In the definition of topological space, we require the intersection of a finite number of open sets to be open while we require the arbitrary union of open sets to be open. why is this?

I'm assuming this has something to do with the following observation: $\cap_{n=1}^{\infty} (-\frac{1}{n},\frac{1}{n}) = \{0\}$ and there is some reason we don't want singletons to be considered open, I am wondering what this reason is. Am I thinking in the right direction here?

Thanks :)

  • $\begingroup$ With your edit, the equation is no longer true. The intersection is empty (and open). $\endgroup$ Sep 6, 2010 at 3:27
  • $\begingroup$ Thanks! I thought I edited it properly but I guess I missed changing the 0. Its fixed now. $\endgroup$
    – WWright
    Sep 6, 2010 at 3:37
  • $\begingroup$ No problem! I guess I better edit my answer to make sense. $\endgroup$ Sep 6, 2010 at 3:39
  • $\begingroup$ @WWright: you are missing curly braces around the $0$ in the equation; you need to put two backslashes before the braces for them to show up. $\endgroup$ Sep 6, 2010 at 3:44
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    $\begingroup$ If we allowed arbitrary intersections, any Hausdorff space automatically becomes discrete. $\endgroup$
    – user899
    Sep 10, 2010 at 5:42

7 Answers 7


You need to think about what the intuition behind open sets are. One way to think about it is through neighborhoods: an open set is a set which is a neighborhood of each of its points. What is a neighborhood of a point? A neighborhood of a point $x$ is a set that contains of all points that are "sufficiently close" to $x$ (what does "sufficiently close" mean? It depends on the situation; you think of different neighborhoods perhaps specifying different degrees of closeness). In particular, any set that contains a neighborhood of $x$ is itself a neighborhood of $x$. And specifying two degrees of closeness specifies another degree of closeness that makes sense (the smaller of the two at any given place, say).

So: if you think about open sets as sets that are neighborhoods of all of the points they contain, then it is natural that the union of any family of pen sets will be open: each point in the union is one of the open sets, and that open set is a neighborhood, and the union contains that neighborhood and so is itself a neighborhood. So the arbitrary union of open sets should still be open.

What about intersection? Well, if you take two open sets $O_1$ and $O_2$, and you consider a point $x$ in $O_1\cap O_2$, then $O_1$ contains all points that are $1$-sufficiently close to $x$, and $O_2$ contains all points that are $2$-sufficiently close to $x$ (with "$1$-sufficiently" and "$2$-sufficiently" describing the two degrees of closeness required), so $O_1\cap O_2$ will contains all points that are both $1$-sufficiently and $2$-sufficiently close to $x$, and so it contains all points that are "sufficiently close" to $x$ for some meaning of "sufficiently close", so it is also an open set. This gives you, inductively, any finite intersection.

But what about arbitrary intersections? Then you run into trouble, because specfying two degrees of "closeness" gives you a degree of closeness (the smaller one), but an infinite number of degrees of closeness may end up excluding everything! (Just as in your example, taking the intersection of all $(-\frac{1}{n},\frac{1}{n})$, which specify all points that are $\frac{1}{n}$-close to $0$, but the intersection excludes everything). So we don't want to require that an arbitrary intersection of neighborhoods be a neighborhood, and so we don't want to require that an arbitrary intersection of open sets be an open set.

  • $\begingroup$ Wow a perfect answer. $\endgroup$
    – Lisramic
    Oct 16, 2020 at 18:16
  • $\begingroup$ “May end up excluding everything”. Wouldn’t this be sensible too because the closeness seems reasonably capped by reflexivity? $\endgroup$ Nov 24, 2022 at 16:28

If you think of it in the case of the topology on the real line - as soon as we leave the restriction from finite intersections we would get singletons as you have seen. However, any set of real numbers is a union of singletons that is $$A= \bigcup_{x\in A} \{ x \}. $$ It would follow that

each set would be open

each set would be closed

only finite sets would be compact

each function $f:\mathbb{R}\to X$ would be continuous, etc...

(Thanks Jonas!)

  • $\begingroup$ Finite sets would be compact. I also mentioned that singletons being open leads to the discrete topology in my post, but your last point helps drive home how bizarre this is. $\endgroup$ Sep 6, 2010 at 4:59
  • $\begingroup$ @Jonas Meyer: Yes, I am just being stupid! :/ $\endgroup$ Sep 6, 2010 at 5:15
  • $\begingroup$ AD., only finite sets would be compact. To supplement your list: every convergent sequence would be eventually constant. $\endgroup$ Sep 6, 2010 at 6:11
  • $\begingroup$ Jonas Meyer: You are totally right. (I have been awake all night with my 8 month baby, so I blame this on her ;) $\endgroup$ Sep 6, 2010 at 7:16

I think that one should have played around with metric spaces before dealing with topological spaces.

The finite intersection of open balls centered at $x$ is again an open ball; the radius is just the minimum of the finitely many radia. But if you take infinitely many balls, the radius is the infimum of infinitely many positive numbers, which will be $0$ in many situations (for example if the radia are $1/n$). A similar argument shows that you cannot prove in general that an infinite intersection of open subsets in a metric space is open again.

