Why Does Apostol Include S Subset A in His Theorem on Set Complements?

[uman]

From page 45 of "Mathematical Analysis" by Tom Apostol:

3-17 Theorem. If S is closed, then the complement of S (relative to any open set containing S) is open. If S is open, then the complement of S (relative to any closed set containing S) is closed.

Proof. Assume S\subset A. Then A-S=E_1-[S\cup(E_1-A)]. (The reader should verify this equation.) If S is closed and A is open, then E_1-A is closed, S\cup(E_1-A) is closed, A-S is open. The converse is similarly proved.

Now I'll prove the part that Apostol leaves to the reader:

Given two subsets A and S of E_1, A-S=E_1-[S\cup (E_1-A)].

Proof: If x\in(A-S), x\in A and x\notin S. Thus x\notin[S\cup(E_1-A)]. So x\in E_1-[S\cup (E_1-A)]. This proves that A-S\subset E_1-[S\cup(E_1-A)].

If x\in E_1-[S\cup(E_1-A)], x\in E_1 and x\notin [S\cup(E_1-A)]. Thus x\notin S and x\notin(E_1-A). But since x\in E_1, this last relation implies x\in A. So x\in(A-S).

I can't see any part of this whole proof that depends on the fact that S\subset A. Am I missing something? If not, why in the world would the author include that hypothesis?

 

[CompuChip]

I think that, if S is not entirely inside A, the proof still works and you can still show that A - S is open. But A - S is not the complement of S in A so you can't conclude that S is closed in A. In other words, it doesn't prove your theorem.

 

[uman]

I'm not sure what you mean by "closed in A". My book defines a "closed" subset of E_1 as a subset that contains all its accumulation points.

 

[CompuChip]

OK, it looks here like they're using the definition: S is closed in A if it's complement in A is open. Is the equivalence of those two definitions proved somewhere?

 

[HallsofIvy]

Also, a set is closed in A if and only if it contains all of its accumulation points that are in A. for example, The set [0, 1) is NOT closed in R but it is closed in (-2,1).

 

[uman]

Okay just to clarify, the book has defined "closed" to mean what apparently CompuChip would call "closed in E_1". CompuChip: That is not the definition they're using, although they do prove that a set is closed if its complement is open and vice versa.