Show that $\mathbb{R^2}$ is not compact.

My Attempt:

First of all, I have tried two ways. First one is very short:

i-) $\mathbb{R^2} \cong ((0,1),\;(0,1)) \cong ((0,\infty),\;(0,\infty))$. Therefore, $\mathbb{R^2}$ is not compact.

As above, it is very easy to see. And below I proved using definitions.

ii-) By definition if a subset is not compact then there exist an open covering of a subset such that for all finite subcover does not cover the subset. Meaning is that:

$$\text{Let A} \subseteq X.\; \text{If} \;\exists \,\lbrace{U_{i}\rbrace}_{i \in I} \; \text{such that} \; A \subseteq \bigcup_{i\in I} U_{i} \;\text{and}\;A\nsubseteq U_{i_{1}} \cup U_{i_{2}}\cup... \cup U_{i_{n}}$$


Let $\lbrace{U_{i}\rbrace}_{i\in I}$ be open covering of $\mathbb{R}^2$ in the form of open ball radius $r$ and centered at $(0,0)$. Obviously,

$$\mathbb{R^2} \subseteq \bigcup \mathcal{B}_{r}((0,0))$$ $$=\bigcup\lbrace{(x,y)\in \mathbb{R^2}: \sqrt{x^2+y^2}<r\rbrace}$$

Now suppose, we have a finite subcover. such that

$$\mathbb{R^2} \subseteq \mathcal{B}_{r_{1}}(0,0) \cup \mathcal{B}_{r_{2}}(0,0)\cup...\cup\mathcal{B}_{r_{n}}(0,0)$$

A contradiction since the while RHS is finite, LHS is infinite. Therefore, we can say $\mathbb{R}^2$ is not compact.

  • 1
    $\begingroup$ RHS is not finite, each open ball is infinite. You mean RHS is bounded. $\endgroup$
    – jjagmath
    Commented May 21, 2022 at 11:46

1 Answer 1


Your route (ii) seems sensible if you want to prove the statement from the definition. But be more precise with your wording (I assume this is for an analysis class, so this is good practice). "The RHS is finite"-- what exactly do you want to say here? Suppose that the $r_i$ are ordered so that $r_i \leq r_n$ for each $i=1,\cdots,n-1$. Then how would you find a point in $\mathbb{R}^2$ that does not lie within the finite subcollection?

Edit: I also want to add that you do not have to write this as a proof by contradiction. You have defined a clever covering, and you are now showing that any arbitrarily chosen finite subcollection cannot cover the whole space. That is enough to conclude that the space is not compact by definition.

I have also replaced "finite cover/subcover" with "finite subcollection" throughout my answers to be more accurate, because of course the job is to prove that it's not a cover.

  • $\begingroup$ "The RHS is finite"-- what exactly do you want to say here? I mean RHS is finite elementwise. Otherwise every open ball is infinite $\endgroup$ Commented May 21, 2022 at 13:31
  • $\begingroup$ But the RHS is not finite element-wise. An open ball contains infinitely many points/elements. There's another concept you want to use. Think in terms of the radii of the open balls in your finite subcollection. Can you argue that there is a maximal radius? (Give it a name, like R). If so, can you write down a specific point (𝑥,𝑦) in $\mathbb{R}^2$ that is not within your finite subcollection? $\endgroup$
    – ithmath
    Commented May 22, 2022 at 0:24
  • $\begingroup$ Of course we can write since we are talking about infinity? $\endgroup$ Commented May 22, 2022 at 13:08
  • $\begingroup$ Give a specific example: (x,y) = ... $\endgroup$
    – ithmath
    Commented May 23, 2022 at 1:19

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