Proof for separable normed space

Question

I want to proof that if the norm space X, has a dense separable subset then X is separable.

I was thinking that if Y is the subset of X then cause it is separable it has a dense,countable subset (for example D). Then D is also subset of X so from the definition of separability, X is separable.

Is it right?

Answer 1

Let $X$ be a normed space and $Y$ a dense separable subset of $X$. Let $x\in X$ and $(x_n)_{n\in\mathbb N}\in Y^{\mathbb N}$ a sequence that converges to $x$.

Let $D=\{d_n\mid n\in\mathbb N\}\subset Y$ countable s.t. $Y=\overline{D}$. For all $n\in\mathbb N$, there is $(d_m^n)_{m\in\mathbb N}\in D^{\mathbb N}$ s.t. $d_m^n\to x_n$ when $m\to \infty $.

• Let $m_1$ s.t. $\|x_1-d^1_{m_1}\|<1$.

• Take $m_2>m_1$ s.t. $\|x_2-d_{m_2}^2\|<\frac{1}{2}$

$\vdots$

• Take $m_k>m_{k-1}>...>m_1$ s.t. $\|x_k-d_{m_k}^k\|<\frac{1}{k}$.

$\vdots$

Therefore $(d_{m_k}^k)_{k\in\mathbb N}$ is a sequence of $D$ that converges to $x$ and thus $X=\overline{D}$. We conclude that $X$ is separable.

Answer 2

What you wrote is true but should maybe be fleshed out more. Let $\overline D$ and $\overline Y$ denote closures in $X$ and $\widetilde D$ denote closure in $Y.$ We have $\overline D\supseteq Y\cap \overline D = \widetilde D = Y.$ It follows that $\overline D\supseteq \overline Y=X,$ hence $\overline D=X$ and $D$ is a countable dense subset of $X.$ Thus $X$ is seperable.