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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Confusion about a theorem

**Physics Forums | Science Articles, Homework Help, Discussion**