# Why do I think Lebesgue’s number lemma is wrong…

Counterexample: $[0,1]\times[0,1]$ with induced subspace topology from $\mathbb{R}^2$ is compact. The open cover $\mathscr{U}$ is just the two circular sectors. When we look at the up-left corner and down-right corner, it fails - there is no such $\delta$, such that let the open ball be only in one of circular sectors... what is wrong?

• There is no way to make those two quarter disks open and to include the corners $(1,0)$ and $(0,1)$. – Thomas Andrews Oct 9 '17 at 23:27

Let $\mathcal U$ be a cover. For any $x$, there is some $\delta$ and some $U\in \mathcal U$ such that $B(x,\delta)\subseteq U$
Note that Lebesgue's number lemma exchanges the quantifiers and says that there is some $\delta$ that works for all $x$. This is what requires compactness.
• Thanks for pointing out. I did not thought in that way, I though the lemma was weirdly wrong, because when I fix a $\delta$ and then moving a small ball towards that two corner, there must be a time point that the small ball is not fully contained in any circular sectors. But yes.. there are another two sectors missing to cover the two corners. – Upc Oct 9 '17 at 23:39