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Let $\mathbb R^\infty$ be the set of all sequence of real numbers endwed with the product topology associated to the distance $d(x,y)= \sum_{i=1}^\infty 2^{-n} \lvert x_n - y_n \rvert (1+ \lvert x_n -y_n \rvert )^{-1}$.

Let $l^2$ be the space of all the sequence $\{x_h\}_h $ of real numbers such that $\sum_{h=1}^\infty x_h^2 < \infty $, endowed with the scalar product $(x,y)=\sum_{i=1}^\infty x_i y_i $ with $x,y \in l^2$.

I have to show that $l^2$ is a Borel set in $\mathbb R^\infty$. Any hint?

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    You should be more accurate when saying that $l_2$ is a set. It is a space. You might be interested if the set of square-summable sequences is a Borel set in $\mathbb{R}^\infty$. But you should prove it lies in $\mathbb{R}^\infty$ first. – 2017-02-21
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    @MichaelFreimann Of course it lies in $\mathbb{R}^{\infty}$. By definition $\mathbb{R}^{\infty}$ is a set of all sequences. I don't see anything inaccurate in OP's question. – 2017-02-21
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    @freakish Yes, I see now. I didn't get the metric right, now it is clear. – 2017-02-21
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    @MichaelFreimann Metrics doesn't matter at all for the definition of $l^2$. He defines $l^2$ as a specific subset of $\mathbb{R}^{\infty}$ satisfying certain conditions which are unrelated to the metric on $\mathbb{R}^{\infty}$. The question is: why this subset is Borel with the given metric on $\mathbb{R}^{\infty}$? – 2017-02-21
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    **Hints:** (1) For each $n$ the map $s_n:x\mapsto \sum_{h=0}^n x_h^2$ of $\Bbb R^\infty$ to $\Bbb R$ is continuous, hence Borel measurable; (2) $s(x):=\uparrow\lim_n s_n(x)$ is also Borel; (3) express $l^2$ as the inverse image of a Borel set under a Borel map. – 2017-02-21
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    @JohnDawkins: You should make that an answer. – 2017-02-25

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Hints: (1) For each $n$ the map $s_n:x\mapsto \sum_{h=0}^n x_h^2$, of $\Bbb R^\infty$ to $[0,\infty)$, is continuous, hence Borel measurable; (2) $s(x):=\uparrow\lim_n s_n(x)$ is also Borel, as a mapping of $\Bbb R^\infty$ into $[0,\infty]$; (3) express $l^2$ as the inverse image of a Borel set under a Borel map.