# Closed subsets of compact sets are compact

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?

• 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 $X$. – Avi Steiner Mar 18 '14 at 1:05
• 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? – user119228 Mar 18 '14 at 1:05
• Next time please choose a more descriptive title for your question. – MJD Mar 18 '14 at 1:08
• I know S-T is open in S, but Why R-T is open ? – python3 Mar 18 '14 at 1:15
• What is the definition of a closed set in $\Bbb R$, can you tell me?? – 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.