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I read Royden's Real Analysis and found a construction proof Urysohn metrization theorem (see here). This construction is interesting and different from our usual attempt to embed the regular spaces with countable base into a cube $Q^{\omega}$.

The property $(1)$ on page 6 is easy to verify. My question is how to prove the property $(2)$ on page 6, i.e. to prove the original topology is finer than the metric topology generated by metric $\rho (x,y)$. The construction of metric $\rho$ makes it hard to find a proper open set $\mathcal{U}_{n}$.

Any help is appreciated. Thanks!

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Property (ii) is $\forall x \in X : \forall \varepsilon>0: \exists n: x \in \mathcal{U}_n \subseteq B_{\rho}(x,\varepsilon)$. So fix $x$ and $\varepsilon > 0$.

The easiest way (I think) to see that this holds, is to note that the function $g(y) = \rho(x,y)$ is continuous from $X$ to $[0,\infty)$, as an absolutely convergent series of continuous functions (we have to use that the $f_{n,m}$ are continuous; here is where we use that, and also continuity of sums and absolute values etc.). Then $O = g^{-1}[0,\varepsilon)]$ is open, contains $x$, so as the $\mathcal{U}_i$ form a base, this $\mathcal{U}_n$ exists.

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