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I started studying functional analysis and I encountered an example of a metrizable space whose general picture I can't fully grasp.

Let $X$ be the set of all sequences of real numbers.

Let $x_n=\{{\xi_i^{(n)}}\}_{i=1}^{\infty}$ and let $x=\{{\xi_i}\}$. We say that $x_n \to x$ whenever $\xi_i^{(n)} \to \xi_i$ as $n \to \infty$ for every $i=0,1,2,3,...$

That way we get a non-metric space $s$. We'll show that the space $s$ can be metrized.

Let $x=\{{\xi_i}\}$ and $y=\{{\eta_i}\}$. We define

$$\rho(x,y)=\sum_{i=1}^{\infty}{\frac{1}{2^i}\frac{|\xi_i-\eta_i|}{1+|\xi_i-\eta_i|}}$$

The book then shows that the distance function satisfies the metric axioms and after that it shows that a sequence $x_n$ converges to $x$ iff the distance $\rho(x_n, x) \to 0$.

I don't quite understand the last part. Isn't it enough to show that the distance function satisfies the axioms for a metric space? Why do we need to show the two definitions of convergences are the same?

My understanding is that we sort of have a topological space with a defined convergence and we want our distance function $\rho$ to preserve the topology and be continuous.

Is this correct? What's the key idea behind this metrization thing?

Nikola
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    You are essentially given a topological space; you want to then construct a metric that generates the same topological space, since you have useful theory available that is specific to metric spaces. Without checking that the notions of convergence are the same, you have no idea whether the topological spaces are the same. – Ian Nov 06 '18 at 22:00

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