Why do we require a topological space to be closed under finite intersection?

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 :)

Solution 1:

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.

Solution 2:

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!)

Solution 3:

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.