a.) $A:= \{(x_n)_{n\in \mathbb{N}} : x_n \in [0,1] \hspace{2mm}\text{for all}\hspace{2mm} n\in\mathbb{N}\}$ in $(l^\infty, \|\cdot\|_{\infty})$
and
b.) $B:= \{f\in C([0,1]) : |f(t)-t|<1 \hspace{2mm}\text{for all}\hspace{2mm} t\in [0,1]\}$ in $(C([0,1]), \|\cdot\|_\infty).$
For a.) We say $|x_n|<\epsilon$, with $\epsilon>0$ being given. Then $A$ is the open ball $B(0,\epsilon)$ in $(l^\infty, \|\cdot\|_{\infty})$. And as open balls are open sets, we say A is open.
Also, Considering $x_m=(\lambda\epsilon, o, o, \ldots)\in A$ for $|\lambda_n|<1$, but if $\lambda_n\longrightarrow 1$, then $x_n\longrightarrow(\epsilon, 0, 0, \ldots) \notin A$.
Would this be correct?
For b.) For this question I'm not entirely sure what to do.
Any help with be appreciated.
Thank you
$A$ is closed, but not open. The sequence $x=(1,1,1,...)$ is not an interior point. If $(x_n)$ is a convergent sequence in $A$ with the limit, then it also converges pointwise. But since $0\le x_{n,m}\le 1$, we also have $0\le \lim \limits_{n\to \infty} x_{n,m}\le1$, so $\lim \limits_{n\to \infty}x_n\in A.$
$B$ is open, but not closed. If $f\in B$, not that $t\mapsto |f(t)-t|$ is continuous as well on a compact intervall. We can use this to conclude $m:=\sup \limits_{t\in [0,1]} |f(t)-t|<1$. Put $\epsilon:=\frac{1}{2}m$. Then for $g\in B_{\epsilon}(f)$ we have: $$|g(t)-t|\le |g(t)-f(t)|+|f(t)-t|\le \|f-g\|_{\infty}+\frac{1}{2}m<m<1$$
To observe that $B$ is not closed, let $f_n(x):=x+1-\frac{1}{n}$ and prove that $\lim \limits_{n\to\infty}f_n \notin B$.