Let $S=\{x\in\mathbb{R}\,: 0 \leq x < 1\}$. $T$ is a topology on $S$: $T=\{Ø,U\}$, where the open sets $U \subseteq S$ is defines as $U=\{x\in\mathbb{R}: 0 \leq x < k\}$, where $0 < k \leq 1$.

I want to see if the topology $T$ stems from a metric in $S$. I know that every metric space is a Hausdorff space. So if the space $(S,T)$ is a Hausdorff space, then $T$ stems from a metric in $S$. In order for the space to be a Hausdorff, then for every distinct points $x$ and $y$ in $S$, there is a pair of disjoint open neighborhoods $V$ of $x$ and $V$ of $y$. So there should exist an open set for $x$ which only contains $x$, and an open set for $y$ which only contains $y$.

All open sets are defined in $U$, so they have the form of $U=\{x\in\mathbb{R}\,: 0 \leq x < k\}$, where $0 < k \leq 1$. But on this form, the open sets are contained on each other. So its not a Hausdorff. Am I right and is that a sufficient argument ?

  • $\begingroup$ But $\displaystyle \cup_{0\leqslant k<1}U=[0,1]$ isn't in $S$. $\endgroup$ – Nosrati Jan 13 '17 at 8:56
  • $\begingroup$ That union is $[0,1)$. $\endgroup$ – user42761 Jan 13 '17 at 8:57
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    $\begingroup$ Since $0$ is in all open, it's can't be haussdorf. $\endgroup$ – user380364 Jan 13 '17 at 8:58

The argument you give is right. In particular any non-empty open subset $U$ contains $0$.


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