Open sets, topology I don't understand why in the definition of a topology, you require the union of an "arbitrary" collection of open subsets to be open and the intersection of a "finite" number of open subsets to be open. That is, I don't get why one of them is arbitrary and the other is finite.
Thanks!!
 A: One motivation for topology in this formal sense is to capture what it means for a function to be continuous when there isn't a metric to which you can apply an $\epsilon - \delta$ approach. If you think about open sets on the real line and how these relate to continuity by the "metric definition" you will note that this is compatible. 
Note that the intersection of an infinite collection of intervals on the real line can be a single point or an arbitrary closed interval. 
The other reason is that the definition is "what works" - i.e. it is what gives fruitful mathematical results.
A: One reason why you don't want to allow arbitrary intersections of open sets is that if you do you can do the following: 
$$ \bigcap_{n \in \mathbb N} (-1, \frac{1}{n}) = (-1, 0]$$
And then $(-1, 0]$ would have to be open (by definition). But this would contradict openness of sets as defined in a metric space (which $\mathbb R$ is): $0$ would not have an open ball around it contained in $(-1,0]$.
A: So, I'm going to answer this question in two parts. Firstly, I will answer a question implicit in the original post: what is the purpose of definitions? Secondly, I will answer the explicit question in the original post: why does the definition of a topology involve an arbitrary union of open sets but only a finite intersection of closed sets? 
In mathematics, we make definitions for many reasons. Perhaps the simplest reason is to expedite the communication of ideas. For example, there is a lot of meaning behind the statement "let $X$ be a separable infinite-dimensional Banach space." However, someone who is familiar with the definitions of these objects quickly understands the properties of the space $X$. On the other hand, the formal description of these properties would be significantly longer and cumbersome to write each time we encounter such a space $X$. 
Another reason we make definitions is to model (or abstract) some specific idea. For example, we intuitively have a notion of distance, but it may vary based on the situation. Some common distance functions are the straight-line distance, or the Manhattan distance (i.e., vertical plus horizontal distance), or the spherical distance (e.g., the distance traveled along the surface of the Earth). Intuitively, these are all distance functions to us and we want a definition that captures all of these (that is, all of these are distance functions according to the definition we create). Now, the definition of a metric tries to abstract the ideas of distance functions.
Yet another reason we make definitions is to guide our intuition. When we have many examples of objects that satisfy a given definition, these serve as landmarks for our intuition. When we see that a particular function satisfies the axioms of a metric we can begin to get a feel for the behavior of a function for which we have a priori little insight. 
These reasons being stated, remember that definitions are created, not given to us from above. That is, at some point, mathematicians created this definition in order to satisfy some of the above properties (and possibly for other reasons).  So, in particular, the definition of a topology was created in order to abstract another idea. A topological space can be thought of as a generalized or abstracted metric space. In a metric space, we have a precise notion of the distance between points. However, sometimes we work in areas where we don't have a precise notion of distance, but we do have a notion of "closeness." A topology tries to capture this idea. So, the definition of a topology that we have today is one that mathematicians have determined best fit the idea they were trying to abstract. 
To answer the explicit question in the original post, if we are trying to abstract the idea of a metric space to one without a distance function, then we need to come up with something in a metric space that doesn't depend on precise distances. In particular, open sets capture this idea. In a metric space, a set $U$ is called open if for every point $x\in U$, there exists an $\epsilon>0$ such that $B(x;\epsilon)\subset U$ (where $B(x;\epsilon)$ is the ball centered at $x$ of radius $\epsilon$). Notice that we only need the existence of such a positive $\epsilon$. Thus we can interpret this as saying a set $U$ is open if for every $x\in U$, all points "sufficiently close" to $x$ are also in $U$. So it seems as if open sets are a prime candidate to describe closeness. However, the definition of open sets in a metric space does use the metric, so we have to try and find the properties that characterize open sets in a metric space. And since we don't necessarily want any other structure on our space, it is natural to try and characterize properties of open sets relating specifically to unions and intersections. 
Thus we ask, in a metric space, under what kinds of unions are open sets closed? It is easy to prove that the answer is arbitrary unions. Again, this makes sense with our idea of open sets capturing the idea of "closeness." For given any point in a union of open sets, $x\in\bigcup_{U\in\mathcal{U}} U$, $x\in V$ for some $V\in\mathcal{U}$. Then, since $V$ is open, it contains all points "sufficiently close" to $x$. Thus all points "sufficiently close" to $x$ are also contained in the union. However, in a metric space, open sets are (in general) only closed under finite intersections. The classic counterexample to the infinite case is $\{0\}=\bigcap_{n\in\mathbb{N}}(-\frac{1}{n},\frac{1}{n})$. All the intervals on the right-hand side are open (in fact, they are balls centered at zero), but the left-hand side is clearly not open. So, the properties that open sets satisfy in a metric space (which don't inherently depend on the notion of a metric) are: closed under arbitrary unions and finite intersections. Thus, this is definition we take for open sets.
Now, the above process gives us the definition of a topology. But how would we know it is the 
"correct" definition (by correct I mean one which correctly captures the essence of the idea we were trying to abstract). The only way to do this is by considering examples. The fact that this definition of topology is the one that stuck around shows that it is the "correct" one in the above sense. 
A: As with many things in mathematics, it's because mathematicians have decided that it be so.
For a slightly less blithe response, recall that topology began as an attempt to generalise notions related to the real line, in particular the notion of continuous function.  The common first-year undergraduate definition of a continuous function (on the reals) is as follows:

