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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3$\begingroup$ Hint: A metric space is separable if and only if it is second countable. Second countability passes to subspaces. $\endgroup$– t.b.Commented Aug 9, 2012 at 15:09
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1$\begingroup$ I see so it is seperable because every open set of a subspace is of the form $Y \cap V$ where V is an open set of $X$ $\endgroup$– clarkCommented Aug 9, 2012 at 15:14
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1$\begingroup$ A metric space is separable IFF it doesn't contain uncountable family of points with pairwise distance greater than some $\delta$. Corollary - subspace of separable space is separable. $\endgroup$– NorbertCommented Aug 9, 2012 at 15:51
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$\begingroup$ My problem was how to pass from the topology of the subspace to the whole topology space $\endgroup$– clarkCommented Aug 9, 2012 at 15:53
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$\begingroup$ Related: math.stackexchange.com/questions/1628253/… $\endgroup$– WatsonCommented Jan 8, 2017 at 16:49
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1 Answer
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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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3$\begingroup$ Exactly. Now intersect your base with the subspace, get again a base and pick a point in each one of the countably many basis sets. +1. (Btw: it's sepArable, not seperable) $\endgroup$– t.b.Commented Aug 9, 2012 at 15:42
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$\begingroup$ Yes yes, I understood from your comment if I proved that how it would go afterwards. Thanks! (P.S. I will not make that mistake again it always slipped my mind, it was because of my wrong accent ) $\endgroup$– clarkCommented Aug 9, 2012 at 15:49
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1$\begingroup$ 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 $\epsilon/2$. $\endgroup$– t.b.Commented Aug 9, 2012 at 15:52