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In the three axioms of topology on $X$, denoted by $\tau_X$, which are,

  1. $\emptyset \in \tau_X$, and $X \in \tau_X$
  2. for elements(open sets) $u_1, u_2, ... \in$ $\tau_X$, $$\bigcup^{\infty}_{i = 1}u_i \in \tau_X$$
  3. for elements(open sets) $u_1, u_2,... \in \tau_X$, $$\bigcap^{n}_{i=1}u_i \in \tau_X$$ What I want to ask is that why the third axiom of topology cannot be "the intersection of infinite number of elements are in topology class", is this due to some properties we want for topology, can somebody tell me why and if so, what will happen?
Rad80
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Ting-Kai Hsu
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    Note that your second axiom (union) is strictly weaker than the usual axiom. This definition will be equivalent to the usual definition of "topology" only for relatively small topologies. – Brian Moehring Aug 25 '23 at 03:09
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    A space where the intersection of open sets is open is known as an Alexandrov topology. All finite topological spaces are of course Alexandrov, so such spaces are not totally uninteresting. But, as CyclotomicField points out, being Alexandrov does not play nice with being a metric space. In fact more is true. Any T1 Alexandrov space has the discrete topology (look up the definition of T1 if you don't know it and see if you can prove this fact). So being Alexandrov does not play nice with even fairly mild separation axioms. – Charles Hudgins Aug 25 '23 at 12:44
  • Also compare https://en.wikipedia.org/wiki/%CE%A3-algebra – Carsten S Aug 25 '23 at 12:52
  • Note that your formulation of axiom 2 int't quite correct. It should also include the union of an uncountable family of open sets. – Rad80 Aug 25 '23 at 13:07
  • Related question. – Arturo Magidin Aug 25 '23 at 15:31

2 Answers2

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Let me first echo Brian Moehring in the comments and remind you that unions can be arbitrary, not merely countable.

Now, to answer your question, the short answer is that if we allow arbitrary unions and arbitrary intersections, then we often* get way too many sets for our topology to have any meaning. In fact, in the case of a metric space (or just the real line $\mathbb R$ if you like), once you decide that balls are open, allowing arbitrary intersections would mean that individual points were open (since each point is an intersection of all the balls centered around it), and then having arbitrary unions would tell you that literally every subset is open. At which point, openness becomes a completely meaningless concept.

So you can have arbitrary unions or arbitrary intersections, but not both. You might wonder, then, why do we do arbitrary unions and not intersections?

Returning to the metric space setting, we pointed out already that allowing arbitrary intersections would make individual points be open. Recall that $x_n\to x$ means eventually $x_n$ lies in every open neighborhood of $x$. So if $\{x\}$ is open, that means $x_n$ has to eventually be identically equal to $x$ just to converge. That's way too strong a requirement to do anything interesting.

*Update - as Charles Hudgins points out in comments, arbitrary unions and intersections are still studied in the form of so-called Alexandrov spaces. But as he also points out, we can’t assume points are closed in an Alexandrov space without the whole thing becoming discrete, so instead we get weird effects like the closure of a point containing other points, and sequences that converge to more than one point simultaneously.

This is, needless to say, a little confusing if you are trying to do analysis, for example.

M W
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  • Note that (at least for IR^n) allowing both countable unions and countable intersections is already sufficient to make every single point open. So one actually needs to resprict to finite intersections to get an interesting notion of a topology. – quarague Aug 25 '23 at 13:19
  • @quarague quite right, and you don’t even need unions at all - just the countable intersections alone will do it!(starting of course from the premise that open balls are to be open) Your comment applies equally to any metric space as well. – M W Aug 25 '23 at 13:36
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We want the intersection of open sets to open. However if you allow infinite intersections I can construct closed sets such as $\cap_{i=1}^\infty(-1-\frac1n,1+\frac1n) = [-1,1]$. This is why only finite intersections are allowed.

CyclotomicField
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