# Prove that $\mathcal{W}(\mathbb{R})$ is not metrizable.

A colection $\mathcal{V}$ of open sets in a topological space $X$ is called a Fundamental System of Open Neighborhoods (FSON) of a point $x\in X$ when:

1. $\forall\ V\in\mathcal{V}$ we have that $x\in V.$
2. If $A\subset X$ is open set containing $x$ then $\exists\ V\in\mathcal{V}$ such that $V\subset A$.

For example in any metric space the set $\{B(x,\frac{1}{n}); n\in \mathbb{N}\}$ is a FSON of $x$.

Let $\mathcal{W}(\mathbb{R})$ the set of continuous functions $f:\mathbb{R}\to\mathbb{R}$ with a topology defined by the following way: Let $f\in\mathcal{W}(\mathbb{R})$ and a continuous positive function $\varepsilon:\mathbb{R}\to\mathbb{R}^+$ and define the set $B(f,\varepsilon)=\{g\in\mathcal{W}(\mathbb{R}); |g(x)-f(x)|<\varepsilon(x)\ \forall x\in\mathbb{R}\}$ which is a basis for the topology.

Prove that $\mathcal{W}(\mathbb{R})$ is not metrizable.

Hint: Show that $f=0$ doesn't have a countable FSON.

Attempt:

Suppose that we have a countable FSON called $\mathcal{R}_{0}$ of $f=0$ then $\mathcal{R}_{0}=\{A_i\}_{i=1}^{\infty}$. Then for each $A_i$ I can choose $\varepsilon_i$ such that $0\in B(0,\varepsilon_i)=\{g\in\mathcal{W}(\mathbb{R}); |g(x)|<\varepsilon_i(x)\ \forall x\in\mathbb{R}\}\subset A_i$. Then I need to find a positive function $\varphi:\mathbb{R}\to\mathbb{R}^+$ such that for all $i\in\mathbb{N}$ we have that $B(0,\varepsilon_i)\nsubseteq B(0,\varphi)$. If were $\varepsilon_i(x)=\frac{1}{n}\ \forall x \in\mathbb{R}$ I have that $\varphi(x)=\frac{1}{1+x^2}\ \forall x \in\mathbb{R}$ satisfies the condition. But I don't know how to find $\varphi$ in the general case.

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For each $n\in\Bbb Z^+$ choose a positive number $r_n<\epsilon_n(n)$. Now construct a continuous $\varphi:\Bbb R\to\Bbb R^+$ such that $\varphi(n)=r_n$ for each $n\in\Bbb Z^+$. (A piecewise linear function is continuous and easy to construct.) Then $B(0,\varphi)$ does what you want.
What topological property of $\mathcal{W}(\mathbb{R})$ allow us to construct this function? – Gastón Burrull Jun 9 '12 at 22:46
@Gastón: I’m not sure what you mean. It’s really just the fact that there are positive real numbers arbitrarily close to $0$ that lets us construct $\varphi$. – Brian M. Scott Jun 9 '12 at 22:51