Proving intersection of finitely many open sets is open

[Eclair_de_XII]

Homework Statement

Prompt in topic title. Proof describes a method in order to find an open ball centered around a point in the intersection. I assert the method will not work if there are infinitely many open sets in the intersection, which will not exclude the possibility that it will work.

Relevant Equations

Ball: A ball centered around some $x$ with radius $r$ is defined to be the set $B(x,r):=\{y:|x-y|<r\}$.
Open: A set $A$ is open if for all $x\in A$, there is a positive number $r$ such that $B(x,r)\subset A$.

Define a collection of open sets to be denoted as $P_i$, $1\leq i\leq N$ where $N\in \mathbb{Z}^+$.

Let $x\in\cap_{i=1}^N P_i$. By definition, $x$ must belong to every single $P_i$.

In particular, $x\in P_1$ and $x\in P_2$. Since $P_1$ and $P_2$ are open, there exist positive numbers $r_1$ and $r_2$ such that $B(x,r_1)\subset P_1$ and $B(x,r_2)\in P_2$. Choose $\inf\{r_2,r_1\}$ and call it $\rho_1$. We prove $B(x,\rho_1)\in P_1\cap P_2$.

Let $y\in B(x,\rho_1)$. Then $|x-y|<\rho_1\leq r_1,r_2$. Hence,

$B(x,\rho_1)\subset B(x,r_1)\in P_1$
$B(x,\rho_1)\subset B(x,r_2)\in P_2$

By definition, $B(x,\rho_1)\subset P_1\cap P_2$.

Now consider $P_3$; there is $r_3>0$ such that $B(x,r_3)\subset P_3$. Define $\rho_2:=\inf\{\rho_2,r_3\}$. Then by similar reasoning above (in other words, change some variable names) $B(x,\rho_2)$ is contained in $P_3$ in addition to $P_1\cap P_2$. Note that $\rho_2=\inf\{r_1,r_2,r_3\}$

Continuing in this fashion, we obtain a decreasing sequence of $\rho_i$, where $1\leq i\leq N-1$. Suppose $P_{N+1}$ is an open set that $x$ also belongs to. Then there is $r_{N+1}>0$ such that $B(x,r_{N+1})\subset P_{N+1}$. Set $\rho_N=r_{N+1}$ and $r=\inf\{\rho_i\}_{i\in\mathbb{N}\cap [1,N]}$. By similar reasoning as above, $B(x,r)\subset \cap_{i=1}^N P_i$ and $B(x,r)\subset P_{N+1}$ which gives us our result.

Note that this choice of $r$ is invalid if there are infinitely many $P_i$ in the intersection, since in this case:

$\inf\{\rho_i\}_{i\in\mathbb{N}\cap [1,N]}=0$

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Example of intersection of infinitely many open sets being open:

$\cap_{n\in\mathbb{Z}^+}(-n,n)=(1,1)$

Note: The method above would fail because the sequence $\rho_i$ in this case would be constant at $\rho_i=1$. So I guess, the method would fail only if the sequence $\rho_i$ were strictly decreasing.

Example of intersection of infinitely many open sets not being open:

$\cap_{n\in\mathbb{Z}^+}(-\frac{1}{n},\frac{1}{n})=\{0\}$

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I'm sure there is a less notation-heavy way of expressing this proof.

 

[Orodruin]

Your method works, but is cumbersome. You can just take finite intersection directly and apply the same reasoning. There is no need to take two sets at a time.

The proof works for a finite number of sets because a finite set of positive numbers has a smallest number. It fails for an infinite set because an infinite set of positive numbers can have inf = 0. However, if inf > 0 the intersection is open.