What is the best way to prove basic set theory statements?

[lockedup]

Homework Statement

I'm working on some set theory stuff to prepare for Topology next semester. I'm actually working out of a Topology book from Dover Publications. I could really use some direction/correction.

1. If S ⊂ T, then T - (T - S) = S.

2. If S is any set, then ∅ ⊂ S.

The Attempt at a Solution

1. Let x ∈ (S ⊂ T). Therefore, x ∉ (T – S). If x ∉ (T – S), then x ∈ T – (T – S). But 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. A set A is a subset of a set S if every element of A is in S. If A is ∅, then A has no elements. Therefore all of its elements are in S.

 

[jedishrfu]

VENN diagram-wise #1 true. S is a subset of T so T - S is T without S which looks like a donut on a VENN diagram. T minus the set (T - S) takes away all of T except for those elements in S hence T - (T - S) is S.

I think your proofs look okay and nicely done too.

For every element x of S you concluded that x is also an element of T - (T - S) and hence S is a subset of T - (T - S). Similarly for every x element of T - (T - S) you concluded that x is also an element of S and hence T - (T - S) is a subset of S. Finally you concluded that when S is a subset of T - (T - S) and when T - (T - S) is a subset of S then these two sets must be equal hence T - (T - S) = S .

 

[lockedup]

Can one use Venn diagrams as proofs? I've heard that's a no-no...

 

[jedishrfu]

I don't know if VENN diagrams are accepted as proof. I would think no because of the issues on set set boundaries like is a given point inside or outside the set.Conceptually though they help to visualize the truth of the statement so you know beforehand that it is in fact provable.

 

[Bacle2]

Ultimately, only rigorous way I know is to use inclusions: Show A=B if every x in A is contained in B, and viceversa.

So, e.g., if x is in S, is x also in T-(T-S)? If y is in T-(T-S) , is y in S?

Maybe a bit tedious, but maybe you'll get used to it after a while. I can't think of a better way. But then start with the parentheses: x is in T-S. Then x is not in S. Then we apply
T-(T-S): elements of T-(T-S) are in T, but not in T-S.

A good way of determining if a suggested equality holds, is to test if it holds for a couple of randomly-chosen elements. Of course, it is not a proof, but , if equality holds, then you should consider a proof first.

 

[Syrus]

1.

The Attempt at a Solution

1. Let x ∈ (S ⊂ T).

To begin, this is a meaningless expression (that is, if the symbol "⊂" is the proper subset symbol- which I expect it is).

Your proof should begin as follows:

Suppose S ⊂ T. {from here, show that T - (T - S) = S by proving both directions (that is, that each set on either side of the equality is a subset of the other}

2. This proof follows by the vacuous truth resulting from the logical form of the statement.

 

[lockedup]

To begin, this is a meaningless expression (that is, if the symbol "⊂" is the proper subset symbol- which I expect it is).

Your proof should begin as follows:

Suppose S ⊂ T. {from here, show that T - (T - S) = S by proving both directions (that is, that each set on either side of the equality is a subset of the other}

I was just going by the pattern in my book. My book proved DeMorgan's laws this way. It starts out "Suppose x [itex]\in[/itex] [itex]\bigcup[/itex](T - S_i)..."

 

[Syrus]

Ah, but that is quite different than what you have above, however. The statement x ∈ ⋃(T - Si) is valid since it asserts that x is an element of the set union of T - Si [⋃(T - Si), which IS a set]. Your previous post contains a nonsense expression since (S ⊂ T) is a statement about the sets S and T, and is not itself a set.

 

[Bacle2]

Ah, but that is quite different than what you have above, however. The statement x ∈ ⋃(T - Si) is valid since it asserts that x is an element of the set union of T - Si [⋃(T - Si), which IS a set]. Your previous post contains a nonsense expression since (S ⊂ T) is a statement about the sets S and T, and is not itself a set.

I imagine it may be shorthand for x is a element of S, which ( meaning S) is a subset of T, but I guess the author should explain so.

Sorry, lockedup, that I did not read your OP, and my suggestion is then exactly what you
were doing initially.

 

[lockedup]

I imagine it may be shorthand for x is a element of S, which ( meaning S) is a subset of T, but I guess the author should explain so.

Sorry, lockedup, that I did not read your OP, and my suggestion is then exactly what you
were doing initially.

This is what I meant. I guess I need to work on my use of notation.