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First of all, rest assured that I have used the search tool and read Why do we require a topological space to be closed under finite intersection? and https://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets.

I'm looking for a more logical approach on this issue. I understand intuitively the reasons why an intersection of infinitely many open sets may fail to be open, but only in the case of a topology induced by a metric. In the general sense, what would be a somewhat rigorous argument? I feel that some counterexamples aren't enough for me to deeply understand the motives. I've read somewhere that the intersection of an infinite family of sets is related with an infinite conjunction, which is not allowed in the usual logical framework we use for mathematics. Is this the issue? Could you explain it in more detail? Are infinite disjunctions permissible, then? Finally, if we were to work within an infinitary logic, would this "issue" disappear, and consequently, would the building of topology become somewhat trivial, or less rich and interesting?

Thanks.

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    @Abel: What "logical issue of long statements"? An arbitrary union may be seen to correspond to an arbitrary disjunction, an arbitrary intersection to an arbitrary conjunction. But you *do* realize that one can equally well define a topology in terms of its *closed* sets, instead of its open sets? If you think "arbitrary union" is fine because it somehow "is" a disjunction and "arbitrary intersection" isn't because it somehow "is" a conjunction, then how would that square with defining topology using closed sets, where you *would* have "arbitrary intersection* but *not* "arbitrary union"?2011-03-31

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If you defined arbitrary infinite intersections of open sets to be open, then in any Hausdorff space, every subset would be open --- thus making the definition completely useless. To see this, note that the intersection of all open sets containing a point $x$ would be the singleton set $\{x\}$ --- thus all singleton sets would be open. But, recalling that arbitrary unions of open sets is open, any set $X$ would be expressible as a union of singleton open sets $X=\bigcup\{\{x\}:x\in X\}$.

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Regarding your points about logic: there is a logic called geometric logic wherein we have finite conjunctions, infinitary disjunctions, existential quantification, infinitary distributivity, and the Frobenius axiom. It is more or less the logic you get your truth-values are the open sets of a topological space.

Geometric logic not quite the same as even, say, intuitionistic first-order logic: there are theories which can be expressed in one but not the other, e.g. the theory of rings with nilpotent elements is geometric, but not first-order. I'm afraid this doesn't really answer your question though — these facts were discovered after the axioms of topology were laid down.

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I'm not sure what you mean by a logical reason, but maybe the following is helpful. Essentially I'm going to suggest that it makes sense if we want to think about limits; otherwise we kind of force any space to either have isolated points or infinitesimally close elements.

We want somehow to capture the idea of closeness without having to refer to distance. Imagine you want to be able to talk about a sequence of points $(p_0, p_1, ...)$ in $X$ approaching a point $x \in X$. The intuition is something like that, for any collection of points which is a collection of points "near to" $x$ that we pick, we expect our sequence to eventually enter this collection and stay in it. That is, we want to talk about some family of "neighborhoods" (nearby areas) of $x$, and we want to say that for any neighborhood of $x$, the sequence eventually stays in it.

Now, suppose we require that every arbitrary intersection of neighborhoods of $x$ still be considered a neighborhood of $x$. Then there are two possibilities. One possibility is that we can take some intersection and get $\{ x\}$ as a neighborhood. In this case, no sequence of other elements can converge to $x$. The second possibility is that there must be a distinct point $y\neq x$ such that $y\in U$ for every neighborhood $U$ of $x$ (because even the intersection over all neighborhoods of $x$ is not just ${ x}$). In this case, if we want to keep thinking of a neighborhood as a collection of nearby points, we're essentially viewing $y$ as infinitely close to $x$.

So, since we might sometimes want to avoid having infinitesimally close elements (i.e. indistinguishable by the topology) and isolated points, we shouldn't require that every arbitrary intersection still be considered a neighborhood.