# Two topologies are equal if they have the same filter convergence

A major drawback with sequential convergence in topological spaces is that two different topologies can have the same convergent sequences e.g. the discrete and cofinite topologies on $\mathbb{R}$.

Filters are meant to be better convergent structures in topological spaces, which leads to my question:

If $\tau_1$ and $\tau_2$ are two topologies on a set $X$ with the same ultrafilter convergence i.e. an ultrafilter $\mathcal{F}\rightarrow x$ in $\tau_1 \iff \mathcal{F}\rightarrow x$ in $\tau_2$, then is it true that $\tau_1 = \tau_2$?

• It's the discrete and co-countable topology that have the same convergent sequences (the eventually constant ones). In the co-finite topology, every sequence with infinitely many points converges to all points, and this is not true in the discrete topology. – Henno Brandsma Jul 29 '17 at 7:48

This is true.

To see why just recall that $x\in \overline{A} \iff$ there is an ultrafilter $\mathcal{F}\rightarrow x$ with $A\in \mathcal{F}$. And show the closures of any subset must be identical in both cases.

For any topological space $X$:

$O$ is open iff for all $x \in O$ and for every ultrafilter $\mathcal{F}$ that converges to $x$, we have $O \in \mathcal{F}$.

Left to right is clear, as $O$ with $x \in O$ is a neighbourhood of $x$, so $O$ must be in any ultrafilter that converges to $x$.

Right to left: suppose $O$ satisfies the condition, but is not open, so some point $p \in O$ is not an interior point of $O$. This means that all neighbourhoods of $x$ intersect $X\setminus O$, and so $\{X\setminus O\} \cup \mathcal{N}_x$ forms a filter base so there is some ultrafilter $\mathcal{F}$ that contains $\mathcal{N}_x$ and $X\setminus O$.
Contradiction, as $\mathcal{F} \to x$ so $O \in \mathcal{F}$ by the condition, but then this ultrafilter contains disjoint sets.

So it follows that if the ultrafilter convergence is the same, the open sets will be the same too.

• Would the statement in the question be true if the limits were not mentioned? i.e. if each convergent filter in one topology converges in the other (not necessarily to the same limit), would it imply that the topologies are the same? – Hrit Roy Sep 13 '20 at 10:12
• @HritRoy no, the limits should be the same. – Henno Brandsma Sep 13 '20 at 10:15
• Just to be clear I do understand that in order for the topologies to be equal the limits have to be equal. I meant to ask if it wasn't mentioned in the statement, if we could have the equality as part of the conclusion. – Hrit Roy Sep 13 '20 at 10:18
• @HritRoy no it wouldn’t be. In the indiscrete topology and Sierpiński topology on $\{0,1\}$ it is the case that all (ultra)filters converge. But the topologies are different. – Henno Brandsma Sep 13 '20 at 10:25
• @HritRoy in that form it’s subtly incorrect. My trivial example works. – Henno Brandsma Sep 13 '20 at 10:46

If $\tau_1\neq \tau_2$, there is a set which is open in one topology but not the other. Suppose (swapping the roles of $\tau_1$ and $\tau_2$ if necessary) $U\in \tau_1$ but $U\notin \tau_2$. Since $U$ is not open in $\tau_2$, there is some point $x\in U$ such that no $\tau_2$-neighborhood of $x$ is contained in $U$. It follows that the set $\{V\mid x\in V\text{ and }V\in \tau_2\}\cup \{X\setminus U\}$ has the finite intersection property, so it extends to an ultrafilter $\mathcal{F}$ on $X$.

Now $\mathcal{F}\to x$ in $\tau_2$, since $\mathcal{F}$ contains all the $\tau_2$-open neighborhoods of $x$, but $\mathcal{F}\not\to x$ in $\tau_1$, since $\mathcal{F}$ does not contain $U$ (which is a $\tau_1$-open neighborhood of $x$).

Let $$\mathcal{N}^1_x$$ be the neighbourhood filter at $$x$$ in $$\tau_1$$ and $$\mathcal{N}^2_x$$ be in $$\tau_2$$. Now $$\mathcal{N}^1_x$$ converges to $$x$$ in $$\tau_1$$ so it must converge in $$\tau_2$$ as well so that $$\mathcal{N}^2_x\subset\mathcal{N}^1_x$$. Similarly we shall get $$\mathcal{N}^1_x\subset\mathcal{N}^2_x$$. So each point in $$X$$ has the same neighbourhoods in either topologies. We conclude $$\tau_1=\tau_2$$.