Show that $\{y^*\}$ separates the points of $Y$.

Question

Let $(X,\|.\|)$ be a Banach space reflexive and $Y$ be the closed linear subspace of $X$. Then $(Y,\|.\|)$ is clearly a separable reflexive Banach space. Hence, the dual space $Y^*$ of $Y$ is certainly separable for the dual norm. Let $\{y^*\}$ be a countable dense subset of $Y^*$.

Why $\{y^*\}$ separates the points of $Y$?

An idea please

Answer 1

Suppose $x_1 \neq x_2$ and $y^{*} (x_1)=y^{*} (x_2)$ for all $y^{*}$ in the countable dense set. By taking limits we get the same equation for all $y^{*}\in Y^{*}$. By Hahn Banach Theorem there exists $y^{*}\in Y^{*}$ such that $y^{*}(x_1-x_2) \neq 0$ so we have a contradiction.

Answer 2

It is a standard corollary of the Hahn-Banach theorem that $Y^*$ separates the points of $Y$. Suppose that $\{y_n^*: n \in \mathbb{N}\}$ is a countable dense subset of $Y^*$ that doesn't separate the points of $Y$. Then, in particular there exist $x,y \in Y$ with $x \neq y$ such that $y_n^*(x) = y_n^*(y)$ for every $n$. Since $\{y_n^*: n \in \mathbb{N}\}$ is dense in $Y^*$, for every $y^* \in Y*$ we can find a sequence $n_k$ such that $y_{n_k}^* \to y^*$ in $Y^*$ as $k \to \infty$. In particular, $y^*(x) = \lim_{k \to \infty} y_{n_k}^*(x) = \lim_{k \to \infty} y_{n_k}^*(y) = y^*(y)$. Since $y^*$ was an arbitrary member of $Y^*$ we conclude that $Y^*$ does not separate points which is a contradiction.