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If the metric space $P$ is separable, then there exists a countable dense subset $X\subset P$.but if P is uncountable, Prove that $P \setminus X$ is not separable.

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    $\begingroup$ It's false. A metric space is separable if and only if it is second countable, and second countability is inherited by subspaces. So every subspace of a separable metric space is separable. $\endgroup$ – Daniel Fischer Mar 28 '16 at 18:36
  • $\begingroup$ Your definition of Seprability is right, but my defination is also right it is written in my University prescribed book. $\endgroup$ – CHANDAN Apr 1 '16 at 18:30
  • $\begingroup$ I haven't stated a definition of separability. Second countability and separability are different properties a space can have. All second countable spaces are separable, but not all separable spaces are second countable. However, all separable metric spaces are second countable. And second countability is inherited by subspaces. Thus for metric spaces, separability is inherited by subspaces. (That's not generally the case, there are separable topological spaces that have non-separable subspaces. But such spaces aren't metrisable.) $\endgroup$ – Daniel Fischer Apr 1 '16 at 18:44
  • $\begingroup$ @DanielFischer finally i got tge proof of my result. And that true $\endgroup$ – CHANDAN Apr 1 '16 at 19:29
  • $\begingroup$ I don't understand what you say there. But just in case: If $P$ is a separable metric space, and $X$ any subset of $P$, then $P \setminus X$ is separable. $\endgroup$ – Daniel Fischer Apr 1 '16 at 19:34
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Just assume $P-X$ is separable and then, by the prerequisits for the metric space you work with, you will get a contradiction. But I dont know what your metric space is! In general the assumption is wrong, think of $(\mathbb{Z},d)$ with any metric(for example the discrete metric).

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    $\begingroup$ $\mathbb{R} \setminus \mathbb{Q}$ is uncountable. However, every subspace of $\mathbb{R}$ is separable, including $\mathbb{R} \setminus \mathbb{Q}$. $\endgroup$ – Carl Mummert Mar 28 '16 at 16:38
  • $\begingroup$ thanks! I was to hasty and already edited it $\endgroup$ – CandyOwl Mar 28 '16 at 18:18
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The statement is false. The reals, for example, have $\mathbb Q$ as a countable dense subset, however $\mathbb R - \mathbb Q$ is separable also. Consider the set $\mathbb Q + \pi = \{q + \pi \, |\, q \in \mathbb Q\}$. We still have that $\overline{\mathbb Q + \pi} = \mathbb R$, but this time we have that $\mathbb Q + \pi \subseteq \mathbb R - \mathbb Q$, thus forming a countable dense subset of $\mathbb R - \mathbb Q$. Therefore the irrationals are separable.

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