Student question about compactness of sets

Question

I'm just learning about compactness and being confused.

Closed sets are not compact?

Take $[a,b]$ and the covering $\{(a+\frac{b-a}n,b-\frac{b-a}n):n\geq3\}$. There is no finite subcover. $n$ must approach infinity otherwise it doesn't cover $[a,b]$.

I'm probably missing something obvious here.

The set of intervals you wrote only covers the open interval $(a,b)$. — Berci, May 15 '14 at 23:28
No. As a limitpoint, yes, but neither $a$ nor $b$ will be an element of the union of these intervals. — Berci, May 15 '14 at 23:29
Does this mean open sets are never compact? Because this applies to open sets. — user144794, May 15 '14 at 23:30
Assuming $T_2$ separation ('Hausdorff') --which of course holds in $\Bbb R$-- every compact set is closed. In $\Bbb R^n$ the compact sets are exactly the closed, bounded sets. — Berci, May 15 '14 at 23:32

score 0 · Accepted Answer · answered May 15 '14 at 23:30

The set of intervals you wrote only covers the open interval $(a,b)$.

Anyway, $[a,b]$ is compact, but e.g. a closed ray $[0,+\infty)$ is closed but not compact in $\Bbb R$.

In $\Bbb R^n$ compact sets are just the closed bounded sets.

By the way, your argument perfectly works to show that the open interval $(a,b)$ is not compact.

Student question about compactness of sets

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