1. Jun 25, 2008

### 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?

2. Jun 25, 2008

### 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.

3. Jun 25, 2008

### 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.

4. Jun 25, 2008

### 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?

5. Jun 25, 2008

### HallsofIvy

Staff Emeritus
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).

6. Jun 25, 2008

### 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.