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Let $X$ be a normed space, i want to show the equivalence of

(i) $X$ is separable

(ii) $B_1(0) = \{ x \in X : \|x\| < 1 \}$ is separable

(iii) $K_1(0) = \{ x \in X : \|x\| \le 1 \}$ is separable

(iv) $S_1(0) = \{ x \in X : \|x\| = 1 \}$ is separable

(v) there exists a countable set $A \subseteq X$ with $X = \overline{\operatorname{span}(A)}$

For every separable space, every subset is also separable, so I got (i) => (ii), (i) => (iii), (i) => (iv), (iii) => (iv) and (iii) => (ii). Furthermore I was able to prove (iv) => (v) and (v) <=> (i). But then there is still something left, for example (ii) <=> (iv). Can you please give me hints what would be the best way to prove (i) <=> (ii) <=> (iii) <=> (iv) <=> (v)?

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Careful: a subset of a separable topological space need not in general be separable (though this does hold for metric spaces), so you may want to check your proof of this. Also, perhaps you could post the proofs of what you have? They might be helpful in suggesting how to handle the remaining parts. –  Nate Eldredge Nov 10 '12 at 14:02
    
Yes, i proved this for metric spaces! –  Stefan Nov 10 '12 at 14:05
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2 Answers 2

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I suggest you to prove the following facts

Fact 1. If $A\subset X$ and $X$ is separable, then $A$ is separable

Fact 2. If for all $n\in\mathbb{N}$ the set $A_n\subset X$ is separable, then $\bigcup_{n\in\mathbb{N}}A_n$ is separable

Fact 3. If $A\subset X$ is separable, $\lambda\in\mathbb{R}$, then $\lambda\cdot A$ is separable.

Fact 4. If $X=\mathrm{cl}(A)$ and $A$ is separable then $X$ is separable

Now the road map to make your homework is the following

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Implications $(i)\to (iii)$, $(iii)\to (ii)$, $(iii)\to (iv)$ follows from fact 1.

Implication $(ii)\to (i)$ follows from equality $X=\bigcup_{n\in\mathbb{N}} n\cdot B_1(0)$ and facts 2, 3.

Implication $(iv)\to (i)$ follows from density of $\bigcup_{q\in\mathbb{Q}_+} q\cdot S_1(0)$in $X$ and facts 2, 3, 4.

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Hint for $(ii) \Leftrightarrow (iv)$: if $Q$ is a countable dense subset of $B_1(0)$, consider $\tilde{Q}= \{ x/||x|| : x \in Q \backslash \{0\} \}$; if $P$ is a countable dense subset of $S_1(0)$, consider $\tilde{P}= \{ \sum\limits_{i=1}^n q_i x_i : n \geq 1, q_i \in \mathbb{Q},x_i \in Q \} \cap B_1(0)$.

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