Show that the set $[0,1]×[0,1]$ is compact in $\mathbb{R}^2$ to standard metric

Question

I have to show that the set $[0,1]×[0,1]$ is compact in $\mathbb{R}^2$ with respect to the standard metric.

I have to show this using only the definition of compactness. The definition I am given is: A set is compact if we have an open cover, we get a finite subcover.

cf. page 39 of Rudin's Principles of Mathematical Analysis (third edition) — J. W. Tanner, Jun 07 '20 at 12:46
Does this answer your question? Cartesian product of compact sets is compact — PinkyWay, Jun 16 '20 at 20:26

score 4 · Answer 1 · answered Jun 07 '20 at 12:33

4

Heine-Borel Theorem states that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and your subset is certainly closed and bounded

answered Jun 07 '20 at 12:33

kkkkk

335

The proof given on Wikipedia (https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem) is quite elegant and directly applicable to the question here. – Alexander Geldhof Jun 07 '20 at 12:35
hello i have to show it with my definition with open cover and a finite subcover – Jun 07 '20 at 13:24

score 2 · Answer 2 · answered Jun 07 '20 at 12:38

2

Let $\{U_i\}_{i=1}^\infty $ a cover of open set of $[0,1]\times [0,1]$. Since $U_i$ is open, there are open cubes $(a_j^i,b_j^i)\times (a_j^i,b_j^i)$ s.t. $$U_i=\bigcup_{j=1}^\infty (a_j^i,b_j^i)\times (a_j^i,b_j^i).$$

Therefore $(a_j^i,b_j^i)_{i,j=1}^{\infty }$ is an open cover of $[0,1]$. Therefore, there is a subcover (denoted $\{(a_j^i,b_j^i)\}_{\substack{i=1,...,n\\ j=1,...,m}}$) of $[0,1]$. Therefore $\{U_i\}_{i=1}^n$ is a finite subcover of $[0,1]\times [0,1]$.

answered Jun 07 '20 at 12:38

Walace

710
4
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i think this is the correct proof because you use my definition, but i never had cubes in my math course. – Jun 07 '20 at 13:10
what are open cubes? open sets? i never had that – Jun 07 '20 at 13:44
@sarahmathmatics67: open cube in $\mathbb R^2$ is a set of the form $(a,b)\times (a,b)$ – Walace Jun 08 '20 at 07:13
is it another name for cartesian product? – Jun 08 '20 at 09:45
no. $(1,2)\times (1,2)$ is a cube whereas $(1,2)\times (1,3)$ is not (the latter is call an open rectangle). – Walace Jun 08 '20 at 16:23
ok thanks i understand! can i let the word out for the proof? because i never learned it – Jun 08 '20 at 19:01
sure :) (just right : "there are sets of the form..." instead of "there are cubes of the form...") – Walace Jun 09 '20 at 07:20
ok thaaanks :)) – Jun 09 '20 at 14:10

score 0 · Answer 3 · answered Jun 07 '20 at 12:33

0

Hints:

Every closed interval in $\mathbb{R}$ is compact.

The product of finitely many compact spaces is compact.

answered Jun 07 '20 at 12:33

Anton Vrdoljak

2,717

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This is not true. Closed, bounded intervals in $\mathbb{R}$ are compact. – Alexander Geldhof Jun 07 '20 at 12:34
Topology, Corrolary 27.2., J. Munkres (2nd Edition) – Anton Vrdoljak Jun 07 '20 at 12:36
That doesn't seem to be available online, @Anton Vrdoljak. $[0, \infty)$ is trivially not compact, using the open cover ${ (n - 2, n + 2) | n \in \mathbb{Z} }$. – Alexander Geldhof Jun 07 '20 at 12:41
@AlexanderGeldhof I think by closed interval he means an interval of the form $[a, b]$ where $a<b \in \mathbb{R}$. – Botond Jun 07 '20 at 12:44
@Botond While that may be true, that's not the canonical definition of a closed interval, and that definition is particularly relevant for someone studying topology. – Alexander Geldhof Jun 07 '20 at 12:48
@AlexanderGeldhof I think it is when you are not dealing with topology (for example, in "early" analysis), and in topology, you can either keep it or modify it. – Botond Jun 07 '20 at 12:52
A real interval is a convex set, i.e. a subset of $\mathbb{R}$ so that for any numbers $a,b \in I$, all numbers $c$ satisfying $a < c < b$, we have $c \in I$. Note that this does not in any way exclude unbounded intervals. – Alexander Geldhof Jun 07 '20 at 12:54
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To @Alexander Geldhof: because the limit in the number of characters for a comment, I am not able to write in more details what is (according to Munkres) closed interval in $X$, where $X$ is a simply ordered set... – Anton Vrdoljak Jun 07 '20 at 12:58
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@AlexanderGeldhof, see this. Such a statement couldn't possibly get 55 upvotes if it were inaccurate. Every closed interval in $\Bbb R$ is bounded, at least that is what they teach us in Croatia, so this can be a weaker , but not a false statement. Furthermore, there is some available literature. – PinkyWay Jun 16 '20 at 20:12
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As a matter of a fact, we even put the accent by calling it a segment. – PinkyWay Jun 16 '20 at 20:17

Show that the set $[0,1]×[0,1]$ is compact in $\mathbb{R}^2$ to standard metric

3 Answers3