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Let $X$ be a separable normed space. Is it true that every subspace is separable? If it was Hilbert space I would take the dense set and then their projections. It sounds trivial but I cannot prove or disprove it...

Thank you.

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    Related: https://math.stackexchange.com/questions/1628253/subspace-of-a-separable-space-is-separable2017-01-08

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Let $X$ be a separable metric space and $x_n$ a dense subset. Let $a_m$ be an enumeration of the positive rational numbers and let $V_{(n,m)}=\{y \in X \mid d(y,x_n) < a_m \}$. This is a countable base of X.

Indeed, take an open set $U$ of $X$. For $y_0 \in U$ there is $\epsilon >0$ such that $B(y_0, \epsilon) \subset U$. Choose $x_{n_0}$ such that $d(x_{n_0},y)< \frac{ \epsilon }{4}$ and $a_{m_0}$ such that $ \frac{ \epsilon }{4}< a_{m_0} < \frac{ \epsilon }{2} $.Then $y \in V_y=V_{ (n_0 , m_0)} \subset U$ and $U= \bigcup_{y \in U}V_y.$

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    Small nitpick: better take all rational radii (instead of only those of the form $1/m$) because there need not be a number of the form $1/m_0$ between $\epsilon/4$ and $\epsilon/2$.2012-08-09