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As I understand the definition of a subspace, I can simply take any subset of a topological space and promote it to a subspace by endowing it with the subspace topology. Then in this case every subset of my Hausdorff space, endowed with the subspace topology for whatever topology the space has, is compact. Therefore every subspace is a closed subset. I would like for this generalization to be true since from here I think I can go on to talk about the topology on my Hausdorff space being the discrete topology and the space being finite.

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  • $\begingroup$ It's indeed true all subsets are closed $\endgroup$ – Henno Brandsma Mar 24 '17 at 6:22
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    $\begingroup$ What is your question, exactly? $\endgroup$ – Eric Wofsey Mar 24 '17 at 6:22
  • $\begingroup$ @EricWofsey If I can take the fact that every subspace is closed to an extreme of sorts by saying every subset of my space endowed with the subspace topology is a subspace and so every subset of my space is closed. $\endgroup$ – jesusbourne Mar 24 '17 at 6:25
  • $\begingroup$ It is still not clear what your question is, but let me tell you that your conclusion that every space satisfying the condition of the title is finite discrete, is correct. $\endgroup$ – MooS Mar 24 '17 at 6:53
  • $\begingroup$ Sorry for being unclear. I just want to make sure that I can say every subset of my space is closed from the assumption that every subspace of my space is compact. As an additional question, since discreteness is incredibly easy to prove if all the subsets of my space are closed, what would you recommend I try to show finiteness? This part actually is not as easy as I thought it'd be. $\endgroup$ – jesusbourne Mar 24 '17 at 7:09
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Let $A \subset X$. Then by assumption, $A$ (in the subspace topology, what else?) is compact. A compact subspace in a Hausdorff space is closed. So $A$ is closed. So all subsets of $X$ are closed, so all subsets have a closed complement as well, so all subsets are open too. Hence $X$ is discrete etc.

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  • $\begingroup$ @MatemáticosChibchas Yes, he already concluded that according to the question. $\endgroup$ – Henno Brandsma Mar 25 '17 at 9:07

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