My thoughts on proving this statement is as follows:

Suppose $G_a$ is an open cover of $Q= [0,1] \times [0,1]$. For each $x$ in $[0,1]$, there is some ball around $x$ with radius $r_x$ such that it covers $[x-r_x, x+r_x] \times [0,1]$. Since $[0,1]$ is closed and bounded, it is compact. Since $[0,1]$ is compact, this vertical strip described above is also compact, i.e., there is a finite collection of these balls of $G_a$ that covers this vertical strip.

Next we do the same thing except for horizontal strips.

This is where I am drawing a blank, how am I to show that the union of these balls are finite, and how would I know that it covers all of the unit square?

  • $\begingroup$ your $r_x$ would have to be at least 1 for what you describe to work. Have you considered sequential compactness? $\endgroup$ – Stefan Smith Nov 4 '13 at 2:23
  • $\begingroup$ What is sequential compactness? This is an intro analysis course so I only have knowledge of basic topology and metric spaces $\endgroup$ – tamefoxes Nov 4 '13 at 2:25
  • $\begingroup$ A metric space is sequentially compact if every sequence has a convergent subsequence. This is equivalent to compactness for metric spaces. $\endgroup$ – Dylan Yott Nov 4 '13 at 2:30
  • $\begingroup$ This seems like a far less tedious way of proving this question however we have not even covered sequences yet so I believe I wouldn't be able to prove it using sequential compactness, I appreciate the advice $\endgroup$ – tamefoxes Nov 4 '13 at 2:33
  • $\begingroup$ I don't understand your argument. However, you should most definitely look for/read a proof of the fact that the product of any two compact spaces is compact. $\endgroup$ – dfeuer Nov 4 '13 at 2:56

You're on the right track to prove this. For each $x$, let $G_x$ be a finite subcover of $G_a$ that covers $\{x\}\times[0,1]$.

Claim: There exists a positive number $r_x$ for which $G_x$ covers the entire strip $[x-r_x,x+r_x]\times[0,1]$. You need to prove this. One way uses the fact - which also requires proof - that you can assume $G$ consists of rectangles.

With $r_x$ defined in this way for each $x\in[0,1]$, consider the collection of intervals $[x-r_x,x+r_x]$. These cover $[0,1]$, which is compact, so there's a finite set $X$ for which $[0,1]\subseteq \bigcup_{x\in X}[x-r_x,x+r_x]$. Then $\bigcup_{x\in X}G_x$ (which is a finite subcover of $G_a$) covers $Q$.

  • $\begingroup$ Thank you, we have proved that the entire strip $[x-r_x,x+r_x] \times [0,1]$ is compact! $\endgroup$ – tamefoxes Nov 4 '13 at 3:12
  • $\begingroup$ Hm. If you proved already that the entire strip is compact, why not let $x=r=1/2$ and conclude immediately that $Q$ is compact? In any case, why I said you need to prove does not follow from the fact that the strip is compact. You need to prove that a finite subcover of the line segment $\{x\}\times[0,1]$ does in fact cover a strip of non-zero width. $\endgroup$ – Steve Kass Nov 4 '13 at 3:15
  • $\begingroup$ Sorry I meant to say in class we have proved that an interval such as this is compact. $\endgroup$ – tamefoxes Nov 4 '13 at 3:20

The closed unit square is a closed and bounded subset of $\mathbb{R}^{2}$, so it is compact by Heine-Borel theorem.

  • $\begingroup$ I appreciate the answer, however this exercise is for marked homework so I believe a thorough proof would only be valid. $\endgroup$ – tamefoxes Nov 4 '13 at 2:30
  • 1
    $\begingroup$ If you have seen the Heine-Borel theorem in $\mathbb R^2$, then this is a thorough proof. $\endgroup$ – Ittay Weiss Nov 4 '13 at 2:31
  • $\begingroup$ Exactly, look for a proof of the Heine-Borel theorem. Proving the Heine-Borel theorem is a proof that closed and bounded subsets of Rn are compact. The closed unit square is then a special case. $\endgroup$ – Lindon Nov 4 '13 at 2:32
  • $\begingroup$ We have covered the Heine-Borel theorem so I guess this will suffice as a proof. Do any of you guys have an idea using an approach similar to the one I was describing? $\endgroup$ – tamefoxes Nov 4 '13 at 2:42

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