Thread: Basic Set Theory Proofs

I'm taking Topology next semester and, to prepare, I'm learning a little set theory. I had bought a Topology book from Dover Publications a couple years ago and am now starting to do some problems in it. The very first section is on Set Theory. Here's the first couple problems:

1. S ⊂ T, then T - (T - S) = S.

Proof:
Let x ∈ (S ⊂ T). Therefore, x ∉ (T – S). If x ∉ (T – S), then x ∈ T – (T – S). And x ∈ (S ⊂ T). Therefore, S ⊂ (T - (T - S))

Let x ∈ (T - (T - S)). This implies that x ∉ (T – S). Which implies that x ∈ S. Therefore, T – (T – S) ⊂ S and T - (T - S) = S.

2. S ⊂ T iff S ∩ T = S.

Proof:
If S ⊂ T, every x ∈ S is in T. That implies that every x ∈ S is in S ∩ T. This implies that S ∩ T = S.

Let x ∈ S. If S ∩ T = S, then x ∈ (S ∩ T). This implies that S is contained in T and S ⊂ T.

I could really use some direction/correction.

First, the symbol ⊂ is somewhat ambiguous because it may mean either proper or improper subset in different sources. To clear the ambiguity, ether use ⊆ for improper subset and ⊊ for proper subset or describe your notation in words.

Originally Posted by aareavis

1. S ⊂ T, then T - (T - S) = S.

Proof:
Let x ∈ (S ⊂ T).

An object x cannot belong to (S ⊂ T) because (S ⊂ T) is a proposition (something that is either true or false), not a set. It is sometimes possible to write x ∈ S ⊆ T as a contraction to "x ∈ S and S ⊆ T," but such shortcuts are better avoided in the beginning.

Originally Posted by aareavis

If x ∉ (T – S), then x ∈ T – (T – S).

Only if x ∈ T, which, granted, is apparently the case here.

Originally Posted by aareavis

Let x ∈ (T - (T - S)). This implies that x ∉ (T – S). Which implies that x ∈ S.

Or x ∉ T.

Originally Posted by aareavis

2. S ⊂ T iff S ∩ T = S.

Proof:
If S ⊂ T, every x ∈ S is in T. That implies that every x ∈ S is in S ∩ T. This implies that S ∩ T = S.

Strictly speaking, it just implies that S ∩ T ⊇ S. The other inclusion is obvious, but at this high level of proof detail maybe it should be said explicitly.

Originally Posted by aareavis

1. S ⊂ T, then T - (T - S) = S.

If you have already studied the concept of universal set , the complementary of a subset , distributive and Morgan's laws, etc you can prove the equality in another way.

Choosing as universal set any set such that (in this case for example ) we have for . So,

Originally Posted by FernandoRevilla

If you have already studied the concept of universal set , the complementary of a subset , distributive and Morgan's laws, etc you can prove the equality in another way.

Choosing as universal set any set such that (in this case for example ) we have for . So,

I have studied the concept of a universal set (Prob & Stat). I just didn't think I could use it here. Now that I think about it, the way my book describes it is this: T - S is "the compliment of S in T" which is the same as saying T ∩ S’. (I use apostrophe instead of C. Habit from Prob & Stat.)

Originally Posted by aareavis

I have studied the concept of a universal set (Prob & Stat). I just didn't think I could use it here. Now that I think about it, the way my book describes it is this: T - S is "the compliment of S in T" which is the same as saying T ∩ S’. (I use apostrophe instead of C. Habit from Prob & Stat.)

Well, in that case you can use the alternative way I provided (if you have previously covered distributive laws, etc).