# An fallacious argument on recursive open cover for the interval [0, 1]

I build the following example that seems to contradict the fact that [0, 1] is compact, but I can't think what I'm getting wrong.

Be $\{G_{\alpha}\}$ a collection of open sets defined recursively as illustrated in the figure bellow:

For the interval $[0, 1]$ define three open sets: \begin{eqnarray} G_{\alpha_1} = \left(-\frac{1}{6}, \frac{1}{6}\right) \\ G_{\alpha_2} = \left(\frac{2}{6}, \frac{4}{6}\right) \\ G_{\alpha_3} = \left(\frac{5}{6}, \frac{7}{6}\right) \end{eqnarray} this leaves two intervals between 0 and 1 uncovered: $\left[-\frac{1}{6}, \frac{2}{6}\right]$ and $\left[\frac{4}{6}, \frac{5}{6}\right]$, so next you do the same thing for each of those intervals, and for the uncovered intervals that will result from these two procedures and so on...

Doing this recursively you get an open cover $\{G_{\alpha}\}$ for the interval $[0, 1]$ and I don't see how a finite subcolection of $\{G_{\alpha}\}$ could completely cover the interval $[0, 1]$. What I'm getting wrong in this construction?

• Why do you cover all of $[0,1]$? – user2154420 Dec 15 '16 at 16:00
• You are effectively claiming that a Cantor set on $[0,1]$ is empty, but that's not true. Your "cover" doesn't cover $[0,1]$ – user160738 Dec 15 '16 at 16:02

## 1 Answer

Your construction is extremely similar to the construction of the Cantor set; we start with $[0, 1]$, and keep removing intervals, leaving less and less of $[0, 1]$ "unremoved" (or in your case, uncovered).

However, eve though the Cantor set - what's left over - is very small, in the sense of not containing an interval (indeed, being nowhere dense), it is definitely not empty; in fact, it has uncountably many points! Similarly, your construction covers "most" of $[0, 1]$, but leaves uncountably many points uncovered.

• This make sense for me! Thank you for the answer! – Antonio Horta Ribeiro Dec 15 '16 at 16:33