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Suppose you have two pseudo-topological spaces $(X,p_1)$ and $(X,p_2)$ where $p_1$ and $p_2$ are relations between the set of ultrafilters on $X$ and the points in $X$.

We can define a topological space using the relation $p_1$ or $p_2$ by saying that $U\subseteq X$ is open if $\forall x\in U $ whenever an ultrafilter $F\rightarrow x$ (i.e. the relation contains $(F,x)$ pair) we have $U\in F$.

If $p_1 \neq p_2$ then are the generated topologies different? Clearly if the pseudo-topological space is topological - that is ultrafilters converge in the pseudotopological space iff they converge in the topological space - then the generated topologies are unique.

But not every pseudo-topological space is topological and I can't seem to show if pseudo-topologies always generate unique topologies. Is it true?

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  • $\begingroup$ For finite sets its unique for sure: there's only one pseudotopological structure. $\endgroup$ – Henno Brandsma Mar 22 '17 at 20:12
  • $\begingroup$ Please provide the proof that we actually get a topology by this definition. It might help get ideas. $\endgroup$ – Henno Brandsma Mar 22 '17 at 22:01
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    $\begingroup$ In what sense you say "if the pseudo-topological space is topological"? You mean that the relation $p$ is generated from the underlying topology by relating every convergent ultrafilter to its limit? I'm not really into pseudo-topological spaces but here's an idea: take pseudo-topological space $(X, p_1)$ which is not topological, then the topology you defined and then generate $(X, p)$ from it. Are topologies generated by $(X, p_1)$ and $(X, p)$ the same? It seems so. But they cannot be equal because the first one is not topological. Does this make sense? Does it work in details? $\endgroup$ – freakish Mar 22 '17 at 22:16
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Let $(X,p_1)$ be any pseudo-topological space that is not topological, and let $p_2$ be the convergence relation for the topology generated by $p_1$. Then $p_1$ and $p_2$ generate the same topology, but $p_1\neq p_2$.

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