A function $f : \mathbb{R} \to \mathbb{R}$ is continuous if for each $x \in \mathbb{R}$ and each $\epsilon > 0$ there is a $\delta > 0$ such that $| f(x) - f(y) | < \epsilon$ for all $y \in \mathbb{R}$ with $| x - y | < \delta$.

What this definition is saying is that given any open interval $I$ centred at the point $f(x)$ we can find a open interval $J$ centred at $x$ such that $f(y) \in I$ for each $y \in J$.  Notice the primacy of open intervals in the above interpretation.
Playing around with this interpretation leads us to two basic facts:
Fact: If $f : \mathbb{R} \to \mathbb{R}$ is continuous, then $f^{-1} [ I ] = \{ x \in \mathbb{R} : f(x) \in I \}$ is a union of open intervals (possibly an empty union) for any open interval $I \subseteq \mathbb{R}$.
Corollary: If $f : \mathbb{R} \to \mathbb{R}$ is continuous, then $f^{-1} [ V ]$ is a union of open intervals for any $V \subseteq \mathbb{R}$ itself a union of intervals.
Moreover, the converse of both of these statements are also true:  A function $f: \mathbb{R} \to \mathbb{R}$ is continuous if $f^{-1} [ V ]$ is a union of intervals for every $V \subseteq \mathbb{R}$ itself a union of open intervals.
So beginning with the basic notions of real analysis, we see that arbitrary unions of open intervals are of some importance.
Why restrict intersections to only finitely many?  I can think of two reasons:


*

*One idea again comes straight from basic analysis.  Note that given any two open intervals $I , J$, we have that either $I \cap J = \emptyset$, or $I \cap J$ is again an open interval.  This dichotomy holds for finite intersections of open intervals, but clearly fails for infinite intersections; e.g., $\bigcap_{n \in \mathbb{N}} ( \frac{-1}{n} , \frac{1}{n} ) = \{ 0 \}$.  If the primacy of open intervals in the characterisation of continuous functions is to be retained, it makes sense to think of these as the smallest kinds of open sets.

*Were we to allow arbitrary intersections (together with arbitrary unions), we would quickly be left in the situation that every subset of $\mathbb{R}$ is open.

A: We can always say that an intersection of finitely-many open sets is open. However, for infinitely-many open sets, we can't necessarily say that their intersection is open. There are certainly cases where infinitely-many open sets have an open intersection. For example, let $I_n=(n,\infty)$ for $n\in\Bbb N,$ so that $\bigcap_{n\in\Bbb N}I_n=\emptyset$ is open.
The reason for the "finite" in the case of intersections is simply because that's as far as we can prove with generality. We can prove for "arbitrary" in the case of unions.
A: I am just a beginning on learning. I think the essense of the definition is on the arbitrary union of open set are still open set. because for closed set, this property doesn't hold. consider a closed interval [a+1/n, b-1/n], when n goes to infinity, the union of the series go to (a,b), which is a open interval. but if you consider (a+1/n, b-1/n), this union will be still open when n goes to infinity, it will never reach a, b, although it will close. This is the property the definition wants to say, for open set, in infinity, it will still keep open.
