General Intersection Property of Topological Space
Contents
Theorem
Let $\left({S, \tau}\right)$ be a topological space.
Let $S_1, S_2, \ldots, S_n$ be open sets of $\left({S, \tau}\right)$.
Then:
- $\displaystyle \bigcap_{i \mathop = 1}^n S_i$
is also an open set of $\left({S, \tau}\right)$.
That is, the intersection of any finite number of open sets of a topology is also in $\tau$.
Conversely, if the intersection of any finite number of open sets of a topology is also in $\tau$, then:
- $(1): \quad$ The intersection of any two elements of $\tau$ is an element of $\tau$
- $(2): \quad S$ is itself an element of $\tau$.
Proof
Proof by induction:
For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
- For any sets $S_1, S_2, \ldots, S_n \in \tau$, it follows that $\displaystyle \bigcap_{i \mathop = 1}^n S_i \in \tau$.
Let $\mathbb S$ be any finite subset of $\tau$.
From Intersection of Empty Set, we have that:
- $\mathbb S = \varnothing \implies \bigcap \mathbb S = S$
From the definition of a topology, we have that $S \in \tau$.
Hence $P(0)$ is true.
From Intersection of Singleton, we have that:
- $\mathbb S = S_1 \implies \bigcap \mathbb S = S_1$
Thus $P(1)$ is trivially true.
Basis for the Induction
$P(2)$ is the case $S_1 \cap S_2 \in \varnothing$, which is our axiom:
- The intersection of any two elements of $\tau$ is an element of $\tau$.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
So this is our induction hypothesis:
- For any sets $S_1, S_2, \ldots, S_k \in \tau$, it follows that $\displaystyle \bigcap_{i \mathop = 1}^k S_i \in \tau$.
Then we need to show:
- For any sets $S_1, S_2, \ldots, S_{k+1} \in \tau$, it follows that $\displaystyle \bigcap_{i \mathop = 1}^{k+1} S_i \in \tau$.
Induction Step
This is our induction step:
Consider the set $\displaystyle \bigcap_{i \mathop = 1}^{k+1} S_i$.
This is $\displaystyle \left({\bigcap_{i \mathop = 1}^k S_i}\right) \cap S_{k+1}$.
But from the induction hypothesis, we know that:
- $\displaystyle \left({\bigcap_{i \mathop = 1}^k S_i}\right) \in \tau$
So from the base case, it follows that:
- $\displaystyle \bigcap_{i \mathop = 1}^{k+1} S_i \in \tau$
So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore, for any sets $S_1, S_2, \ldots, S_n \in \tau$, it follows that $\displaystyle \bigcap_{i \mathop = 1}^n S_i \in \tau$.
$\Box$
The converse follows directly from the above.
In particular, the fact that $S$ is itself an element of $\tau$ follows from the definition of Intersection of Empty Set.
$\blacksquare$
Sources
- 1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): $\S 1.1$
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3.1$: Topological Spaces: Definition $3.1.1$