Why for unions we can have arbitrary number of sets and for intersections it's only finite?

Question

In the definition of a topology it says that only a finite intersection of a open sets is open. What if we have the interval $(-1,1)$ and then for $a_n=\{1,2,3,...\}$ we have the intersection $\bigcap_{n\in \mathbb{N}}(-1-a_n,1+a_n)=(-1,1)$. So what is the problem with this? Why for unions we can have arbitrary number of sets and for intersections it's only finite? I have seen this and this but then I don't see how this disproves the point I made.

It also states that the union of finite number of closed sets is closed. Again, what if $a_n=\{1,2,3,...\}$ and then we have that $\bigcup_{n\in\mathbb{N}}\{a_n\}$ is also closed?

Answer 1

Your problem is with the placement of the word "only". You wrote

In the definition of a topology it says that only a finite intersection of [...] open sets is open.

But what's actually true is that

In the definition of a topology it only says that a finite intersection of [...] open sets is open.

In particular, the definition says nothing about whether infinite intersections might or might not be open, or whether some of them might be open and others not open, etc.

Getting definitions right is really difficult; reading them correctly is, if anything, harder.

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Perhaps you're wondering what happens if, in the axiom about intersections, we remove the word "finite."

If you do that, then for the standard topology on the real line (i.e., one in which all open intervals are "open"), we have, for any number $c$, that $$ \bigcap_{n = 1, 2, 3, \ldots} (c-\frac1n, c + \frac1n) = \{c\} $$ is an open set, so that every singleton is open, hence every set is open (by the union axiom), so the topology you get isn't very interesting. (It's called the discrete topology, and it's fun to play with a little bit, but...it'd be a pity if it were the only topology allowed. Because then every function from the reals to any space would be continuous.)

Answer 2

What is $1-a_n=1-\{1,2,3,...\}$? I guess you wanted to write $\bigcap_{n\in \mathbb{N}}(-1-n,1+n)$ but that is $(-2,2)$, not $(-1,1)$ (assuming $\mathbb N$ starts with $1$).

Now, consider instead $\bigcap_{n\in \mathbb{N}}(-1-\frac1n,1+\frac1n)=[-1,1]$ which is not open, and illustrates that intersection of infinitely many open sets need not be open.

Your question is not clearly stated anyway. The title is clear enough, but the body seems to be confused (and not necessaily related to the title, as I interpret it). If you take the title on its own:

"Why for unions we can have arbitrary number of sets and for intersections it's only finite?"

then the above asks why do we define topology the way we do, and not some other way. E.g. Why do we not want to allow and require in the definition of topology that intersection of arbitrary many open sets is open. That would certainly be a fine definition, but the (topological or not) spaces that it describes would not be that interesting to study. So one could interpret your question as asking: Why the definition of topology that we use (with finite intersections) is good and interesting to study? I don't feel like this is what you meant to ask, but nevertheless let me point it out. The answer to that would be a bit long and open ended. One intuitive explanation, if we restrict to metric spaces, a set $U$ is open if given any point $x\in U$ we could move some positive distance away from $x$, in any possible direction, and remain in $U$ all the time. (Direction may not be defined in every metric space, but think of the plane as an illustration.) So there is some "freedom" that we step a bit away from that point, yet that would not take us out of the set $U$. Now, if we intersect infinitely many open sets, all containing some fixed point $x$, note that the infimum of infinitely many positive distances may be $0$. So, then we are "free" to move distance $0$ away from $x$, but this is not really moving, we just have to stay at $x$. But, if we do not have the freedom to move away from $x$ at least a little bit, then the intersection set is not open. (So, then, we have no intuitive justification to impose such a requirement that any intersection of infinitely many open sets must be open.)

So, think of "open" as "the possibilities are open" (or, "we do have the freedom") to move away from $x$ a little bit, yet remain in the open set $U.$ Under this intuitive definition of open sets, we could not require that the intersection of infinitely many open sets is open, since the possibility to move a little bit away from $x$ in the intersection need not hold.