Define a topological space X that is not compact and define a set A ⊂ X that is compact. Use the definition of finite open subcovers to show that A is compact.

Ok so I think that a topological space that would not be compact could be the set of integers Z on standard topology. A subset could be [-3,3]. But I am not sure how to so that [-3,3] is compact using the definition of open subcovers.

  • $\begingroup$ Try to show that a discrete space is compact iff it is finite. Then you can take your example (provided you use the notation $[-3,3]$ to mean $[-3,3] \cap \mathbb{Z}$). $\endgroup$ Dec 20, 2015 at 2:38
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    $\begingroup$ Use say the reals with the usual topology for $X$. And now let's cheat, use a one point set for $A$, or even the empty set. $\endgroup$ Dec 20, 2015 at 2:38
  • $\begingroup$ @MattSamuel I've never seen a convention where "compact" implied "nonempty." Is this sometimes assumed? $\endgroup$ Dec 20, 2015 at 3:44
  • $\begingroup$ @Noah Not as far as I know. I was actually making a joke...I also find it a bit silly to say the empty set isn't connected. $\endgroup$ Dec 20, 2015 at 3:48

1 Answer 1


In $\mathbb{Z}$, the interval $[-3, 3]$ - that is, $\{-3, -2, -1, 0, 1, 2, 3\}$ - is compact because it is finite, and any finite subset of any space is compact. Do you see why? HINT: If $A$ is finite, write $A=\{a_1, . . . , a_n\}$; now if $\mathcal{U}=\{U_i: i\in I\}$ is an open cover of $A$, how many elements of $\mathcal{U}$ does it take to cover $\{a_1\}$? $\{a_1, a_2\}$? . . . All of $A$?

Side note: in $\mathbb{R}$, the interval $[-3, 3]$ is also compact, but this is harder to show.


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