What is the correct way to denote empty n-ary intersection of sets?

Question

I'm trying to prove the statement:

Show that a finite intersection of open subsets in a metric space is open.

If I'm able to enumerate the sets as $U_1, \ldots, U_n$ and consider $U := \bigcap\limits_{i=1}^n U_i$, I have no problem writing the proof. But "finite" can also mean empty, so by listing the sets out like this, I think I'm sacrificing generality. What I'm really doing is defining $U := \bigcap\limits_{i \in I_n} U_i$, where $I_n$ is an indexing set of size $n \geq 0$. If $n = 0$, $U = X$ by definition, which is open. But the notation $\bigcap\limits_{i=1}^n U_i$ doesn't make sense in the case where $n = 0$, unless $\bigcap\limits_{i=1}^0 U_i$ means "empty intersection."

Is there a way to get around this subtlety?

Answer 1

The empty set is an open set so that does not pose a problem. In fact you can begin by saying if the intersection is empty then we are done. Then let $x$ be a point in the intersection. Since $x$ is in each of the $U_i$ and each of the $U_i$ are open there exists $\epsilon_1, ..., \epsilon_n$ such that $B_{\epsilon_i}(x)\subset U_i$. Take the minimum of the $\epsilon_i$ to get that $B_{\text{min}(\epsilon_i)}(x)$ is a subset of the intersection.

Answer 2

Even if you have one set, there is no "intersection". There has to be at least two sets and this is understood when you interpret the "intersection" part.

Answer 3

To prove that all finite intersections of open sets are open it is necessary and sufficient that "$U,V$ open implies $U \cap V$" open. You don't need a "general case" at all. It follow by induction on general principles. This yields the simplest proof IMO. If you insist you can note that the empty intersection is just $X$ (the whole space) by common logical convention (and this is open anyway) and a one-set intersection is just the set itself (so a set is open if a set is open, a tautology). Just do the binary case and that's the only relevant one.

Answer 4

Let me give a full proof.

Take two open sets $U$ and $V$ .

Consider the set $U\cap V$.

Let $x\in U\cap V$.

Then as $x\in U\,\exists\,r_{1}>0$ such that $B(x,r_{1}(x))\subset U$. and correspondingly $r_{2}>0$ such that $B(x,r_{2})\subset V$.

Take $r_{0}(x)=\frac{\min(r_{1}(x),r_{2}(x))}{2}$. So You have $B(x,r_{0})\subset B(x,r_{1})\subset U$ and

$B(x,r_{0})\subset B(x,r_{2})\subset V$.

So $B(x,r_{0}(x))\subset U\cap V$.

Thus $\displaystyle U\cap V=\bigcup_{x\in U\cap V} B(x,r_{0}(x))$ is open as it is union of open balls.

Now that you have intersection of two sets is open. you can extend it for finitely many using induction.

for example $U_{1}\cap U_{2}\cap U_{3}$ is open as $(U_{1}\cap U_{2})=U_{0}$ is open and hence $U_{0}\cap U_{3}$ is also open.

Now the only degenerate case is when the intersection is empty . In that case it is open as the null set is always open by definition of topology. If the intersection is non-empty we proceed like I did above.

Answer 5

Suppose you have $U = \bigcap_{j=a}^b U_j$ defined as:

$$ \forall x: [x \in U \iff x \in X \land \forall j \in \mathbb{Z} : \underbrace{a \le j \land j \le b \Rightarrow x \in U_j}_{(*)}] $$

If you set $a = 1$ and $b = 0$, the implication (*) becomes vacuously true for all $x \in X$ and $j \in \mathbb{Z}$. The condition $1 \le j \land j \le 0$ may look silly, but it’s a perfectly well-formed logical formula; it just happens to be always false. As such, $\bigcap_{j=1}^0 U_j$ is likewise perfectly well-defined as the whole space $X$.

One could take the awkwardness of this notation as an argument for using zero-based indexing and half-open intervals even in non-CS mathematics. While I am personally quite sympathetic to the idea, good luck convincing your fellow mathematicians to accept your alternative definition of $\bigcap_{j=a}^b$ where the upper bound is excluded. The practice is just too entrenched.⁰

If it really bothers you, a more practical option would be to give up the $\bigcap_{j=a}^b$ notation, and instead come up with a succinct enough notation for the interval $[0, n) \cap \mathbb {Z}$ (the von Neumann ordinal of $n$); it is common to use $[n]$ for a set like this. Then you can use $\bigcap_{j \in [n]}$ everywhere instead.

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⁰ Well, other than in notations like $\sum_{j=1}^\infty a_j$, which somehow don’t include $a_\infty$ as the last summand.