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I’ve just begun studying about compact spaces and encountered a part which I don’t get.

In the note, it says (0,1] is not a compact space because {(1/k,1) : k=1,2,3,...} is an open cover of (0,1] with no finite subcover.

But I don’t see how the union of such intervals, which all have an open end at 1, can be closed at 1.

I’d be really grateful for any help. Thank you.

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    $\begingroup$ Yeah that’s definitely not a cover of $(0,1]$. $\endgroup$ – User May 15 '18 at 22:28
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    $\begingroup$ Perhaps they had a typo with $]$ instead of $)$. $\endgroup$ – Alex R. May 15 '18 at 22:30
  • $\begingroup$ I think you want something like $\{(1/k, 2)\}_{k=1}^\infty$ as your open cover.(?) $\endgroup$ – MMASRP63 May 15 '18 at 22:30
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    $\begingroup$ Nonetheless, you can add one more set to the union $(1-\delta, 1+\delta)$ and you have an open cover of the interval, yet is still does not have a finite sub-cover. $\endgroup$ – Doug M May 15 '18 at 22:30
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The given set is not a cover of $(0,1]$. There are two possible corrections. If “compact” is to be understood as “compact subset of $\mathbb R$”, then you can consider e.g. the open cover $\{(1/k, 1+1/k)\}$. If “compact” is to be understood as “compact as a topological space by itself” (which I think is the intended one), then the small fix would be to consider $\{(1/k, 1]\}$. The intervals $(1/k, 1]$ are open sets in $(0,1]$ when given the subspace topology.

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  • $\begingroup$ Thanks! That really helped a lot. $\endgroup$ – June15 May 15 '18 at 22:35
  • $\begingroup$ @June15 If the answer settled your question, please accept it. $\endgroup$ – Henno Brandsma May 19 '18 at 7:33

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