Hello everyone. In this video, we're going to do another if and only if proof.
This time we're going to use sets.
Prove that if…
or prove that A is a subset of B, subset or equal to B,
if and only if A intersect B is equal to A.
So again with an if and only if proof, we're going to prove
that the hypothesis implies the conclusion, and that the conclusion implies
the hypothesis. So the forward direction and the reverse direction.
For the forward direction,
assume
that A is a subset or equal to B.
Now we're trying to prove a set equality
and remember to prove a set equality, you need to show that every element
in the left-hand set is in the right-hand set, and every element in the right-hand set
is indeed a member of the left-hand set.
So to show…
what am I doing? So to show
A intersect B
is a subset of A, we note
that A intersect B is the set
of all elements
contained
in both
A and B,
and so
all elements
are contained
in A.
That's strictly by the definition. If you're in both A and B,
then you're most certainly inside A. So that's true,
A intersect B is definitely a subset or equal to A.
To show
A is a subset of A intersect B,
we're going to do this by showing an arbitrary element of A
must be inside A intersect B.
So I'm going to say let x be an element of A.
You could have done this for the other part as well, but the other part really does just follow
by writing out the sentence of what the set is.
So I'm going to take this set - I'm going to take an arbitrary element in A,
and I'm going to show that it lives in this right-hand set.
So how do I do that? I can't do that on its own because on its own that's not true,
but with the hypothesis I can get there.
So this statement was true no matter what the hypothesis was,
but this statement we need to use hypothesis.
Then since
A is contained inside B, we have
that
x is an element of B.
As x is inside A
and x is inside B, we have
that x is inside A intersect B, by definition.
Remember this is the set of all elements that are in A and B.
Well x is in A and B, so we're done.
So since
A intersect B is a subset or equal to A
and A is a subset of A [intersect] B,
we have that
A is equal to A intersect B, alright.
I guess I wrote it in the flipped direction, but that's fine. Symmetry
of equality, that's still okay.
Alright, so the forward direction we're okay. For the reverse direction,
we're going to assume
that A intersect B is equal to A.
Alright, and so what are we trying to show? We're trying to show therefore that
A is a subset or equal to B.
Let's look at this set right here A intersect B.
Notice that before we showed that A intersect B
is a subset or equal to A, right,
and we use nothing about the hypothesis. So in fact,
this same argument is going to hold to show that A intersect B is a subset or equal to B.
So remember as mathematicians, we try to be as lazy as possible.
Sort of a fun little joke,
but we've already really done the same amount of work here.
So really all I have to do is just copy the work except I change this A to a B.
So I'm going to say as before,
we have that
A intersect B is a subset or equal to B
since...
since A intersect B is all elements
contained
in both A and B, alright.
Great, so that's true.
As A is equal to A intersect B,
for any x inside A,
x is inside A intersect B
and since…
oops, and since A intersect B is contained inside B, we have that
x is an element of B.
So we've shown that any arbitrary element of A
must live inside B,
that's precisely the - that's precisely the statement that A is contained inside B.
So therefore,
always start sentences with words don't start sentences with symbols, it's bad form.
There we go. A is inside B.
That's it, that's a full if and only if proof using set notation.
So hopefully this helps and all the best.