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Could you give me some hint to prove this assertion? I've found a proof when the closed subset doesn't have isolated points: If $X$ is a separable space, then it admits a numerable dense basis. Now consider the internal points of a closed subset $C$, since it is open, it is a numerable union of elements of the numerable basis. Now, let we consider $C$ a closed subset of $X$; $C$ doesn't have isolated points, so the closure of the interior of $C$ is the closure of $C$, since we can write the interior of $C$ like a numerable union of elements in the numerable basis, there exists a subset of $C$ with cardinality $\leq \aleph_0$ such that its closure is all $C$, so $C$ is separable.

Another approach to prove the assertion is make the projection of $N$ (the numerable dense subset of $X$ whose closure is all $X$), in the closed subset $C$, but I doubt this thing will works in Hilbert spaces only, because in a Banach space I don't know if $\forall m\in N$ $\arg\inf_{c\in C}{d(m,c)}$ is unique.

-The problem is that I can't visualize how $\mathbb{R}\setminus\mathbb{Q} $ is a separable space... who is its numerable dense subset whose closure is all $\mathbb{R}\setminus\mathbb{Q} $?-

Elia
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    Every subset of separable metric space is separable. Read this: https://math.stackexchange.com/questions/516886/prove-that-a-subset-of-a-separable-set-is-itself-separable?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa The particular case of $\mathbb{R}\backslash\mathbb{Q}$ is quite simple: the dense countable subset is the subset of all (non-rational) algebraic numbers, or even simplier ${r\sqrt{2}\ |\ r\in\mathbb{Q}}$. – freakish Apr 30 '18 at 09:58
  • It would be even a bit better to consider ${r + \sqrt{2} | r \in \mathbb{Q} }$ above, since ${r \sqrt{2} | r \in \mathbb{Q} }$ is not a subset of $\mathbb{R} \setminus \mathbb{Q}$. But this is just a very minor inaccuracy. – pcp Apr 30 '18 at 11:08

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Let $X$ be any topological space $X.$ The weight $w(X)$ is the least infinite cardinal $W$ such that $X$ has a base (basis) $B$ with cardinal $|B|\leq W.$ The density $d(X)$ is the least infinite cardinal $E$ such that $X$ has a dense subset $D$ with cardinal $|D|\leq E. $

(Of course if $X$ does not have a finite base then $X$ has a base $B$ with $|B|=w(X).$ And if $X$ does not have a finite dense set then $X$ has a dense subset $Y$ with $|Y|=d(X).$)

(I). We always have $w(X)\geq d(X).$ For if $B$ is a base for $X$ with $|B|\leq w(X)$ then for $\phi \ne b\in B$ let $f(b)\in b.$ Then $D=\{f(b): \phi \ne b\in B\}$ is a dense subset of $X,$ and $|D|\leq |B|\leq w(X).$

(II). If $Y$ is a sub-space of $X$ then $w(Y)\leq w(X).$ For if $B$ is a base for $X$ with $|B|\leq w(X)$ then $C=\{b\cap Y: b\in B\}$ is a base for $Y$, and $|C|\leq |B|\leq w(X)$.

(III). For a metric space $(X,d)$ we have $w(X)=d(X)$: From (I) we have $w(X)\geq d(X).$ To show $w(X)\leq d(X),$ let $Y$ be a dense subset of $X$ with $|Y|\leq d(X).$ Then the set of open $d$-balls $B=\{B_d(y,q):y\in Y\land q\in \Bbb Q^+\}$ is a base for $X$ [Exercise]..... and we have $|B|\leq |Y\times \Bbb Q^+|=|Y|\cdot \aleph_0 \leq d(X)\cdot \aleph_0=d(X)$.

(IV). So if $Y$ is a sub-space of a separable metric space then $d(Y)=w(Y)\leq w(X)=d(X)=\aleph_0.$

A normed linear space is a metric space with the metric $d(x,y)=\|x-y\|.$ By (IV) any sub-space of a separable normed linear space is separable.

DanielWainfleet
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