# Compactness of the set $(0, 1)$ as a subset of $\mathbb{R}^2$?

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?

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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.

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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.

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