I am actually having an introduction to filters.

Today I was trying to prove that the collection of open sets of a convergence space satisfy the axioms of a topology: O $\subset$ X is open iff $lim \mathcal{F} \cap O \neq \emptyset \Rightarrow O \in \mathcal{F}$

I wanted to verify that the collection of $O_i$ are a topology., and I am having some difficulties.Here is my try:

1) X is in everyfilter and $\emptyset$ is in the topology by definition (not sure abou this, but I can't apply the definition of open on this set)

2)I take $O_1 $ and $ O_2$ convergence opens such that $lim \mathcal{F} \cap (O_1 \cap O_2) \neq \emptyset$. This implies that $(O_1 \cap O_2) \in \mathcal{F}$ because it is closed under intersection.

3)I take a family of $O_i$ convergence opens such that $lim \mathcal{F} \cap (\cup_i O_i) \neq \emptyset$. (Here start some doubts). Since $\cup O_i \supset O_j \forall j$ and each $O_i$ is in the filter, the arbitrary union is in the filter.

Having tried in this way, I have 2 doubts :

1) I am proving it in a wrong way, and if that is the case what am I supposed to prove?

2) By this last point, shouldn't filters be close under arbitrary union? In fact, if $O_i \in \mathcal{F} \ \forall i$, then $\cup_i O_i \supset O_j \forall j$ and since filters are closed under superset operation, the union should always be in the filter.

Thank you for your help


The empty set is open in this definition, because the condition on the left (the non-empty intersection with $O$) is never fulfilled for the empty set, of course. And an implication whose left clause is false, is a true statement (ex falso totum).

The intersection argument is correct, but could be more explicit: suppose that $O_1$ and $O_2$ are open according to this definition. Then suppose $\mathcal{F}$ is a filter such that $\lim\mathcal{F} \cap (O_1 \cap O_2) \neq \emptyset$. As $O_1 \cap O_2 \subset O_1$, in particular $\lim\mathcal{F} \cap O_1 \neq \emptyset$, and so $O_1 \in \mathcal{F}$, and similarly $O_2 \in \mathcal{F}$, and so $O_1 \cap O_2 \in \mathcal{F}$. It follows that $O_1 \cap O_2$ is also open in this definition.

For the union, suppose that $O_i$ is open for all $i \in I$. Suppose that $\mathcal{F}$ is such that its set of limits $\lim\mathcal{F}$ intersects their union $O=\cup_{i \in I} O_i$. Then there is some $j \in I$ such that $\lim\mathcal{F}$ intersects $O_j$. This then implies that $O_j \in \mathcal{F}$ and so $O \in \mathcal{F}$, as $O_j \subset O$. This implies that the union is also open in this definition as well.

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  • $\begingroup$ Oh, I understood what I was missing about the union. Thank you. But then, is the filter property of being "closed under supersetting" a more general property of being "closed under arbitrary union"? i.e. for a collection of sets, being close under superset $\Rightarrow$ being close under arbitrary union? Maybe it was trivial to you but having just started filters and have never used this fact have never made me think of it in that way. thank you for your help $\endgroup$ – User Mar 9 '14 at 10:18
  • $\begingroup$ @user132951 The set of open sets with respect to the convergence are closed under arbitrary unions, and this follows from filters being closed under supersets, as the proof shows. But the filters don't contain just the open sets. So there is a relation, but filters are not "more general", they are just different kind of objects. $\endgroup$ – Henno Brandsma Mar 10 '14 at 19:33
  • $\begingroup$ I was referring to a signle filter. Is a single filter close under arbitrary union of sets in the filters?( just as a topology) $\endgroup$ – User Mar 11 '14 at 8:28
  • $\begingroup$ @user132951 Yes. A filter is closed under finite intersections and all enlargements, so all unions. It also contains $X$ but not $\emptyset$, so a filter is not a topology, but it almost is. You can take the union of a filter and the empty set to get a topology (it's easy to check that adding this one set fulfills all the topology axioms). These are called filter spaces, and are sometimes used as examples. $\endgroup$ – Henno Brandsma Mar 11 '14 at 11:58

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