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A pseudo-topological space is defined as a non-empty set $X$ together with a relation between the set of all ultrafilters on X and the points of X such that the principal filter $F_x = \{A\subseteq X: x\in A\}$ 'converges' (is sent) to the point $x$ for every $x\in X$.

My question is: why is the property that every principal filter converges to its point important for defining a topological space?

It seems that it is important when trying to define a topological space from the pseudo-topological one, in that if we have an $x\in X$ but the principal filter does not converge to $x$, then in the topological space (that is generated from the convergence relation) no set containing $x$ is open, but then $X$ is not open.

Is this the correct intuition here?

John C. Baez
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fosho
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  • Yes, this is because in the topological setting every neighborhood of $x\in X$ is required to contain $x$. If you think in terms of sequences, you want at least constant sequences to converge to their values. – Moishe Kohan Mar 22 '17 at 16:02
  • @HennoBrandsma Can someone give a reference where this is defined? This notion is mentionned without definition in this article. The definition given in the answer of Henno has some similarity with sequential opens – Noix07 Mar 22 '19 at 14:11
  • The notion of pseudo-topology was introduced by Gustave Choquet in Convergences p.79 – Noix07 Mar 26 '19 at 10:54

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The standard way to try to define a topology on a space $X$ with such a convergence structure $(\mathcal{U}(X), c)$, where $\mathcal{U}(X)$ is the set of ultrafilters on $X$ and $c: \mathcal{U}(X) \rightarrow \mathscr{P}(X)$ (which sends each ultrafilter to the set of its limits (possibly empty)) is the following: $O$ is open iff

$$\forall x \in O: (\forall \mathcal{F} \in \mathcal{U}(X): (x \in c(\mathcal{F}) \rightarrow O \in \mathcal{F}) $$

If you think about it, the condition that $x \in c(\mathcal{F}_x)$ for all $x$ is needed to avoid trivialities. What would happen if all $c(\mathcal{F}) = \emptyset$ e.g. ? The idea is the same as the fact that constant sequences always converge. If you define closure via convergence, you want $A \subseteq \overline{A}$ etc.

Henno Brandsma
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  • Could you please expand on how it is the same as the fact that constant sequences converge? Is this because we are forcing all open sets containing $x$ to be contained in the set of all sets containing $x$? (Reading that back sounds rather silly) – fosho Mar 22 '17 at 19:37
  • @Dman constant sequences or nets define the fixed ultra filter as its tail filter. Convergence of nets iff convergence of tail filter etc. – Henno Brandsma Mar 22 '17 at 19:40
  • Would I be correct to say if there is no filter in the pseudotopology that converges to a $x\in X$ then ${x}$ is always open? – fosho Mar 22 '17 at 19:56