Since topological spaces are the natural generalization of metric spaces (replacing concrete distances by the concept of neighborhoods), the same is true for topological spaces and in the definition only finite intersections of open sets are allowed.


It is not that we require open sets to be closed under finite intersection...

Definitions are not invented out of thin air: people had topological spaces long before anyone thought of coming up with the definition of topological spaces and in the examples they had available, open sets were closed under finite intersection and that property was in fact quite useful to get things done. In other words, the definition of topological spaces (and the requirement you mention in the title along with it) was abstracted from examples, not imposed on them.

  • 2
    $\begingroup$ So in other words, "it seemed like a good idea at the time"? $\endgroup$
    – SamB
    Nov 28, 2010 at 0:31
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    $\begingroup$ @SamB: no, not really: that is not at all what I mean. $\endgroup$ Nov 28, 2010 at 1:16
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    $\begingroup$ Note to self: try to make jokes that are actually funny and/or true $\endgroup$
    – SamB
    Nov 28, 2010 at 1:38

Minor correction: $[0,1/n]$ is not open in the usual topology of the real line. But you could get the same thing with The example $\cap_{n=1}^\infty (-1/n,1/n)=${0} is a good one. And yes, points are not open in the usual topology on the line. If they were, then every set, being the union of one-point sets, would be open. This is called the discrete topology.

I think this definition can be fairly well motivated by what happens in metric spaces, of which the real line is a special case. Call a set $U$ open if around each point in $U$ there is a ball (with respect to the metric) centered at that point and contained in $U$. Then you can show as an exercise that intersections of finite collections of open sets are open, and that arbitrary unions of open sets are open. However, the example above shows that you can't take arbitrary intersections. (In the case of the real line, "ball" just means open interval.)

But why does it work so well to define general topological spaces to satisfy this property of metric spaces? I don't know. For more on the motivation of defining topologies in terms of "open sets", you may be interested in this MathOverflow question.


In my opinion, our idea of topological space comes not from open sets, but from neighbourhoods. Actually, I think that the hole thing comes from the idea of convergence. A neighbourhood of a point $a$, is a set that is "big enough" to be considered a neighbourhood. I know, this is a cyclic definition. The family of all neighbourhoods of a point $a$ describes what are the sequences (nets, if you prefer) that converge to $a$.

A sequence $x_n$ converge to $a$ if given a neighbourhood $V$, there is an $N = N_V$ such that $x_n \in V$ for all $n \geq N$. If $x_n \rightarrow a$, and $V$ and $W$ are neighbourhoods of $a$, then you might as well consider $V \cap W$ as being a neighbourhood of $a$, where $N_{V \cap W} = \max(N_V, N_W)$.

I said that a neighbourhood is a "big enough" set. When you talk about convergence, you say:

Given a "big enough" set, no matter how small it is, there is an $N$...

The "no matter how small it is" is usually omitted, since it has no mathematical use. When you want a "no matter how small", you use $\varepsilon$ and $\delta$. Whereas when you want a "no matter how big", you use $M$ or $N$. ;-)

Suppose that you want to define convergence (of nets, to be correct) in terms of neighbourhoods. You choose a family of sets $\mathcal{S}$ that will define convergence just as in the previous paragraph. Now, what is the largest family $\mathcal{V}$, that includes $\mathcal{S}$, and such that the convergence defined by this new family is exactly the same. What sets should you include into $\mathcal{V}$?

The answer is:

  1. First, build the family $\mathcal{B}$ of all finite intersections of sets in $\mathcal{S}$. This new family is a neighbourhood base for $a$. The reasoning for allowing finite intersections is that reasoning of the second paragraph of this post.

  2. Now, if you are considering a set $V$ as a neighbourhood of $a$, you might as well consider any $W \supset V$, since you can use for $W$ the same $N$ you would use for $V$. This is the "no matter how small"... it means that sets that are too big will not make much of a difference.

As you have already pointed out, infinite intersections will not work. This is just a way of saying that $\sup N_V$ might be $\infty$, so you might not be able to get an $N$ for infinite intersections of neighbourhoods.

Open sets are just sets that are neighbourhood of all its points. Notice that what is described above is exactly how one usually construct the topology for metric spaces. First, you define a family $\mathcal{S}$ (actually, $\mathcal{B}$) that is the set of all balls centered at $a$. Then you define open sets as being those that are neighbourhood of all its points!!! People give to much attention to open sets... in general topology, one usually gets introduced to open sets as if they were something really natural. But actually, neighbourhoods seems much more natural for those coming from metric spaces...

Now, you have a good motivation for defining filters and nets, as well... :-)

An for nets, you have now a motivation for the fact that given two indexes $\alpha$ and $\beta$, you need to have a $\gamma$ such that $\gamma \geq \alpha$ and $\gamma \geq \beta$. This is the finite intersection property for neighbourhoods!!!


I think that finite intersection is necessary for a topological space because when you are dealing with infinite intersection then you are leaving almost everything. Lots of open sets in topology will disappear and that's alarming.

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    $\begingroup$ Closed under infinite intersections would mean MORE open sets. In particular, in the reals every singleton would become open, making the space discrete. $\endgroup$ Jul 15, 2013 at 18:13

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