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

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.

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

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 $$[0,1]\times [0,1]$$.

• i think this is the correct proof because you use my definition, but i never had cubes in my math course. – sarahmathmatics67 Jun 7 '20 at 13:10
• what are open cubes? open sets? i never had that – sarahmathmatics67 Jun 7 '20 at 13:44
• @sarahmathmatics67: open cube in $\mathbb R^2$ is a set of the form $(a,b)\times (a,b)$ – Walace Jun 8 '20 at 7:13
• is it another name for cartesian product? – sarahmathmatics67 Jun 8 '20 at 9:45
• 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 8 '20 at 16:23

Hints:

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

The product of finitely many compact spaces is compact.

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