Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

An open ball in $\mathbb{R}^2$, centered at the point $(1/2, 0)$ and of radius $1/2$ covers the segment $(0,1)$. The open ball thus forms a finite cover of $(0,1)$, implying that $(0,1)$ is a compact set. But that is wrong, for compact sets are closed. Can some one help me find the fallacy?

share|cite|improve this question

A set $A$ is compact if every open cover of it has a finite subcover. You can always cover a set with one open set: the whole space is always an open set. Here, for instance, you could have covered the set with the single open set $\Bbb R^2$.

For $n\ge 3$ let $B_n$ be the open ball of radius $\frac12-\frac1n$ centred at $\left\langle\frac12,0\right\rangle$; then $\{B_n:n\ge 3\}$ is an open cover of the interval that does not have a finite subcover.

share|cite|improve this answer
Hmpf. You win this round! :-P – Asaf Karagila Aug 15 '13 at 16:56
@Asaf: Yes, but you just had a whole week when I was visiting friends $375$ miles from home and was able to be online much less than usual! :-) – Brian M. Scott Aug 15 '13 at 17:03
But I was busy with other things for the past three weeks (I was relatively inactive, too)! – Asaf Karagila Aug 15 '13 at 17:06
you both deserve gold stars – Lubin Aug 15 '13 at 17:28
@Brian. Thanks to both. I was trying to find an open cover for both (0, 1) and [0, 1] hoping to show that it stayed being infinitely large for the former case and finite for the latter. I am yet to get it. Can you help me find it? – Amey Joshi Aug 16 '13 at 8:08

Congratulations. You have found a finite cover of the set. But the condition for compactness is that every open cover has a finite subcover.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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