I was reading this interesting post about compactness. I understood that the outcome of the story was that finiteness is the same as compactness+discreteness. I understand that discreteness+compactness $\implies$ finiteness. Also it is clear that finiteness implies compactness. But why finiteness $\implies$ discreteness? If I consider the finite space $X = \{x_1,x_2,x_3,x_4,x_5\}$ with the coarse topology $T=\{\emptyset,X\}$ it is not discrete no?
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It seems they were working on $\mathbb{R}^{n}$, according to the OP of that post. – Rigid AOE2 Nov 21 '20 at 16:54
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1It holds in $T_1$ spaces, and $T_ 1$ separation is a minimum requirement for a great many topological theorems. – Brian M. Scott Nov 21 '20 at 18:04
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When I wrote that answer I had in mind a “pre-theoretical” understanding of finiteness, and also as Law Nagi says in the comments the OP’s question was about subspaces of $\mathbb{R}^n$. You can interpret me to have been restricting attention to metric spaces or Hausdorff spaces if you want a precise statement.
Qiaochu Yuan
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