Question

Prove or disprove any metric defined on X(#0) induces a topology on X

Answer 1

Let $(X, \rho)$ be a metric space with metric $\rho$. Let's define the following:

$B_{\varepsilon}(x) := y \in X | \rho(x,y) <\varepsilon$. Let's show, that for any metric space $(X, \rho)$ : $B_{\varepsilon}(x) := y \in X | \rho(x,y) <\varepsilon, x \in X, \varepsilon > 0$ is a base for some topology. If we prove this, it will be enough, since any base generates topology on $X$, that has, as open sets, all unions of elements of the base.

So, to prove that the defined above set is a base we will check if base criterias are true, namely:

1) $\bigcup_{x,\varepsilon}B_{\varepsilon}(x) = X.$

2) If $z \in B_{\varepsilon}(x) \cap B_{\delta}(y),$ then there exist such element of our base $B_0$, that $z \in B_0$ and $B_0 \subset B_{\varepsilon}(x) \cap B_{\delta}(y)$.

1) For a given $x \in X$, $X = \bigcup_{n=1}^{\infin}B_n(x)$. Therefore, $\bigcup_{x,\varepsilon}B_{\varepsilon}(x) = X.$

2) Let $z \in B_{\varepsilon}(x) \cap B_{\delta}(y)$. Let $r := min(\varepsilon - \rho(x,z), \delta - \rho(y, z)).$ Then, $B_r(z) \subset B_{\varepsilon}(x) \cap B_{\delta}(y)$.

So, we proved that the defined set is a base for some topology for $(X, \rho).$