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I just stumbled upon a proof of topological lemma that I don't understand: it would be great if anyone could give me some advices.

To be blunt, I am convinced that the proof does work but to me it seems like the author quotes some results that are not so trivial without any explanations about them: chances are that I am missing something.

Also please note that connectedness is the first topological invariant I ever studied, so I only have some basic background knowledges about metric spaces/closure/openness of sets etc. ie. no topological argument makes sense to me.

Lemma: Let $X$ be connected (metric space), and let $S$ and $T$ be closed subsets of $X$ such that $X=S\cup T$. Suppose that $S\cap T$ is also connected. Then both $S$ and $T$ are also connected.

Proof: $S$ and $T$ cannot be disjoint since otherwise $X$ is not connected. Thus let $x\in S\cap T$. Suppose that $C\subseteq S$ is both open and closed in $S$ and $x\in C$. Our aim is to show that in fact $C=S$.

  • so far so good. My confusion starts from here

(1) Then since $x\in C\cap T$, $C\cap T$ is a non-empty closed and open subset of $S\cap T$

  • I see that the set in the question is non empty, but how can one be sure that this is closed and open in $S\cap T$? The only argument that seems to convince me is: since $C\cap T \subseteq S \cap T \subseteq S$ and $C$ is closed and open in $S$, it follows that $C\cap T = C\cap (S \cap T)$ is closed and open in $S\cap T$. Is this really the case? I doubt that author would explained this in words if it was the case. It took me some time to figure this..

(2) Thus since $S\cap T$ is connected, then this implies that $C\cap T=S\cap T$. This is equivalent to $S\cap T \subseteq C$. Hence $S\setminus C$ and $C\cup T$ partitions $X$ into non empty disjoint closed subsets of $X$.

  • I see that $S\cap T \subseteq C$ implies that $S\setminus C$ and $C\cup T$ are disjoint subsets but I have no idea how one can conclude immediately that the both sets are closed in $X$. Only convincing argument I could find was:

  • $S\setminus C \subseteq S \subseteq X$ and $S\setminus C$ is closed in $S$, $S$ is closed in $X$ so $S\setminus C$ is closed in $X$.

  • $C$ closed in $S$, $S$ closed in $X$ so $C$ is closed in $X$. Then as $T$ is closed in $X$ their union is closed in $X$

Then of course we have $S \setminus C = \emptyset$ and so $S=C$.

Sorry for rather long question, but it would be greatful if someone could give me some hints about understanding points made in (1),(2)

user160738
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    Everything you have done seems perfectly fine to me. Great work! – Clayton Oct 22 '14 at 23:36
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    My question is, why do you not have confidence in your reasoning? Why do you feel there is a flaw in what you have said? You're doing the right thing by proving the unproved assertions given by the author. The only thing I can think of is that the author may have already proved the facts that he's using, and perhaps they are earlier in the chapter and somehow you haven't read them yet. It can happen, sometimes, that we pick up in the middle of a chapter, and start reading proofs where we encounter seemingly "unproved assertions", when actually they're at the beginning of the chapter. – Rustyn Oct 22 '14 at 23:43
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    @Clayton is correct: you've essentially answered your own question. – Brian M. Scott Oct 22 '14 at 23:43
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    Oh, thanks, that's very assuring :). When I was reading the proof for the first time I couldn't understand those claims; I had many experiences in the past in which I ocassionally would not see a simple way of arguing and take a 'roundabout' which only complicated things - and consequently they dont help me to understand theorems. Since author seemed to be taking them trivial and I couldn't, I doubted maybe I did the same thing again. I will check if those assertions are in earlier part of the book. – user160738 Oct 22 '14 at 23:49

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