Given: 1) is (-2,4) compact by applying Heine-Borel Property? and 2) is [-2,4] compact by applying Heine-Borel property?

I've followed the proof of Heine-Borel property and i have attempted to apply the proof to answer the following and i am not quite sure how to prove? it...

i know that [a,b], every open cover has a finite subcover and a set of real numbers is compact if and only if it is closed and bounded.

So for 1) no and 2) yes?

  • $\begingroup$ $(2, -4)$ is not compact since it is not closed, while $[-2,4]$ is compact since it is closed and bounded. $\endgroup$
    – Crostul
    Nov 6, 2014 at 7:00
  • $\begingroup$ could you show me an example for both $\endgroup$
    – chris
    Nov 6, 2014 at 7:02
  • $\begingroup$ You are correct. $\endgroup$
    – copper.hat
    Nov 6, 2014 at 7:09

1 Answer 1


If you know Heine-Borel Theorem, then its fairly obvious wich is compact and wich isnt, $(-2,4)$ is not closed so it cannot be compact, and $[-2,4]$ is both closed and bounded, so it is compact. If you want to do it more 'hands on' you can try the following:

For $1)$: Try to construct a sequence wich doesnt have any subsequence converging in the space $(-2,4)$

For $2)$ How do you know $[a,b]$ is compact? Hint: Let $I_n$ be an open cover of $[a,b]$, then take

$$C = \{x : \text{ such that $[a,x]$ has a finite subcover} \}$$

Prove $\sup C$ exists and that $\sup C = \max C = b$

This will yield a general result for $[-2,4]$ you can even replace $a = -2, b = 4$ and use the same hint.

  • $\begingroup$ would you like to show everyone how you know? $\endgroup$
    – chris
    Nov 6, 2014 at 7:37
  • $\begingroup$ @chris Did you tried to do anything of what I suggested? I just used what I know about the characterization of compacts sets, you can read more on wikipedia en.wikipedia.org/wiki/Compact_space $\endgroup$
    – aram
    Nov 6, 2014 at 7:39

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