I want to show by definition that $\mathbb{R^{n}}$ is not compact, so I did this:

Let ${\Omega}$ be an open cover of $\mathbb{R^n}$ and suppose that $\mathbb{R^n}$ is compact, then there exists a finite subcollection of ${\Omega}_{i=1}^{k}$ such that $\mathbb{R^{n}}$ is covered by $\Omega_{i=1}^{k}$. So, ${\cup}_{i=1}^{k}\Omega_{i}$ $\supseteq \mathbb{R^{n}}$.

From this part, I don't know how to continue the proof, I think the last argument is a contradiction but I'm not quite sure. Can you help me? Please.

  • 5
    $\begingroup$ You don't want a proof by contradiction. Come up with a particular open cover of $\mathbb{R}^n$ that has no finite subcover. $\endgroup$
    – nkm
    Feb 18, 2019 at 2:14

2 Answers 2


Hint: If you want to show that $\mathbb{R}^{n}$ is not compact, you just need to provide a specific example of an open cover that has no finite subcover.

Try with the open cover $\{\Omega_{k}\}_{k=1}^{\infty}$, where $\Omega_{k}$ is the open ball of radius $k$, centred at $\mathbf{0}$.

(By the way, this is an example of the general principle that to show that something is not always true, giving a specific counterexample is sufficient.)

  • $\begingroup$ If I was asked to show it by definition then using a counterexample is not by definition, right? I'm a little bit confused. $\endgroup$
    – SMath
    Feb 18, 2019 at 2:22
  • 1
    $\begingroup$ Well the definition of "$X$ is compact" is "every open cover of $X$ has a finite subcover". Therefore, the definition of "$X$ is not compact" is "there exists an open cover of $X$ that has no finite subcover". Hence to show $\mathbb{R}^{n}$ is not compact by definition, you just need to show that there exists an open cover of $\mathbb{R}^{n}$ with no finite subcover. To show an existence like this, an explicit example is sufficient. $\endgroup$ Feb 18, 2019 at 2:25
  • 4
    $\begingroup$ You are asked to prove that the definition of compactness does not hold for $\mathbb R^n$. To do this, you prove that the negation of the definition of compactness is true. When you negate the definition, you come up with the statement "there exists an open cover that has no finite subcover". To prove something exists, you simply have to find an example. $\endgroup$
    – Lee Mosher
    Feb 18, 2019 at 2:26
  • $\begingroup$ Open cover in the definition assumes the set you give is countable. But the set R is not countable so how you create a cover. $\endgroup$
    – user960654
    Dec 1, 2021 at 3:25

How about the open cover consisting of the open balls of radius $1$: $\mathscr C=\{B(r,1)\mid r\in\Bbb R^n\}$?

I don't think there's a finite subcover.

For suppose we have a finite subcollection: $\mathscr F=\{B(r_k,1)\mid k=1,\dots,n\}$.

Take a point $x\in\Bbb R^n$ with distance from the origin $d(x,0)\gt\max \{1+d(r_k,0)\mid k=1,\dots,n\}$. You can do this because $\Bbb R^n$ is unbounded. (I guess technically you need the Archimedean principle for this.)

Then $x\not\in B(r_k,1)$ for any $B(r_k,1)\in\mathscr F$.


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