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Suppose $X$ is a normed space. Denote $B_X$ as closed unit ball in $X$. Let $D$ be a dense subset of $B_X$ and $Y$ contains $D$ where $Y$ is a subspace of $X$. Show that $Y$ contains a dense subset $E$ of $X$.

My attempt:

Suppose that $E$ is a dense subset of $X$. We want to show that $E \subseteq Y$.

Let $x \in E$. Choose $m \in \mathbb{N}$ such that $m > \| x \|$ (exists by the Archimedean property). Then we have $$\frac{x}{m} < \frac{x}{\| x \|} <1 \Rightarrow \frac{x}{m} \in B_X.$$

Since $\overline{D}=B_X$, there exists a sequence $(x_n)_{n \in \mathbb{N}}$ from $D$ such that $$\dfrac{x}{m} = \lim_{n \rightarrow \infty}{x_n} \Rightarrow x = m \lim_{n \rightarrow \infty}{x_n}$$

Since $D \subset Y$ and $Y$ is a subspace, we have $m x_n \in Y$. But we can only conclude $x \in \overline{Y}$ from here instead of $x \in Y$.

Can anyone guide me?

Idonknow
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    You're assuming what you need to prove :) — "Suppose $E$ is a dense subspace of $X$". Instead, consider $E = {r x \mid x \in B_X, r \in \mathbb{R}}$. – BrianO Oct 28 '15 at 02:50

1 Answers1

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The first sentence is not a good starting point: you would end up proving that $Y$ contains every dense subset of $X$, which might not be the case.

Consider instead $W = \text{span}(D)$. Since $Y$ is a vector space we have that $W \subset Y$. Now all is left to prove is that $W$ is dense in $X$, which is not hard to show. Indeed, fix $\epsilon > 0$ and take $x \in X \setminus \{0\}$. Then $\frac{x}{\|x\|} \in B_X$, hence there exists $d \in D$ such that $\Big\|\frac{x}{\|x\|} - d\Big\| < \frac{\epsilon}{\|x\|}.$ This shows that $\|x\|\cdot d \in W$ satisfies $\|x - \|x\|\cdot d\| < \epsilon.$

Giovanni
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