Is the unit ball of a dense set dense?

Question

Let $(X,||\cdot||)$ be a normed vector space, and let $Y \subset X$ be a dense subset of $X$. Does it follow that $\{y: y \in Y, ||y|| \leq 1\}$ is dense in $\{x: x \in X, ||x|| \leq 1\}$?

Answer 1

Let $B$ stand for the unit ball. Then the answer is yes if $closure(interior(B)) = B$.

Let $x \in B$, and let $\varepsilon > 0$. We have to show that there is a $y \in Y \cap B$ within $\varepsilon$ of $x$.

Since $Y$ is dense in $X$, there is a $z \in Y$ with $\delta$ of $x$ for any $\delta > 0$. Thus, if $x \in interior(B)$, then we can pick $\delta = \min \{{1-||x||,\varepsilon\}}$ small enough so that $z \in B$, and hence $z \in B \cap Y$.

If $x$ has $||x|| = 1$, then we will reduce it to the first case as follows: Pick another point $x' \in interior(B)$ within $\varepsilon /2$ of $x$, which exists because $closure(interior(B)) = B$. Now since $x'$ is in the interior, we can use the first case to find a point $z' \in B \cap Y$ within $\varepsilon /2$ of $x'$. By triangle inequality, this is within $\varepsilon$ of $x$.

Note that we did not need to use fully that $Y$ was dense in $X$, just that its dense in $B$.

Answer 2

Yes. Let $x \in X$ such that $\lVert x \rVert \le 1$. Consider $\epsilon > 0$.

Define $x_n = \frac{n-1}{n}x$. Then $\lVert x_n \rVert = \frac{n-1}{n}\lVert x \rVert < 1$ and $\lVert x - x_n \rVert < \epsilon/2$ for $n \ge N$. Since $Y$ is dense, there exists $y_N \in Y$ such that $\lVert y_N - x_N \rVert < \min(\epsilon/2, 1 - \lVert x_N \rVert$. Then $\lVert x - y_N \rVert \le \lVert x - x_N \rVert + \lVert x_N - y_N \rVert < \epsilon$ and $\lVert y_N \rVert \le \lVert y_N - x_N \rVert + \lVert x_N \rVert < 1 - \lVert x_N \rVert + \lVert x_N \rVert = 1$.