2
$\begingroup$

I'd like to make a topology $T$ (on a fixed set, say $X$) such that if $T'$ is metrizable topology on X, then $T \subset T'$

let $A=\{T : T$ is topology on X which is metrizable$\}$
then maybe $ \cap T$($=\cap_{T\in A} T$) satisfy those conditions.
for any collection of topology defined on fixed set, its intersection is topology. so $\cap T$ is topology.

I want more information about $\cap T$. is it T4? or metrizable?

I've tried some non-equivalent metric on $\mathbb R$ but got no important result.

1 Answers 1

3

This is just the cofinite topology on $X$. In particular, it is $T_1$, but not $T_2$ (hence not metrizable) unless $X$ is finite (in which case it is the discrete topology on $X$).

Clearly every cofinite set in a metric space is open. Conversely, given any nonempty set $A\subseteq X$ which is not cofinite, we can find a metrizable topology on $X$ so that $A$ is not open in that topology.

How? Well, since $A$ is not cofinite, let $\{x_i: i\in\mathbb{N}\}$ be distinct elements of $X\setminus A$, fix $a\in A$ (since $A$ is nonempty), and define a metric so that $x_i\rightarrow a$. For instance, the map given by

  • $d(y, y)=0$ for all $y$,

  • $d(a, x_i)=d(x_i, a)=2^{-i}$ for all $i$,

  • $d(x_i, x_j)=\vert 2^{-i}-2^{-j}\vert$ for all $i, j$, and

  • $d(y, z)=1$ otherwise

is a metric which does the job.