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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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    Yes, i proved this for metric spaces!2012-11-10

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 $\tilde{P}= \{ \sum\limits_{i=1}^n q_i x_i : n \geq 1, q_i \in \mathbb{Q},x_i \in Q \} \cap B_1(0)$.