The intersection of $T_1$ topologies.

A topological space $$(X, \tau)$$ is a $$T_1$$ space if $$\forall$$ $$x,y\inX$$, $$x\neqy$$ 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 $$x\inX$$. 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. Sep 9, 2016 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. Sep 9, 2016 at 22:59