open/closed ball in the normed space $l_{\infty}$ is open/closed set in the locally convex space $l_{\infty}$ Let $\tau$ be the topology of all neighborhoods at zero $N_n(r)= \{ x \in l_{\infty}: q_n(x) < r\}$ where $q_n(x)=|x_n|$ is a seminorm on $l_{\infty}$.
I need to show that

*

*if $B(x,r)$ is open ball in the normed space $(l_{\infty},||.||_{\infty})$ then $B(x,r)$ is open set in $(l_{\infty},\tau)$.


*Is it true that if $E$ is closed ball in $(l_{\infty},||.||_{\infty})$ then $E$ is closed set in $(l_{\infty},\tau)$?
My thought:
For (1), let $y\in B(x,r)$ then $||y-x||_{\infty}<r$ and since $q_n(y-x) \leq ||y-x||_{\infty}<r$ for all $n$ then $y \in N_n(r)$. Does that show that every element in the ball has a neighborhood and so it is open?
For (2), I think it is not necessarily true, but not sure.

Actually 1) is false and 2) is true.
Convergence of a sequence  in $\tau$ is equivalent to convergence of each coordinate.
Consider $B(0,1)$. If $e_n$ is the sequence with $1$ in position $n$ and $0$ in other positions then $ne_n \to 0$ in $\tau$ and $\|ne_n\|=n$ for all $n$. This shows that no neighborhood of $0$ in $\tau$ is contained in $B(0,1)$.
For 2) take  sequence $(x^{n})$ in the closed unit ball and suppose this sequence converges to $x$ w.r.t. $\tau$. Then $|x^{n}_k| \leq 1$ for all $n$ and $k$. Letting $n \to \infty$ we get $|x_k| \leq 1$ for all $k$, so  $x$ belongs to the closed unit ball.
