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I want to prove $f(x) = \sum\limits_{n=1}^\infty x_{n}\sin(n)$ is a continuous function from $(\ell^1,||.||_1 )$ to $R$ with the usual metric. I have an idea to prove this, I have to show that the pre-image of every open/closed set in R is open/closed in my norm space. My proof: let $\epsilon$>0.

I have the set $B_{R}(f(x),\epsilon)$ = $\{ f(y)\in R\ | d_R(f(y), f(x)) \leq \epsilon\} $ is closed. How do I show it's pre-image is closed? I know because sin(n) $\leq 1$, then f(x) $\leq \sum x_n = ||x||$. So, $d_R(f(y),f(x)) = |f(y)-f(x)| \leq ||y|| - ||x||$. I'm stuck here

Then there exits a $\delta$ such that B(x, $\delta) \subset f^{-1}(f(x), \epsilon) $ and then f(B(x,$\delta) \subset B(f(x), \epsilon). $

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  • $\begingroup$ $x=(x1,x2,x3,...) \in l^1 | \sum_{n=1}^\infty |x_n| < \infty $ Does this make sense? $\endgroup$
    – user522122
    Feb 20 '18 at 18:44
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It may be easier to use pure sequential characterisation of continuity.

For $(x^{n})\subseteq l^{1}$, $x\in l^{1}$ are such that $x^{n}\rightarrow x$, where $x^{n}=(x_{1}^{n},x_{2}^{n},...)$, $x=(x_{1},x_{2},...)$, then $\displaystyle\sum_{k=1}^{\infty}|x_{k}^{n}-x_{k}|\rightarrow\infty$ as $n\rightarrow\infty$.

Then $|f(x^{n})-f(x)|=\left|\displaystyle\sum_{k=1}^{\infty}x_{k}^{n}\sin k-\sum_{k=1}^{\infty}x_{k}\sin k\right|=\left|\displaystyle\sum_{k=1}^{\infty}(x_{k}^{n}-x_{k})\sin k\right|\leq\displaystyle\sum_{k=1}^{n}|x_{k}^{n}-x_{k}|\rightarrow 0$ as $n\rightarrow\infty$, so $|f(x^{n})-f(x)|\rightarrow 0$ as $n\rightarrow\infty$, so $f(x^{n})\rightarrow f(x)$.

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  • $\begingroup$ I'm kinda new to this. What exactly do you mean by $x^n$. $\endgroup$
    – user522122
    Feb 20 '18 at 19:20
  • $\begingroup$ $x^{1}$ is a string in $l^{1}$ so is $x^{2}...$. $\endgroup$
    – user284331
    Feb 20 '18 at 19:21

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