some confusion about the topological space A topological space is pair (X, $\tau$) where $X$ is a set and $\tau$ is the set of subsets of $X$ satisfying certain axioms.


*

*$\emptyset$ and space $X$ are both in $\tau$

*the union of any collection of sets in $X$ is contained in $\tau$

*the intersection of any finitely many sets in $X$ is contained in $\tau$


My Question:
Why we need "finitely" many sets?
What is wrong with "infinitely" many sets?
 A: The following is a heuristic picture of the situation.
An open set has "room" around each point which is still in the set. So you can do whatever unions you want; each point from each of the sets comes with its own "room" which is automatically in the union. But you can't do whatever intersections you want, because in an infinite process you may collapse all the "room" around a certain point into nothing, leaving behind just the point itself. An example in $\mathbb{R}$ is with $U_n=(-1/n,1/n)$.
The situation is reversed for a closed set. A closed set contains all of its limit points (for general topological spaces this takes a slightly different definition of "limit point" than the one in $\mathbb{R}$). This makes it so you can take arbitrary intersections, because if a point is a limit point of all of the sets in the intersection, then by closedness it is in each of them, and so it is in the intersection. On the other hand, in an infinite union you may wind up introducing more limit points without including them in the union, so such a union may not be closed. An example in $\mathbb{R}$ is with $F_n=\{ q_n \}$ given an enumeration of the rationals $\{ q_n \}_{n=1}^\infty$.
A: For example, consider the standard topology $\tau$ on $\mathbb R$.  The intervals $(-1/n, 1/n)$ are in $\tau$, but their intersection is $\{0\}$, which is not.
A: Let's think about the case of $\Bbb{R}$. Consider the sets $S_n=(0,1+\frac1n)$.
Then $$\bigcap_{n\in \Bbb{n}} S_n=(0,1]$$ 
which is surely not open.
