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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?

Thank you in advance!

Community
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Marios Gretsas
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  • 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 09 '16 at 21:49
  • Yes this is indeed true but your observation is about the proof of the two topologies? – Marios Gretsas Sep 09 '16 at 21:53
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    @DanRust: Are you sure these conditions are not equivalent? – hardmath Sep 09 '16 at 21:53
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    They are equivalent, but you'd need to show that (or at least quote that result if you've already seen it). – Dan Rust Sep 09 '16 at 21:53
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    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 09 '16 at 22:59
  • A hint about your "general question": Although the intersection of an arbitrary family of open sets in a topology is not guaranteed to be open, the intersection of an arbitrary family of topologies on $X$ will be a topology on $X$ (but possibly the trivial topology). So it should "the same argument" once that preliminary is settled. – hardmath Sep 10 '16 at 01:12

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Each of your topologies contains the co-finite topology ( a set is open if it is the empty set or if its complement is finite.) So the intersection of your topologies contains this topology, and so points are closed in this topology as well.

Airymouse
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