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I was reading Bourbaki's first topology book, specifically, its definition of a topology, when I came across what appears to be an important logic mistake. Rather than the usual set of three axioms, it defines a topology by the use of only the two axioms regarding union and finite intersection of the subsets of the topological space. The authors say that these two axioms imply the entailment of the empty set and the topological space by the topology, what is, obviously, false (we can give many counter-examples only using simple finite sets). So we still need a third axiom regarding the entailment of the empty set and the topological space by the topology, right? Am I missing something here? Because I know Bourbaki is known for its rigorous approach. Thanks in advance!

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J. W. Tanner
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Davi1399
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    What are you claiming the mistake is, exactly? This is completely standard. What counterexample do you have in mind? – Naïm Camille Favier Jun 26 '24 at 15:12
  • The author, basically, says that the fact that both the empty set and the topological space X belongs to every topology on X is implied by axioms OI and OII, what is not true. Take for example the set X={a,b,c,d}, it is obvious that S={{a,b},{a,b,c}} fulfill the properties of OI and OII, but the empty set and X does not belong to S. If what Bourbaki says were truth, there would not be the need of presenting a third axiom, as usual, for the fact that a topology need contain the empty set and itself. – Davi1399 Jun 26 '24 at 15:28
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    @Davi1399 In your example, $S$ is not closed under arbitrary unions (because $\varnothing = \bigcup \varnothing$ is not in it) nor finite intersections (because $X = \bigcap \varnothing$ is not in it). The only reason many authors include the "usual" third axiom is because people are often confused by logical reasoning about entities like empty intersections and unions. Bourbaki, in their usual minimalistic style, are not concerned about hand-holding the reader in this way (although they are kind enough to make the point explicit in the section you highlighted in yellow). – Alex Kruckman Jun 26 '24 at 15:33
  • Ok, thanks! I wasn't even aware of the concept of union and intersection of no subsets. Now it makes sense. – Davi1399 Jun 26 '24 at 15:48

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It seems that you have doubts concerning the text marked in yellow.

Bourbaki is correct, but perhaps his wording is not perfect. So let us reformulate the axioms.

(O$_I$) says that for each subset $\mathfrak S \subset \mathfrak O$ one has $\bigcup \mathfrak S = \bigcup_{S \in \mathfrak S} S \in \mathfrak O$.

(O$_{II}$) says that for each finite subset $\mathfrak F \subset \mathfrak O$ one has $\bigcap \mathfrak F = \bigcap_{S \in \mathfrak F} S \in \mathfrak O$.

In (O$_I$) Bourbaki takes $\mathfrak S = \emptyset$. Then $\bigcup \mathfrak S = \emptyset$. See Empty intersection and empty union.

In (O$_{II}$) Bourbaki takes $\mathfrak F = \emptyset$. Then $\bigcap \mathfrak F = X$. See again Empty intersection and empty union.

Kritiker der Elche
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