Compactness of $\mathbb{R^2}$ and its relation between open balls

Question

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:

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

Proof:

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.

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.