- #1

Chicago_Boy1

- 6

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**First Prove**

{} is a subset of {}, where {} refers to an empty set

My professor told me to do this by contradiction.

So I assume that {} is not a subset of {}. That would imply that there exists an element "x" in {} that is NOT in {}.

As far as I can tell, there are two contradictions in that statement. First, {} is, by definition, empty, so there cannot exist an element "x" in {}. Second, an element cannot both be and not be in a set.

So I am just wondering whether it's the first contradiction, the second one, or perhaps both (?) that allows me to claim that I've solved this via contradiction.

**Second Proof**

A union B = {} iff A = X and B = X

The way I see it, there are two parts to this problem:

Part #1:

Assume A union B = {} and prove that A = {} and B = {}

Part #2:

Assume A = {} and B = {} and prove that A union B = {}

Part #1 Proof:

Assume A union B = {}, prove that A = {} and B = {}

Let x be a part of A union B = {} <=> x is a part of {}, but that can't happen (empty set does not have elements by definition), so it has to follow that there are no elements in A or in B.

Part #2 Proof:

Assume A = {} and B = {}, A union B = {}

Let x be a part of A, which would imply that x is a part of {}, but that can't happen, so A is empty

Let x be a part of B, which would imply that x is a part of {}, but that can't happen, so B is empty

So since A and B are both empty, it must follow that A OR (i.e. union) B = Empty or empty = {}.

Basically, for this second proof, I just need to make sure that I am doing it right.