Hi all I think I found a new proof that $[0, 1]$ is compact but I am not 100% if it is correct, could you help me check? In the usual proof we just take the sup{x in [0, 1] : [0, x] is covered by finitely many intervals} etc.

My proof goes like this:

First we show that compactness for $[0, 1]$ is equivalent to countable compactness: if we have an uncountable open cover of $[0, 1]$ then let us first notice that if a point is covered by more than 2 intervals one of the intervals is contained in the union of the other two so we may as well assume that any point in $[0, 1]$ is covered by at most two intervals.

Next, select a rational from inside every interval in the cover and note that since every rational is contained in at most two intervals by the above and since there are countably many rationals we get a countable cover.

Finally, to show that $[0, 1]$ is countably compact, we can do the following : suppose $I_1, I_2, ..., I_n, ...$ is a countable cover such that $[0,1]$ is not covered by $I_1, ..., I_k$ for any natural k. Now $[0,1] - I_1$ is a union of at most two closed intervals. It must be the case that at least one of these two intervals is never covered by $I_1,...,I_k$ for any natural k. Select this interval, wlog let's call it $J_1$. Next, $J_1-I_2$ is also a union of at most two closed intervals, one of which cannot be covered by $I_1,...,I_k$ for any natural k.Let's call this interval $J_2$. Inductively we get $J_1 \supset J_2 \supset J_3$ ... a descending sequence of closed intervals whose intersection is non-empty and not covered by any $I_n$, contradiction.

Is this correct? I find this proof much more intuitive than the usual one, don't you agree?

N.B. actually a student of mine found this proof

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    $\begingroup$ Your proof reminds me of Lindelof covering theorem. Any set with an open cover admits a countable subcover. If the set is closed and bounded then the countable subcover can be reduced to a finite one (Heine Borel). You should have intersection of $J$ as empty. $\endgroup$ – Paramanand Singh Oct 15 '18 at 13:04
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    $\begingroup$ You lost me at "so we may as well assume that any point ... is covered by at most two intervals." If a point $x$ is covered by finitely many intervals, I see how you can throw away all but two of them without losing anything. But what if $x$ is covered by infinitely many intervals $I_1\subset I_2\subset I_3\subset\cdots$? How do you decide which ones to keep and which to throw away? $\endgroup$ – bof Oct 16 '18 at 0:26
  • $\begingroup$ @bof the more points there are in the intersection of the $J_n$ the better because $J_n$ lies in the complement of the union of $I_1$ up to $I_n$ and therefore if a point lies in all of the $J_n$ it will lie in none of the $I_n$ $\endgroup$ – Sorin Tirc Oct 16 '18 at 4:22
  • $\begingroup$ Oh, right. Seems to me it would be simpler (and more intuitive) to just define $x_n$ to be the least element ofthe closed set $[0,1]\setminus(I_1\cup I_2\cup\cdots\cup I_n)$ and let $x$ be the limit of the nondecreasing sequence $x_n$. $\endgroup$ – bof Oct 16 '18 at 5:15

I don't see how your 2nd step is guaranteed to produce a countable cover of all the irrationals in $[0,1].$

I suggest this: Let $C$ be a cover of $[0,1]$ by open subsets of $\Bbb R.$ For each $x\in [0,1]$ choose a bounded open real interval $j(x)$ with rational end-points and such that $x\in j(x)\subset c$ for some $c\in C.$

Now $D=\{j(x):x\in [0,1]\}$ is a subset of the family of open real intervals with rational end-points, so $D$ is countable. And $D$ is a cover of $[0,1].$

So if there exists a finite $E\subset D$ such that $\cup E\supset [0,1]$ then for each $e\in E$ choose $c_e\in C$ such that $e\subset c_e.$ Then $\{c_e:e\in E\}$ is a finite subset of $C$ and a cover of $[0,1]$.

In general: (i). A space whose topology has a countable base (basis) is called second-countable. And the family of all real open intervals with rational end-points is a countable base (basis) for the topology of $\Bbb R.$ If $Y$ is any subspace of a second-countable space then $Y$ is second-countable. (ii). A space for which every open cover has a countable sub-cover is called a Lindelof space. A second-countable space is always Lindelof. (iii). To prove that a space $X$ is compact, it suffices to find a base $B$ for $X $such that any cover of $X$ by a subset of $B$ has a finite sub-cover.

So it DOES suffice to prove that any cover of $[0,1]$ by a countable family of open real intervals has a finite sub-cover, and your proof of that is good.


It's not clear to me how the selection procedure in your third and fourth paragraphs would work.

Let $\alpha = 1/\sqrt{2}$, and consider the cover of $[0, 1]$ by the uncountable collection consisting of (i) the open interval $K = (-1, 2)$, (ii) all the open intervals $I(\beta) = (\beta, \beta + 1)$, where $\alpha < \beta < 1$, and (iii) all the open intervals $J(\gamma) = (\gamma - 1, \gamma)$, where $0 < \gamma < \alpha$.

Every rational in $[0, 1]$ belongs to $K$, and either infinitely many intervals $I(\beta)$ or infinitely many intervals $J(\gamma)$.

For rational $q > \alpha$, I cannot see why your procedure, whatever it is, might not be tricked into considering only three of the intervals $I(\beta)$, for example $I((3\alpha + q)/4)$, $I((\alpha + q)/2)$, and $I((\alpha + 3q)/4)$, and then choosing $I((3\alpha + q)/4)$ and $I((\alpha + 3q)/4)$, perhaps because their union contains $I((\alpha + q)/2)$. Similarly for rational $q < \alpha$, with $J$ in place of $I$.

If this were to happen, then $K$ would not be chosen (even when a rational number is selected from it), and the selected countable subcollection of open intervals would not cover $[0, 1]$.

Have I got the wrong end of the stick? Quite likely! But even so, some clarification seems to be needed.

Everything depends on exactly how your words are to be interpreted. I hope I have not put too twisted an interpretation on them. I have tried not to enforce any single detailed interpretation, by mere "nitpicking".

Can you spell out in more detail exactly what would happen to the uncountable cover that I defined above?

  • $\begingroup$ I don't see how the 2nd step in the proof in the Q is guaranteed to produce a cover of all of $0,1]$ . $\endgroup$ – DanielWainfleet Oct 16 '18 at 0:16

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