# How to prove that a set, by the open cover definition, is not compact

For an exam I have to be able to prove whether certain sets are open, closed or neither and, by extension, (ab)using the Heine-Borel theorem to prove if these sets are compact or not.

Because I prefer not to use the Heine-Borel theorem I would like to use the definition of a compact set to prove whether the following sets are compact or not (spoiler: they are not): $$V = \lbrace x \in \mathbb{R}^2 : -1 \leq x_1 \lt 2, -3 \lt x_2 \leq 4 \rbrace$$ $$W = \lbrace x \in \mathbb{R}^2 : x_1^2 + x_2^2 \lt 1 \rbrace$$

Obviously, $W = B(0;1)$ and by definition, open balls are open and therefore it is not compact. But how would you prove this with the open cover definition

Could anyone give me some hints how to do these kind of problems?

• Give an open cover that does not have a finite subcover. For $W$, a natural candidate would be $U_r = \{ x : x_1^2+x_2^2 < r\}$ for $0 < r < 1$. – Daniel Fischer Jan 23 '14 at 15:10
• for $W$ take disks of increasing radius centered at the origin: $$D_{\frac{n+1}{n}}(0,0),$$ where $n\in\Bbb{N}$ – janmarqz Jan 23 '14 at 15:10
• Compact subsets of Hausdorff spaces are closed. – Pedro Tamaroff Oct 4 '14 at 6:27

The basic idea for a metric space is (usually) to find a set of open sets that cover more and more of a sequence of points that lie within the set but have limit outside. In the case of $W$, others have already pointed out that choosing balls of increasing radius but never reaching $1$ will work, and can be seen as covering more and more of a sequence that heads towards the boundary of the open set $W$. For $V$ there are a number of options, and we can choose for example the sequence $( (2-\frac{1}{n},0) : n \in (1,2,...) )$ with limit $(2,0)$. A suitable open cover is $\{ (-2,2-\frac{1}{n}) \times (-4,5)) \}$ which gradually covers more and more of the sequence but not all at any time.

• Could you explain the $\lbrace (-2, 2-\frac{1}{n}) \times (-4,5) \rbrace$ notation to me? – Nigel Overmars Jan 23 '14 at 15:50
• After some drawing I think get it. For all $n \in \mathbb{N}$, $V_n=\lbrace (-2, 2-\frac{1}{n}) \times (-4,5) \rbrace$ denotes the surface on $\mathbb{R}^2$ from $x_1 = -2$ to $x_1 = 2- \frac{1}{n}$ and $x_2=-4$ to $x_2 = 5$. Since you need an infinite amount of $V_n$'s this set doesn't have a finite cover. – Nigel Overmars Jan 23 '14 at 16:22
• Just a word of warning about your wording. It's not entirely sure what you mean by "since you need an infinite amount of $V_n$s". It's better to say "since no finite number of $V_n$s can cover the whole set" (this wording is better because it directly references what the definition of compactness is). – 5xum Jan 23 '14 at 19:01
• Yes that notation is the conventional one for cartesian product of sets. You can say that any subset of the cover that covers the original set is infinite, but to prove it you still need to use the equivalent "any finite subcover does not cover...". – user21820 Jan 24 '14 at 1:05

Use an infinite cover of $W$ by a family of balls of radius $r$ for all $r<1$. Here it does not matter whether the balls are open or closed: one obviously can't choose a finite subcover.

For $W$, take sets $A_n$ defined as $A_n=B(0, 1-1/n)$. You can see that $$\bigcup_{n\in\mathbb N}A_n = W.$$ Can you find a finite subcovering?

When you understand why such a covering does not exist, you can try and prove $V$ is not compact in the same way.

• I understand why such a covering does not exist, but I am having a hard time writing it down rigorously. And for $V$, should I also use balls? – Nigel Overmars Jan 23 '14 at 15:20
• It's not really that hard to write it down rigurously. Since for $m<n$, you have $A_m \subseteq A_n$, any finite union of sets $A_{n_1}\cup A_{n_2}\cup \dots \cup A_{n_k}$ is equal to $A_N$ where $N=\max\{n_1,n_2,\dots, n_k\}$. Since $A_N\neq W$, this finite covering is not good. Since this is true for any covering, $W$ is not compact. – 5xum Jan 23 '14 at 18:40