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?

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    $\begingroup$ Because that's the definition? $\endgroup$ – Asaf Karagila Aug 21 '14 at 11:12
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    $\begingroup$ No. The definition of open sets require only finitely many intersections of open sets to be open; and arbitrary unions of open sets to be open. If the definition would have required countable intersections of open sets to be open, then this would have been the case as well. But it doesn't. So you can't have that. $\endgroup$ – Asaf Karagila Aug 21 '14 at 11:15
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    $\begingroup$ If you allow the intersection of infinitely many open sets to be open, then you have to accept, say, $\{x\}=\bigcup_{n=1}^\infty \bigl(x-\frac1n, x+\frac1n\bigr)$ as open for any $x\in\mathbb R$, and from there it's a hop, skip, and jump to having every possible subset of $\mathbb R$ be open. $\endgroup$ – user856 Aug 21 '14 at 11:21
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    $\begingroup$ Topology really began from the study of the real line, and then realising that similar approaches work in various "spaces". Since in the real line the family of open sets is closed under arbitrary unions, finite intersections, but not necessarily infinite intersections, it makes sense that the definition of a topology only demands that the family of open sets be closed under finite intersections. There are classes of topological spaces that a closed under longer intersections: for example, in a P-space the family of open sets is closed under countable intersections. $\endgroup$ – user642796 Aug 21 '14 at 11:21
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    $\begingroup$ Previously: Why can the intersection of infinite open sets be closed? $\endgroup$ – user856 Aug 21 '14 at 11:24

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.

  • $\begingroup$ I'm accepting this answer even though the other answers are more mathematically rigorous, because the example really clears up my doubts and helps visualize the concept. $\endgroup$ – sprajagopal Aug 22 '14 at 9:34
  • $\begingroup$ rigour schmigour. understand and rigour takes care of itself. $\endgroup$ – John Lawrence Aspden Aug 22 '14 at 12:53

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.


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.


Why is that.. only finitely many intersections of open sets is open?

In the context of metric spaces, this is because:

  1. Every finite set $X$ of positive real numbers has a positive lower bound.

  2. An infinite set $X$ of positive integers may not have a positive lower bound. For example, $\{\frac{1}{1+n} : n \in \mathbb{N}\}$ is a set of positive numbers with no positive lower bound. (By definition, zero is not positive.)

Therefore, when we try proving that the intersection of a family $\cal O$ of open sets is itself open, we end up needing to assume that $\cal O$ is finite. Here's the details.

Definition. Let $X$ denote a metric space. Call $O \subseteq X$ open iff there exists a radius function on $O$, by which I mean a function $f : O \rightarrow (0,\infty]$ such that for all $x \in O$, the ball of radius $f(x)$ about $x$ is a subset of $O$. (This definition is non-standard, but equivalent to the usual definition.)

Proposition. Let $X$ denote a metric space and suppose $\mathcal{O}$ is a family of open subsets of $X$. Write $S$ for the intersection. Then $S$ is open.

Proof. Since every $O \in \mathcal{O}$ is open, hence there is a radius function $f_O$ on $O$ for every $O \in \mathcal{O}.$

Now consider $x \in S$. The goal is to find a positive real number $r$ such that the ball of radius $r$ centered at $x$ is a subset of $S$. Since $x$ is an element of $S,$ and since $S$ is the intersection of $\mathcal{O},$ hence $x$ is an element of $O$ for every $O \in \mathcal{O}$. Hence $f_O(x)$ is well-defined for every $O \in \mathcal{O}$. Since $\{f_O(x) \mid O \in \cal O\}$ is a finite set of positive real numbers, hence it has a positive lower bound. Define $r$ to be any such lower bound.

The goal is to show that the ball of radius $r$ centred at $x$ is a subset of $S$. For every $O \in \mathcal{O}$, since the ball of radius $f_O(x)$ about $x$ is a subset of $O$, and since $f_O(x) \geq r$, hence the ball of radius $r$ about $x$ is a subset of $O$. Hence the ball of radius $r$ about $x$ is a subset of $S$, as required.


Consider the set of all open sets containing just one particular point in Euclidean space. If you like, call that point the origin, so we're talking about the set of all open sets that contain the origin.

Their intersection contains just that one point. That's not an open set.

An arbitrary union of open sets includes as a subset some open neighborhood of each of its points. Hence it makes sense to define things in such a way that such a set is open.

PS: It is not strictly necessary to use an index set every time you talk about a union or an intersection. Notice that I don't refer to any index set in my first sentence above. Mentioning an index set in that particular example would just be clutter that makes things appear more complicated than they really are.


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