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A friend of mine has this statement, that I cannot establish wether it is true or false.

Let us consider $\mathbb{R}$ with the euclidean topology. Let us consider $A \subset \mathbb{R}$ such as the topology induced on it is the discrete topology, so $A$ is numerable.

Actually he states that the statement holds whenever $A$ is closed, but I ignore his proof.

I tried to demonstrate, recreating something like the Cantor set, that it is false proved that there exists a non separable set $A$ with that induced topology, but I cannot prove the existence of such $A$.

With discrete topology I mean that for every $x \in A$ there exists an open neighborhood of $x$, that I'll call $I(x)$ such that $I(x) \cap A = {x} $.

Any suggestions?

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  • $\begingroup$ A separable subspace of a separable metric space is itself separable. So yes, existence of a non-separable subspace of reals implies that statement (whatever it might be), albeit vacuously so. $\endgroup$ – tomasz Jan 3 '17 at 18:23
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For each $p\in A$ there is a $U_p$ such that $A \cap U_p = \{p\}$; the collection $\{U_p\}$ is uncountable. In each $U_p$ we can find an interval $V_p$ with rational endpoints containing $p$. Since also $A \cap V_p = \{p\}$, the $V_p$ are distinct and the collection $\{V_p\}$ is uncountable. But the collection of intervals $(q,r)$ with rational endpoints is in bijection with ordered pairs of rational numbers, of which there are only countably many.

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  • $\begingroup$ I wouldn't say the $V_p$ are unique (they are not), but what you meant to say was that the map $p \rightarrow V_p$ is 1-1, so we have an injection from $A$ into the countable set. $\endgroup$ – Henno Brandsma Jan 5 '17 at 4:39
  • $\begingroup$ yea distinct was probably the word $\endgroup$ – juan arroyo Jan 5 '17 at 4:46
  • $\begingroup$ Agreed. (type some stuff for minimum...) $\endgroup$ – Henno Brandsma Jan 5 '17 at 4:50

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