# The intersection of $T_1$ topologies.

A topological space $(X, \tau)$ is a $T_1$ space if $\forall$ $x,y$$\in$$X$, $x$$\neq$$y$ we can find an open set that contains $x$ and not $y$ abd an open set that contains $y$ and not $x$.

A useful proposition about $T_1$ spaces is that in a $T_1$ space, every singleton is a closed set.

Suppose we are given two $T_1$ spaces $(X,\tau_1)$ and $(X,\tau_2)$.Prove that $(X,\tau_1 \cap\tau_2 )$ is a $T_1$ space.

I did this:

Let $x$$\in$$X$. We know that $\{x\}$ is closed in in $(X,\tau_1)$ and in $(X,\tau_2)$ so $X$\ $\{x\}$ is open in both spaces. Hence $X$\ $\{x\}$$\in$ $\tau_1\cap \tau_2$. It is easily proved that the intersection of topologies in $X$ is again a topology in $X$ and finally we have that $\{x\}$ is closed in $(X,\tau_1 \cap\tau_2 )$.

My general question is that can we prove with the same argument that the intersection of an arbitrary collection of $T_1$ topologies is a $T_1$ topology?

Or is there a tricky counterexample to this last statement?

• You need to prove slightly more. $T_1$ only implies that every singleton is closed. You can't assume the converse implies $T_1$ though. – Dan Rust Sep 9 '16 at 21:49
• The proposition to be used is: a space is a $T_1$ space if and only if every singleton is closed. Based on that you can indeed prove smoothly that the intersection of a collection of $T_1$-topologies is a $T_1$- topology. – drhab Sep 9 '16 at 22:59