Verifying a Topological Property

Question

Let (X,T) be any Topological Space. Verify that Intersection of any finite number of members of T is a member of T. I tried to prove using that intersection of any two sets of T belongs to T. So the result of the intersection can Intersect with any other subset of T, the result of which will also belong to T. This will continue for any finite number of sets of T. Hence, verified. Please Help Definition of Topological Space:$$1. X \in T\ and\ \emptyset \in T$$$$2.Union\ of \ any(finite\ or\ Infinite)\ members\ of\ T\ belongs\ to\ T$$$$3.Intersection\ of\ any\ two\ members\ of\ T\ belongs\ to\ T$$

Answer 1

The base case for induction is the intersection of $n=2$ members of $T$ is also in $T$. This is in your definition

Inductive hypothesis is: Given $n \geq 2$ members of $T$, call them $A_1 , ... , A_n$, whose intersection $\cap_{k=1}^{n} A_k$ is also a member of $T$. Then given another member $A_{n+1}$ of $T$, $\cap_{k=1}^{n+1} A_k$ is a member of $T$

Now to prove it. By assumption we have the set $B = \cap_{k=1}^{n} A_k$ is a member of $T$. Now by definition of a topology, intersections of $2$ members are a member. Namely, $$B \cap A_{n+1}$$ is a member of $T$. Notice that $B \cap A_{n+1} = (\cap_{k=1}^{n} A_k) \cap A_{n+1} = \cap_{k=1}^{n+1} A_k $ as desired

Answer 2

Given $(X,T)$ is a topological space. Here $T$ is nothing but a collection of open sets of $X$. So we need to prove that any finite intersection of open set is open. We use induction to prove this.

Suppose $U_1~ and~ U_2$ are open sets.Let $x\in U_1\cap U_2$ then $x\in U_1$ and $x\in U_2$ . Because of openness there exist basis elements $B_1$ and $B_2$ containing $x$ such that $x\in B_1\subset U_1$ and $x\in B_2\subset U_2$. Thus by the definition of basis there exist a basis element $B_3$ containing x such that $x\in B_3\subset B_1\cap B_2$. It follows that $x\in B_3\subset U_1\cup U_2$. Hence $U_1\cap U_2$ is open. Now to end this $U_1\cap U_2\cap\dots \cap U_n=(U_1\cap U_2\dots \cap U_{n-1})\cap U_n$.