Show that the closed interval $[1,5]$ is compact.

Question

Definition of a compact set: A set $S$ in $\mathbb{R}^n$ is said to be compact if, and only if, every open covering of $S$ contains a finite subcover, that is, a finite subcollection which also covers $S$.

The Heine-Borel theorem states that every closed and bounded set in $\mathbb{R}^n$ is compact. Now we prove the converse result.

I know that $4$ is the bounded set for $[1,5]$ for $1<x<5$ but I got little confused on getting if it is compact or not since there is no finite subcover. Can you help me here please?

I am confused, what do you mean by $4$ is the bounded set for $[1,5]$. — Siong Thye Goh, Sep 06 '21 at 14:16
Well I got confused , maybe I did something wrong in my process — soysoy hilay, Sep 06 '21 at 14:37
Does this answer your question? How to prove every closed interval in R is compact? — Giorgos Giapitzakis, Sep 06 '21 at 14:41

score 1 · Answer 1 · answered Sep 06 '21 at 18:28

Welcome to Stack Exchange and topology :)

I suppose you want to use the Heine-Borel theorem in this case. Then you shouldn't ask about compactness but you should check boundedness and closeness of the given set.

The given set is bounded below by 1 (or any number less than 1) and above by 5 (or any number greater than 5). Since the interval $[1,5]$ is bounded below and above, we conclude that it is bounded.

On the other hand, $[1,5]=\mathbb{R} \backslash\left( (-\infty ,1)\cup (5,\infty)\right)$, hence it is closed.

So, we showed that $[1,5]$ is bounded and closed, thus by Heine-Borel theorem it must be compact.

I understand , thank you for clarification. – soysoy hilay Sep 06 '21 at 19:44 — soysoy hilay, Sep 06 '21 at 19:44

Show that the closed interval $[1,5]$ is compact.

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