In this video, we're going to talk about
a proof, but instead of
typing the proof…well discovering it
we're going to read a proof, that I had just written.
So the statement that we're going to prove is the following:
so we're going to prove that A intersect B…so let me make this a little bigger.
We're going to prove that A intersect B is equal to
A minus - or A set difference with A set difference B. Okay?
So again we have a proof involving set equalities, so in order to show this,
we're going to first show
that the left-hand set is contained inside the right-hand set, and then
we'll show that the right-hand set is contained inside the left-hand set.
So first, we show that A intersect B is contained inside
A set difference with A set difference B.
So to do this, we start with an element from the left-hand set.
Let x be inside A intersect B,
and what does that mean? So now we unwind the definition,
so if [x] is inside the intersection, then it must be inside both the sets,
right? So then x is in A and x is in B.
By definition, what do we know? Well if x is in A and x is in B,
then we know that x can't be in the set difference of A minus B,
right, because it's in the intersection…
or the set difference. So
what does that tell us? Well however,
we already know that x is inside A
and we know that x is not inside A set difference B,
and so by definition again we have that
x must be inside A set difference with A minus B.
Again that’s…this is just definition pushing.
So you just have to remember what these definitions mean,
right? A intersect B means that each element of A is contained in B,
and A set difference with B means that
x is an element of A and x is not
an element of B. That's what the set difference set is.
Now we try the reverse inclusion, right?
So let x be inside A set difference with bracket A set difference B.
By definition, we know that this means that, okay well we know that this means that
x is in A and that x is not inside
A set difference B.
That is the definition of x belonging to this set.
Now the next question is for x to not be inside this set, what does that mean? Well
this means one of two things,
it either means that x is not inside A,
in which case would be done, but we already know that's false
so we know that this part can't be true,
but there's an “or”, right? So either x was not inside A or x is inside A,
and x is inside B, right, because if it's inside both of these sets,
then it's not in the set difference because it's common, okay?
Since we know the first part is false, right, we already have that x was in A,
what can we conclude? Well we conclude that the second part has to be true, so then
thus x is an element of A and x is an element of B,
and then hence, by definition, x must be inside the intersection.
That shows the reverse containment,
so now we've shown the double containment,
and so maybe I should make a concluding sentence…
therefore…
the claimed statement is true. So let’s just…
easy to do on a computer when you can copy and paste.
So what should our concluding sentence be? Therefore A intersect B is equal to
A set difference with A set difference B.
Okay, hopefully this helped. Thank you very much.