Why is that *any* union of open sets is open but only *finitely many* intersections of open sets is open?

I understand that when we talk about union of open sets, we introduce an index set which can be countable or uncountable. But could I not do the same for the intersection of open sets too?

Solution 1:

Generally this goes back to the underlying concept of convergence of a sequence and the so called Hausdorff'sche Umgebungsaxiome (see: Felix Hausdorff. Grundzüge der Mengenlehre. Veit & Comp., Leipzig, 1914).

Intuitively the convergence of a sequence to a point means, that in each neighbourhood of the point can be found almost all parts of the sequence.

The mathematical version of that concept is to name for each point $x$ of a set $X$ certain subsets as neighbourhoods of this point satisfying the following so-called Hausdorff'sche Umgebungsaxiome (neighbourhood = germ. "Umgebung"):

1. $x$ belongs to each of its neighbourhoods.
2. Any superset of a neighborhood of $x$ is again a neighborhood of $x$ (esp. $X$ is a neighborhood of $x$).
3. The intersection of two neighborhoods of $x$ is again a neighborhood of $x$.
4. Each neighborhood $U$ of $x$ contains a neighborhood $V$ of $x$ such that $U$ is a neighborhood of each point of $V$.

On this basis, one defines a subset of $X$ to be open if it is a neighborhood of each of its points (f.i. a circular disk in $R^2$ without its edge).

Open are exactly those subsets, taht contain only "internal" points (and neither "outer points" or "edge-points") in the sense that at any point in the open set a neighbourhood can be specified, which is completely contained in the open set.

From this can be derived following theorems:

1. The empty set and the space itself are open.
2. The intersection of two open sets is open.
3. The union of any number of open sets is open.
4. A subset $U$ is exactly then a neighborhood of $x$ if there exists an open set $O$ with $x \in O \subset U$.

If instead of Hausdorffs axioms one takes the theorems 1 to 3 as axioms and theorem 4 as a definition for neighbourhood then one comes out with the usual definition of a topology:

Let $X$ be a set. A system $T$ of subsets of $X$ is called a topology on $X$, if:

1. $∅ \in T, X \in T$,
2. $O_1, O_2 \in T ⇒ O1 \cap O2 \in T$,
3. $S \subset T ⇒ (\bigcup_{S'∈S} S') ∈ T$.

This shows that "finite intersections aswell as arbitrary unions of open sets are open" is a generalization of the properties of "open sets" defined in terms of "neighbourhoods" that were defined to get a precise meaning of the idea of the convergence of a sequence.

Solution 2:

Proof by example!

Consider the real numbers, and a sequence of ever-smaller open intervals around zero: (-1,1), (-1/2,1/2), (-1/3,1/3), ...

If we take any union of any of these sets, then any point in that union will have a little neighbourhood around it (can you see why?).

For intersections, it's not quite so simple.

If we take a finite number of the sets and take their intersection, then we'll still get an open set (hint: why is the intersection of two open sets still open?)

But if we take the infinite intersection, the only point that is in all the intervals is 0.

And 0 on its own isn't an open set, because 0 doesn't have a little neighbourhood around it.

If you understand why that's true, you should have no trouble seeing it more generally.

Solution 3:

Perhaps this would be easier if it is motivated from the metric space context. If $(X,d)$ is a metric space, a subset $U \subset X$ is open if for all $x \in X$, there is some ball $B(x,r)$ of some radius $r > 0$, around $x$ such that $B(x,r) \subset U$. Now suppose we have a family of open sets $U_i$ for $i \in I$ some indexing set. The matter is fundamentally one of logical quantifiers.

(Unions) If we let $U$ be the union of the $U_i$, then if $x \in U$, we must have $x \in U_i$ for some single $i$. There is a ball $B(x,r) \subset U_i$. And by definition of a union $B(x,r) \subset U$. Crucial is that everywhere here we only need to verify "there exists some $i$ such that ..."

(Intersection) On the other hand $U$ is the intersection of $U_i$. If $x \in U$ then $x \in U_i$ for all $i$. We can find a ball $B(x,r_i) \subset U_i$ for each $i$. Let us assume this ball is chosen with maximal possible radius. If there was a ball $B(x,r) \subset U$ around $x$ then $B(x,r) \subset B(x,r_i)$ for each $i$ by the maximal radius assumption. Thus $r < r_i$,and $r \leq \inf_{i \in I}(r_i)$ would work. But now we have a problem because unless the index set $i$ is finite it is entirely possible the family of $\{r_i\}$ has $0$ as it's infimum. This is because to check $B(x,r)$ was contained in $U$ we needed to check $B(x,r) \subset U_i$ for all i.