Today I Learned

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February 11th

Today I learned (finally) the proof that $[0,1]$ is compact in the standard topology on $\RR.$ Of course, the proof generalizes to show that $\RR$ is locally compact, but we don't bother with this.

We begin by presenting the proof with no motivation. Suppose that $\{U_\alpha\}_{\alpha\in\lambda}$ is an open cover of $[0,1].$ The trick is to consider\[S=\{\varepsilon\in(0,1]:[0,\varepsilon]\text{ can be covered with a finite subcover}\}.\]Very quickly, we remark that if $\varepsilon_0\in S,$ then each $\varepsilon \lt \varepsilon_0$ is also in $S.$ Indeed, if $[0,\varepsilon_0]$ can be finitely covered, then the same cover can be used to cover $[0,\varepsilon]\subseteq[0,\varepsilon_0].$

We would like to use the least upper-bound property on $S.$ Note that some open set $U_\bullet$ covers $0$ and so contains an interval like $[0,2\varepsilon)$ for some $\varepsilon \gt 0.$ (Formally, dissolve $U_\bullet$ into intervals because by the basis of our topology; one such interval contains $0.$) This implies that $[0,\varepsilon]\subseteq U_\bullet$ can be finitely covered, so $\varepsilon\in S$ inhabits $S.$

Additionally, $S$ has an upper bound: by definition, $\varepsilon\in S$ implies $\varepsilon\le1.$ Thus, we may apply the least upper-bound property and let $\varepsilon_0$ be the least upper bound of $S.$ Note $\varepsilon_0\le1$ because $1$ is an upper bound of $S.$ Further, we remark that, for $\alpha\in(0,1],$ we see $\alpha\notin S$ implies nothing above $\alpha$ is in $S,$ so $\alpha$ is an upper bound of $S,$ so $\varepsilon_0\le\alpha.$ Thus, $\alpha \lt \varepsilon_0$ implies $\alpha\in S.$

We now do casework on $\varepsilon_0.$

  • If $\varepsilon_0=1,$ then we can finish quickly. As before, we can pick up $U_\bullet$ containing $1,$ and then extract an interval $(\alpha,1]$ inside of $U_\bullet,$ where $\alpha \lt 1.$ In particular, $\alpha \lt \varepsilon_0,$ so $\alpha\in S.$ Thus we have a finite subcover \[\{V_k\}_{k=1}^N\] covering $[0,\alpha].$ To finish, we note \[\{V_k\}_{k=1}^N\cup\{U_\bullet\}\] which now finitely covers $[0,\alpha]\cup(\alpha,1]=[0,1]$ in full.

  • If $\varepsilon_0 \lt 1,$ then we derive contradiction. To start, we pick up $U_\bullet$ containing $\varepsilon_0,$ and then extract an interval $(\alpha,\beta)$ inside of $U_\bullet$ which contains $\varepsilon_0.$ Now, again, $\alpha \lt \varepsilon_0,$ so $\alpha\in S,$ and we get a finite subcover \[\{V_k\}_{k=1}^N\] which covers $[0,\alpha].$ But then \[\{V_k\}_{k=1}^N\cup\{U_\bullet\}\] finitely covers $[0,\alpha]\cup(\alpha,\beta)=[0,\beta).$ In particular, we can cover $\left[0,\frac{\varepsilon_0+\beta}2\right]$ finitely, but $\frac{\varepsilon_0+\beta}2 \gt \varepsilon_0,$ breaking our least upper bound.

Having covered all cases, we conclude that we can always extract a finite subcover of $[0,1].$

We remark with some motivation because the given proof is quite magical. The argument is somewhat motivated by pieces I heard of it a year ago but mainly (I would like to think) by the following framing. Imagine we are given an open cover $\{U_\alpha\}_{\alpha\in\lambda}$ of $[0,1],$ and we need to extract a finite subcover.

We can codify this situtaion as a game: I am allowed to query an "interval genie'' with a real number $r\in[0,1],$ and I will receive some $U_\bullet$ containing $r.$ Effectively, this is all of the power I have over my cover. The naive approach, now, to extract an efficient subcover, is the following.

  1. Query the genie with $0.$ I get back some $U_\bullet$ containing the interval $[0,\varepsilon_0).$

  2. Query the genie with $\varepsilon_0.$ I get back some (maybe other) $U_\bullet$ containing the interval $(-,\varepsilon_1)$ containing $\varepsilon_0.$

  3. Query the genie with $\varepsilon_1.$ Rinse and repeat.

This, of course, doesn't work because the genie can do things like giving me\[\left[0,\frac14\right),\left(\frac18,\frac38\right),\left(\frac5{16},\frac7{16}\right),\cdots,\left(\frac{2^n-3}{2^{n+1}},\frac{2^n-1}{2^{n+1}}\right),\cdots.\]Here I keep querying numbers of the form $\left(2^n-1\right)/2^{n+1},$ off to infinity, never even going to cover $\frac12$ with my generated subcover.

The key observation is that, after I realize what the genie is doing to me, I can query the genie for $\frac12.$ This gives me some open interval containing $\frac12$ with a little space to the left, which means that I really only need finitely many of the genie's garbage intervals from before.

As I understand, the argument presented above is more or less a formalization of this trick. Notably, this trick alone is not enough to get a finite subcover, for the genie could do the same thing trapping me below $\frac34,$ and then below $\frac78,$ and so on. I would have to iterate the trick on top of itself, and maybe more times on top of that.

What we have to do is consider covering, in the abstract, various intervals\[[0,\varepsilon]\]for various values of $\varepsilon.$ Notably, the genie can always play tricks up to any particular value of $\varepsilon,$ but we can always overcome the genie at this single $\varepsilon.$ The way to formalize the idea that we can "always push farther'' up to $1$ is via the least upper-bound property, fixing $\varepsilon$ the farthest we can push, as done above.