If S is a compact subset of R and T is a closed subset of S,then T is compact. (1) Prove this using the definition of compactness.

Can somebody prove it? I think we should select a open cover of S randomly, and then we should think about the set S-T. Is S-T open in R? I don't know how to continue?

  • $\begingroup$ It doesn't matter if $S-T$ is open in $R$. The definition of compactness is internal: A topological space $X$ is compact if any cover of $X$ by sets which are open in $X$ contains an open subcover of $X$. $\endgroup$ – Avi Steiner Mar 18 '14 at 1:05
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    $\begingroup$ what can you say about $\lbrace T^c \rbrace \cup \lbrace U_n ;\,n\in {\rm I\!N\,}\rbrace$ where $(U_i)_{\scriptsize {i \in I}}$ a family of open (open for the topology of S) whose union contains S? $\endgroup$ – user119228 Mar 18 '14 at 1:05
  • $\begingroup$ Next time please choose a more descriptive title for your question. $\endgroup$ – MJD Mar 18 '14 at 1:08
  • $\begingroup$ I know S-T is open in S, but Why R-T is open ? $\endgroup$ – python3 Mar 18 '14 at 1:15
  • $\begingroup$ What is the definition of a closed set in $\Bbb R$, can you tell me?? $\endgroup$ – Ishfaaq Mar 18 '14 at 1:17

Consider any open cover $G_{\lambda}$ of $T$. Then if $S \subseteq G_{\lambda}$ too there is a finite covering of $S$ using sets from $G_{\lambda}$ which also contains $T$ and hence is a finite covering of $T$. Suppose $S \not \subseteq G_{\lambda}$. Then consider $ G_{\lambda} \cup T^C $ which is an open covering of $S$ since $T$ is closed and $T^C$ is an open set. Then again since $S$ is compact we have that there is a finite covering of $S$ using sets in $G_{\lambda} \cup T^C $. Removing $T^C$ if it was part of this finite covering we have a finite covering of $T$. Hence $T$ is compact.

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    $\begingroup$ The question was a good example of a question for which it easy to give useful hints, and probably considerably better and/or more useful than writing down the complete solution $\endgroup$ – user119228 Mar 18 '14 at 1:09
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    $\begingroup$ @Julien: True. This is the first time I came across the problem too. Guess I got excited after solving it. $\endgroup$ – Ishfaaq Mar 18 '14 at 1:11
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    $\begingroup$ For those of us reading later having the whole answer spelled out is much more useful than parsing a bunch of hints. This is a Q and A site not a homework coaching site. $\endgroup$ – Bob Woodley Jul 14 '15 at 19:37

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