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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.

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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

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  • $\begingroup$ The proof given on Wikipedia (en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem) is quite elegant and directly applicable to the question here. $\endgroup$ – Alexander Geldhof Jun 7 '20 at 12:35
  • $\begingroup$ hello i have to show it with my definition with open cover and a finite subcover $\endgroup$ – user786835 Jun 7 '20 at 13:24
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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 $[0,1]\times [0,1]$.

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  • $\begingroup$ i think this is the correct proof because you use my definition, but i never had cubes in my math course. $\endgroup$ – user786835 Jun 7 '20 at 13:10
  • $\begingroup$ what are open cubes? open sets? i never had that $\endgroup$ – user786835 Jun 7 '20 at 13:44
  • $\begingroup$ @sarahmathmatics67: open cube in $\mathbb R^2$ is a set of the form $(a,b)\times (a,b)$ $\endgroup$ – Walace Jun 8 '20 at 7:13
  • $\begingroup$ is it another name for cartesian product? $\endgroup$ – user786835 Jun 8 '20 at 9:45
  • $\begingroup$ no. $(1,2)\times (1,2)$ is a cube whereas $(1,2)\times (1,3)$ is not (the latter is call an open rectangle). $\endgroup$ – Walace Jun 8 '20 at 16:23
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Hints:

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

The product of finitely many compact spaces is compact.

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    $\begingroup$ This is not true. Closed, bounded intervals in $\mathbb{R}$ are compact. $\endgroup$ – Alexander Geldhof Jun 7 '20 at 12:34
  • $\begingroup$ Topology, Corrolary 27.2., J. Munkres (2nd Edition) $\endgroup$ – Anton Vrdoljak Jun 7 '20 at 12:36
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    $\begingroup$ 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... $\endgroup$ – Anton Vrdoljak Jun 7 '20 at 12:58
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    $\begingroup$ @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. $\endgroup$ – Invisible Jun 16 '20 at 20:12
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    $\begingroup$ As a matter of a fact, we even put the accent by calling it a segment. $\endgroup$ – Invisible Jun 16 '20 at 20:17

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