From page 45 of "Mathematical Analysis" by Tom Apostol:(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]S\subset A[/tex]. Then [tex]A-S=E_1-[S\cup(E_1-A)][/tex]. (The reader should verify this equation.) If S is closed and A is open, then [tex]E_1-A[/tex] is closed, [tex]S\cup(E_1-A)[/tex] is closed, [tex]A-S[/tex] 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 [tex]E_1[/tex], [tex]A-S=E_1-[S\cup (E_1-A)][/tex].

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

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

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

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# Confusion about a theorem

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