understanding closed under finite intersection notation

Question

"Closure under finite intersection" is often stated as following: If $O_1$ and $O_2$ are in $T$, so is there intersection, $O_1 \cap O_2$. How is this indicative of finite intersection, instead of countable or arbitrary intersection?

Answer 1

• Assume that for all $O_1$ and $O_2$ in $T$, $O_1\cap O_2$ is in $T$, then for all $(\Omega_i)_{1\leqslant i\leqslant n}$ elements of $T$, one has: $$\bigcap_{i=1}^n\Omega_i\in T.$$ The result is proved using induction.

• If for all $(\Omega_i)_{1\leqslant i\leqslant n}$ in $T$, their intersection is in $T$, then certainly for all $O_1,O_2$ in $T$, $O_1\cap O_2\in T$.

Finally, a countable intersection of elements of $T$ is not necessarily in $T$ as it is shown with: $$\bigcap_{n=1}^{\infty}\left]-1-\frac{1}{n},1+\frac{1}{n}\right[=[-1,1],$$ for the usual topology on $\mathbb{R}$.

Answer 2

From $2$ to all finite is a standard argument using induction:

Suppose that $\mathcal{T}$ is closed under the intersection of two elements.

We want to prove the statement

$$\phi_n : \forall O_1, O_2 \ldots, O_n \in \mathcal{T}: \cap_{i=1}^n O_i \in \mathcal{T}$$

for all $n \in \mathbb{N}$. Well, $n=1$ is trivial as $\cap_{i=1}^1 O_i = O_1 \in \mathcal{T}$.

Suppose that $\phi_n$ holds and we want to show $\phi_{n+1}$ so we take $O_1, \ldots, O_n, O_{n+1} \in \mathcal{T}$. Then by $\phi_n$ (the induction hypothesis) we get that $\cap_{i=1}^n O_i \in \mathcal{T}$ and as we have closure under the intersection of $2$ elements: $$\cap_{i=1}^{n+1} O_i = (\cap_{i=1}^n O_i) \cap O_{n+1} \in \mathcal{T}$$

as required, so $\phi_{n+1}$ holds.

So we get $\forall n \phi_n$ but this does not reach the infinite case. Only arbitrarily large finite cases.

Note that we need an associative operation to have this argument work, so if a set is closed under unions of two, it is for all finite unions, ditto for sums, products, concatenation, etc. etc.

Answer 3

It may help to see another property like this. For example, the property of a set "being finite" is closed under pairwise unions: if $A$ and $B$ are finite sets, then so is $A\cup B$.

You can prove that "being closed under pairwise unions" is equivalent to $(\iff)$ being closed under all finite unions. This gives us a rigorous reason to consider closure under pairwise unions and closure under all finite unions as meaning the same thing.

(And more generally, in any situation where you have a binary operator $\oplus$ and a set of things that are closed under pairwise $\oplus$, you can prove that the set of things is also closed under all finite applications of $\oplus$.)

But the property of being closed under pairwise/finite unions does not imply being closed under infinite unions. To prove this, all you need is an example: "being finite" is closed under pairwise unions, but it's not closed under infinite unions, as the example $\bigcup_k \{k\}$ shows.

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• Let's call the property "Closure under pairwise intersection". This means that if $O_1$ and $O_2$ are open, then so is $O_1\cap O_2$.

• The reason that closure under pairwise intersection means closure under finite intersections is that we can prove that they're logically equivalent. After all, if you have closure under all finite intersections, you have closure under pairwise intersections as a special case $(\Leftarrow)$. And if you have closure under pairwise intersections, then for any finite $n$, there's a proof that goes like this:

  Suppose we have closure under pairwise intersection. If $O_1, \ldots, O_n$ are any open sets, then:

  • $O_1\cap O_2$ is open, so
  • $(O_1\cap O_2)\cap O_3$ is open, so
  • $((O_1\cap O_2) \cap O_3) \cap O_4$ is open, so
  • $\vdots$
  • so $O_1\cap \cdots \cap O_n$ is open.
  • So we have closure under $n$-ary intersections.

  Because we can produce this short proof for any $n$ and any collection of $n$ open steps, it follows that we have closure under all finite intersections $(\Rightarrow)$.

• Closure under pairwise intersection is provably equivalent to closure under finite intersections. From this point of view, one reason why we closure under pairwise intersections doesn't imply closure under infinite intersections is that in some sense proofs must have a finite number of steps.

  Imagine you wanted to prove that if you have closure under pairwise intersections and $\{O_1,O_2,\ldots\}$ is an infinite family of open sets, then the intersection $\bigcap_k O_k$ is also open. The proof would correctly state:

  • $O_1 \cap O_2$ is open, so
  • $O_1 \cap O_2 \cap O_3$ is open, so
  • $\vdots$
  • $O_1 \cap \ldots \cap O_k$ is open
  • $\vdots$
  • $\infty$ So $\bigcap_k O_k$ is open (?).

  Unfortunately, the proof strategy we used before does not work because it does not yield a proof in a finite number of steps — in fact you can be closed under pairwise intersections, which means that you are closed under all finite intersections, without being closed under infinite intersections of any kind.