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In Bourbaki's Topology, a filter is defined by the following:

A filter on a set $X$ is a family of subsets $F$ of $X$ which has the following properties:

  1. Every subset of $X$ which contains a set of $F$ also belongs to $F$.

  2. Every finite intersection of sets of $F$ belongs to $F$.

  3. The empty set does not belong to $F$.

It's an easy observation that $F$ can be the empty family. However, Bourbaki do not note this. This seems useful because even the empty set has a filter - the empty filter. Thus we can say all sets have filters without exception.

However Bourbaki further on note that:

let $(F_i)$ be any nonempty family of filters on a set $X$ (which therefore must be nonempty) ...

This implies that Bourbaki assume that the empty filter cannot be a filter because we can take $(F_i)$ to be just this empty filter and this does not force $X$ to be nonempty. Only if filters must be nonempty and recalling that this means they do not include the empty set, is $X$ forced to be nonempty.

Q. Thus is it an oversight on Bourbaki's part to not exclude the empty filter as a filter in their definition above? More evidence for this is that when Bourbaki define a filterbase they explicitly exclude empty filterbases.

Q. Categorically, it's often important to include trivial cases to get a good category. Is this the case here? Should we include the empty filter? I've already noted above, for example, that every set has a filter when we accept the empty filter as a filter.

José Carlos Santos
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Mozibur Ullah
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  • You always assume filters are non-empty, yes. – Jakobian Mar 12 '25 at 22:40
  • This question is similar to: Unclear part of Filter definition. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Alex Kruckman Mar 12 '25 at 23:15
  • Regarding your second question, the trivial case is not the empty filter, but the filter ${X}$. – tomasz Mar 12 '25 at 23:36
  • @AlexKruckman: I don't think the answer to your proposed duplicate answers this question. – Rob Arthan Mar 12 '25 at 23:56
  • @tomasz: I'm not sure I follow you - I usually think of the trivial filter on $X$ as either any filter including the empty set and then which is then the powerset $P(X)$ by reason of up closure. Thus its the powerset $P(X)$ which to my mind is the trivial filter. For every inhabited set $A$, then ${A}$ is a filter and it is the minimal filter that includes $A$. This also characterises ${X}$. – Mozibur Ullah Mar 13 '25 at 00:01
  • @tomasz: Why do you consider ${X}$ to be trivial? – Mozibur Ullah Mar 13 '25 at 00:04
  • @MoziburUllah: A filter is a largeness notion. To consider only the whole space large is a trivial largeness notion. The power set is not trivial, it is degenerate. – tomasz Mar 13 '25 at 00:07
  • @tomasz: Ok, got you. – Mozibur Ullah Mar 13 '25 at 00:09
  • @RobArthan Yes, it does. Quoting from the answer to the duplicate: 'The "empty intersection" is the whole set X, so X is a finite intersection.' So $X$ is in every filter, so there is no empty filter. The core issue in both this question and the proposed duplicate is that the OP has forgotten to consider the empty intersection. – Alex Kruckman Mar 13 '25 at 00:10
  • @AlexKruckman: I am not convinced., given that answer is commenting on very different question. But I'll leave it to the OP on this question to defend it if they wish to. – Rob Arthan Mar 13 '25 at 00:14

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No, $\emptyset$ is not a filter on $X$. That's because of condition 2: “Every finite intersection of sets of $\mathcal F$ belongs to $\mathcal F$.” So, if $\mathcal A$ is a finite set of elements of $\mathcal F$, then $\bigcap_{A\in\mathcal A}A\in\mathcal F$. In particular, $\bigcap_{A\in\emptyset}A\in F$ (since $\emptyset$ is a finite subset of $\mathcal F$). But $\bigcap_{A\in\emptyset}A=X$. So, $X\in\mathcal F$.

José Carlos Santos
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  • @tomasz: to clarify your comment, what I think you mean is that if $X = \emptyset$, then José's argument correctly shows that in that case, axiom 2 implies that $X = \emptyset \in F$, but then axiom 3 fails, so you don't get a filter. Is that what you meant? – Rob Arthan Mar 12 '25 at 23:52
  • I do not have the source text at hand, so I don't know the exact context, but normally empty intersections are not defined: you say $\bigcap_{A\in\emptyset}A$ equals $X$, but since the former expression doesn't mention $X$ in any way, one would be justified to ask "what $X$?". It might be that there a convention is adopted here that in the context of taking intersections of parts of $X$ the intersection of the empty family is defined to be (the local universe) $X$, but I see no trace of such a convention. If there is none, "finite intersection" would mean "finite non empty intersection" – Marc van Leeuwen Mar 13 '25 at 08:28
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    @MarcvanLeeuwen It is explicitly written in Bourbaki's Theory of Sets (chapter II, §4), right after the definition of intersection, that, for a family $(X_i){i\in\emptyset}$ of parts of a set $E$, $\bigcap{i\in\emptyset}X_i=E$. – José Carlos Santos Mar 13 '25 at 09:00
  • @JoséCarlosSantos Thank you for that reference that I certainly would not have found myself. It is still enigmatic, in that they first exclude the empty family from the intersection definition, clearly aware of a problem, then go on to define another notation mentioning $E$ for the intersection of a family of subsets of $E$, and then revert to the notation $\bigcap_{i\in I}X_i$ (without $E$) setting (for a fam. of subsets of $E$) its value to $E$ if $I=\emptyset$. But they define being a family of subsets as $Y\subset E$ for every $Y$ in the codomain of the family (which is a map)... – Marc van Leeuwen Mar 13 '25 at 11:49
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    ... which implies that whenever $E\subset E'$, any family of subsets of $E$ is also a family of subsets of $E'$. And this breaks their attempt to properly define the intersection of an empty family of subsets of $E$. The frustrating part of this is that they were so close to actually doing this right: if one would define a "family of subsets of $E$" to be a map (thought of as $i\mapsto X_i$) with domain some set $I$ and codomain the powerset of $E$ (which is not the powerset of any larger $E'$) then this codomain would have carried th information necessary for defining the empty intersection. – Marc van Leeuwen Mar 13 '25 at 11:55