$\{(x,y) \in \Bbb R^2 \mid 0\le x \lt 1, 0\le y \le 1\}$ is not compact

Question

Claim

$A = \{(x,y) \in \Bbb R^2 \mid 0\le x \lt 1, 0\le y \le 1\}$ is not compact.

I want to prove above claim. I might need to find out finite sub-cover of open cover of given set A. It requires me two step simultaneously, first think about open cover (which is a union of infinite open balls) second, then reduce the open cover but to be still infinite.

How could I construct like that example? Is there any easier or alternative way to show the claim not actually construct some specific examples?

Hint. The set is not closed. How can you tell it is not closed? Base your open cover on thet. Eg. Let you open cove be all the with some distance of (1,1) — fleablood, Jun 02 '17 at 07:09

score 2 · Accepted Answer · answered Jun 02 '17 at 06:12

2

For $c\in [0,1)$, let $U_c=\{\,(x,y)\in X\mid x<c\,\}$. Then $U_c$ is open, $\bigcup_{c\in[0,1)} U_c=A$, but no finite subcover $U_{c_1}\cup\ldots\cup U_{c_n}$ suffices as it does not contain the point $(\frac{1+\max_i c_i}{2},0)\in A$.

answered Jun 02 '17 at 06:12

Hagen von Eitzen

374,180

score 0 · Answer 2 · answered Jun 02 '17 at 06:05

0

Heine-Borel theorem: the compact sets are precisely the closed and bounded sets. Your set is bounded; is it closed? (Hint: if it were closed, then you could use this lemma. Or "just do it".)

answered Jun 02 '17 at 06:05

Patrick Stevens

36,135

but compact sets are not always closed and bounded but it depends on the metric space. isn't it? – Beverlie Jun 02 '17 at 06:08
@jackerysmith The question as stated is using the Euclidean metric, where Heine-Borel is true. (In general, the Euclidean topology is assumed when no other topology is given.) – Patrick Stevens Jun 02 '17 at 06:09
@PatricStevens. Got it – Beverlie Jun 02 '17 at 06:10

Michael Rozenberg · Answer 3 · 2017-06-02T06:13:07.043

0

For $A=\{(x,y)|0<x\leq1,0\leq y\leq1\}$

Let $a_n=\left(\frac{1}{n},0\right)$.

Hence, $\lim\limits_{n\rightarrow+\infty}a_n\notin A$, but $a_n\in A$.

For $A=\{(x,y)|0\leq x<1,0\leq y\leq1\}$

Let $a_n=\left(\frac{n}{n+1},0\right)$.

Hence, $\lim\limits_{n\rightarrow+\infty}a_n\notin A$, but $a_n\in A$.

The given was changed some times.

edited Jun 02 '17 at 06:13

answered Jun 02 '17 at 06:05

Michael Rozenberg

194,933

1

$(0,0) \in A$. You want $1-\frac{1}{n}$. – Patrick Stevens Jun 02 '17 at 06:07
@Patrick Stevens The given was changed. Now $(0,0)\notin A$. – Michael Rozenberg Jun 02 '17 at 06:08

$\{(x,y) \in \Bbb R^2 \mid 0\le x \lt 1, 0\le y \le 1\}$ is not compact

3 Answers3