I found a question which has asked to prove or disprove any closed set of $\mathbb{R}$ is compact.

My Attempt

Since $\mathbb{R}$ is a set of itself and it is closed ($\mathbb{R}$ is open and closed too. So I though I can get it as closed one here) it is closed. That means any closed set of $\mathbb{R}$ is closed. But the problem is that Any set of $\mathbb{R}$ is not bounded since $\mathbb{R}$ itself is not bounded. So I think that statement is false.


But I don't know whether my proof is true or false. If someone can please help me to figure this out.

Note : Problem has clearly stated that "Any set of $\mathbb{R}$"

  • 1
    $\begingroup$ You're right, you have produced a counterexample. $\mathbb R$ is not compact, yet it is a closed subset of itself. Similarly, $\mathbb Z$ is a closed subset of $\mathbb R$ which is not compact. $\endgroup$ – MPW Nov 3 '16 at 14:36
  • $\begingroup$ $\mathbb R=\bigcup_{n\in\mathbb N}(-n,n)$. This is a cover of open sets, but a finite subcover does not exist. So $\mathbb R$ is not compact. $\endgroup$ – drhab Nov 3 '16 at 14:37

You are correct. The Heine-Borel Theorem states that a subset of $\mathbb{R}$ is compact if and only if it is closed and bounded. So, you a specific counter-example you just need to exhibit a subset of $\mathbb{R}$ which is closed and unbounded. For instance $[0,\infty)$.

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  • $\begingroup$ Thanks. I know that $\mathbb{R}$ is both open and closed. So can I get it as either as Open or as closed whenever I want? $\endgroup$ – Samitha Nanayakkara Nov 3 '16 at 14:42
  • $\begingroup$ Yes. So you may also used the fact that $\mathbb{R}$ is not compact as your counter-example. $\endgroup$ – ervx Nov 3 '16 at 14:44

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