If $Y$ is a dense linear subspace of a normed space $X$, how to show $B_Y$ is dense in $B_X$.

Question

Let $X$ be a normed space and $Y$ be a proper dense linear subspace of $X$. Let $B_Y$ be the closed unit ball of $Y$ and $B_X$ be the closed unit ball of $X$. Then how to show $B_Y$ is dense in $B_X$.

Thank you in advance!

Answer 1

Let $x\in B_X$ and let $y_n\to x$ with $y_n\in Y$. Then, $\lVert y_n\rVert\to \lVert x\rVert$. If $\lVert x\rVert<1$, then $y_n$ is eventually in $B_Y$ by the previous remark. If $\lVert x\rVert=1$, then $\frac{1}{\lVert y_n\rVert} y_n\to x$ and $\frac1{\lVert y_n\rVert}y_n\in B_Y$ because $\left\lVert\frac1{\lVert y_n\rVert}y_n\right\rVert=1$.

Answer 2

Note: I am using the sequential definition of dense.

If $Y$ is dense in $X$, that means that for all $x \in X$ there exists $\{y_k\} \in Y$ such that $y_k \rightarrow x$ as $k \rightarrow \infty$. Then to show your claim Let $x \in B_X$ then by definition there exists $\{y_k\} \in Y$ such that $y_k \rightarrow x$. Note that due to that there exists a subsequence $y_{kj} \in B_X$ that has that same property ie $y_{kj} \rightarrow x$. Now you also know that $ B_T\subset B_X$ therefore you have that there exists $y_{kj} \in B_Y$ that converges to $x$. This way you avoid using normalization arguments.