Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
3  
Hint: A metric space is separable if and only if it is second countable. Second countability passes to subspaces. – t.b. Aug 9 '12 at 15:09
1  
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 $X$ – clark Aug 9 '12 at 15:14
1  
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. – Norbert Aug 9 '12 at 15:51
    
My problem was how to pass from the topology of the subspace to the whole topology space – clark Aug 9 '12 at 15:53
up vote 5 down vote accepted

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 $$U= \bigcup_{y \in U}V_y.$$

share|cite|improve this answer
2  
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) – t.b. Aug 9 '12 at 15:42
    
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 ) – clark Aug 9 '12 at 15:49
1  
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 $\epsilon/2$. – t.b. Aug 9 '12 at 15:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.