haven't expanded it at all. I just got what I, I still have all,
every ping pong ball is already a ball. So I haven't added anything extra to the set
of balls. Can you say what O means in words? this is asking, is every basketball
a non ping pong ball? Okay. Okay. And in fact yes, there are no basketballs that
are ping pong balls. Okay. Okay. Alright, get in, over my head you're talking about
sports, you know it's just. . P, okay, is everything, every none ball a none ping
pong ball? Yes, okay that's true, right. If you get every ping pong ball was a ball
so every none ping pong ball, every non ball is a none ping pong ball. right. That
was what we were just doing? Okay. Not every ball is a basketball. So q is false.
not every ball is a ping pong ball so far is true. Right? Not every ball is a
basketball, so S is false. Every basketball is a non-ping pong ball. There
are no basketballs that are not non-ping pong balls. So T is false. T is saying,
saying that there are basketballs. That are. Ping pong balls, yes? Do I got this
backwards? That are not, not ping pong balls. So it's saying that there are
basketballs that are ping pong balls. Right? . Okay. . Then, If I take ping pong
balls and basket balls, have I got anything that isn't a ball? No. No. So
this is true. Everything that's either a ping pong ball or a basketball is a ball.
U is true. The. v is saying, that n-, not every ball is a basketball. That's true,
right? This is saying every ball is either a ping pong ball or a basketball. That's
false. Right? And this is saying that there are ping pong balls that are not
basketballs. So, that should be true. All right, we're going to run back through
these once more before we press, check answer. No. Enough's enough, okay. All
right. Okay. Alright. There's more. Well, I, I mean, does it help to think through
this as a group? Yes. Yes? Okay. So, let's, let's, let's do it. You must like
to code this stuff up. Okay, so e is the set of integers, f is the set of even
integers, g is the set of odd integers. Do the sets f and g exhaust the set e? So if
I look at the even integers and the odd integers, is every integer either even or
odd? Yes. Yes. Okay, so that should be true. Oops. Except that I should click
true if I say true. Okay. F and G are disjoint. Are ther e any numbers that are
both even and odd? No. Okay. So then, do the sets F and G, partition the set E. So,
every integer is either an even integer or an odd integer, it cannot be both. And
there are no numbers, there are, there are, there are no numbers besides integers
that are either even integers or odd integers. Right? We haven't included
anything beyond the set of integers, so this is true. Okay, all right, now we'll
blow this up again with the cool new trick that I've been taught. Except I need to go
sideways a little bit first. How much can I blow this up without? Running off the
screen. Okay, so now this is taking these set relations and expressing them in words
instead of in symbols. And again, a lot of this stuff is just about going back and
forth between words and symbols. And that's where it gets hard. So, if every a
is a b, what is that in symbols? A is a subset of B. Okay. What is that as a Venn
diagram? , right? That's what we're looking at okay? That every non-A is a
non-B. Are there, non A's that are B's? There is. Yeah. There's one. Right? Okay.
So A is false. So if x is in a, and x is in both b and c. Sorry, is in either b or
c. Then, either x is in a and b or, x is in a and c. Huh? No. Alright, so what,
what, what, what are we saying in symbols? It's. X is in. A intersect B, union C,
right? Okay, and what's going on, on the right hand side, is we're saying, does it
follow that X is in. A intersect B or A intersect C. Okay. And, and, and it does,
right? That's exactly how. Intersection distributes over a union. Picture. . Okay
So, here is. A, B, and I might as well make C here. C and, we're, we're being
told that x's and a intersect, via union c. So, what should I be shading? . Yes,
okay, so right in there. If I'm in there, then I'm I I'm either in a intersect b.
What's a intersect b? That's this. Or I'm in eight or sect c. That's this. Okay. I
could be in both, but I'm in at least one of those two things. Okay? So we got B is
true. . Alright. C effects isn't in A or B. That it isn't in A, it isn't in B. Yes?
If X isn't in both A and B, then X isn't in A and X isn't in B. No, it could fail
to be in A and B just because it isn't an A, or just because it isn't in B. It
doesn't have to fail to be in both. If X is in both A and B, or X is in both A and
C, then X must be in A. Yep. Oops, except that I have to click the right one. Okay,
if X isn't in A, or X isn't in B. Then x isn't in. A or b. Okay. This is false. . .
If X isn't in A and X isn't in B, then it's not in A union B. Oh, . If it X, if
it isn't in A or it isn't in B, then isn't in A intersected. Okay? But, for example,
X might. Let's say, X isn't in A, but X is in B. Then the first part satisfied, but
the last part isn't. Because is in A union B. Okay. So we're going to run out of time
momentarily. if X is in A intersect B or X is in A intersect C, then X is in B
intersect C. Mm, I, I don't see why it should be. If every A is a B, every non-B
is, is a non-A. Yep. That's true. That's just the reversal of set inclusion under
compliments. The X is in A, or X is in both, is in either B or C then X is in a
union B and X is a union C. No. If this were an or that would be true. If X is not
in both A and B, then either X is not in A or X is not in B. That's true? If X is in
A or X is in both B and C, then X is in either A or B and X is in either A or C. .
Okay, let's suppose that it's an a. Then it's in both sides, here. Suppose that it
isn't in a, but it's in both b and c. Then, it's in this side because it's in B
and this side because it's in C. Yes? Okay, so that would be true. Okay, see you
guys next time.