A locally convex space is metrizable if and only if its topology is determined by a countable set of seminorms.
In the proof of Conway's book, I have a trouble in understanding => direction:
Assume that $X$ is metrizable with metric $\rho$. Let $U_n = \{x: \rho(x,0)<1/n \}$. Because $X$ is locally convex, there are continuous seminorms $q_1, \cdots, q_k$ and $\epsilon_1, \cdots, \epsilon_k>0$ such that $\cap_{j=1}^k \{x: q_j(x)<\epsilon_j \}\subset U_n$. Let $p_n := \epsilon_1^{-1} q_1 + \cdots + \epsilon_k^{-1} q_k$. Then it is proved that $x_j \to 0$ in $X$ if and only if for each $n$, $p_n(x_j) \to 0$ as $j \to \infty$. This shows that $\{ p_n \}$ determines the topology on $X$.
I can't understand the last sentence. Why does it hold?
My trial: $x_j \to 0$ in $(X,\rho)$ iff for each $n$, $p_n(x_j) \to 0$, and I proved that this is equivalent to $x_j\to 0$ in $(X,\{ p_n \})$. Since translation is a homeomorphism, this means that $x_j \to x$ in $(X,\rho)$ iff $x_j\to x$ in $(X,\{ p_n \})$. Now if $C$ is a closed set in $(X,\rho)$ and $x_j \in C$, $x_j \to x$ in $(X,\{ p_n \})$, then $x_j \to x$ in $(X,\rho)$, so $x \in C$. So $C$ is a closed set in $(X,\{ p_n \})$. By the symmetry, this means $(X,\rho)$=$(X,\{ p_n \})$.
