Confused by proof in Hocking and Young

[yossell]

Am trying to learn topology (again) and failing.

Lemma 2.8, p43 of Hocking and Young states: Let $a$ and $b$ be distinct points of a compact Hausdorff space $S$ and let $\{H_\alpha \}$ be a collection of closed sets simply ordered by inclusion. If each $H_\alpha$ is contains both $a$ and $b$ but is not the union of two separated sets,one containing $a$ and the other containing $b$,then the intersection $\cap_\alpha H_\alpha$ also has this property.

Proof: Let $H = \cap_\alpha H_\alpha$ and suppose that $H$ is the union of two separated sets $A$ and $B$, with $a$ in $A$ and $b$ in $B$. Since H is closed and $A$ and $B$ are closed in $H$, it follows that $A$ and $B$ are closed in the space $S$ and...'

It's that last 'it follows' that is losing me -- I can't see why it follows that they're closed in $S$. I'm missing something trivial, but I don't know what. Any suggestions would be appreciated.

MENTOR note: replaced all $ with double # to render properly under mathjax

 

[Gavran]

This is trivial. There are two closed sets $A$ and $B$ and they will always be closed, no matter where they are placed, in $H$ or in $S$. The only important thing is that the sets $A$ and $B$ are closed because of the Axiom C₁.

 

[fresh_42]

Why are $A,B$ closed in $H$?

If so, then they are closed in $S,$ too:

The boundary of $A$ in $S$ is contained in $H$ and the boundary of $H$ in $S$. But $H\subseteq S$ is closed, so its boundary in $S$ is fully contained in $H$. Therefore, the boundary of $A$ in $S$ is fully contained in $H$ and since $A\subseteq H$ is closed, fully contained in $A.$ This means that $A$ is closed in $S.$ For short: you cannot gain additional boundary points of $A$ from searching in $S$ since all those points are already in $H$ and its boundary in $S,$ and thus in $A.$

 

[mathwonk]

I took a quick look at that book and wondered why you were interested in such a tedious approach to topology. Then I noticed some very nice theorems on pages 54-55 with simple characterizations of a circle and an interval. E.g. apparently any compact connected metric space which is disconnected by the removal of any point, except for two of them, is homeomorphic to a closed interval. And such a c.c. metric space that is disconnected by the removal of any two points, is apparently homeomorphic to a circle. nice.

 

[WWGD]

By " Simply ordered by inclusion", you mean a total order , i.e., for $O, O'$ closed, either $O \subset O'$ or $O' \subset O$?