3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $Q$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.