# Proving that the intersection of any finite number of members of T belongs to T when (X,T) is a topological space.

I know this is part of the definition of a topology. But the book I am using which is 'Topology without Tears' ask to prove this using mathematical induction.I hate to ask questions but proofs I have seen for this question in this site were different than mine.

Here it is...

The statement "Intersection of any two sets in $$T$$ belongs to $$T$$ " is given by the definition.

Suppose $$T_1,T_2,T_3,...$$ are members of $$T$$ and $$\bigcap_{n=1}^{k} T_n\in T \label{a}\tag{1}$$ is true. If $$\,\bigcap_{n=1}^{k+1} \,T_n\in T$$ is also true, the proposition can be proven by mathematical induction.

Since the element in the the left side of (\ref{a}) is a also set, let $$M=\bigcap_{n=1}^{k} T_n$$.

$$\bigcap_{n=1}^{k+1} T_n = \bigcap_{n=1}^{k} T_n \cap T_{k+1}$$ $$\bigcap_{n=1}^{k+1} T_n = M \cap T_{k+1}$$

Since $$M\in T$$ by (1) and $$T_{k+1}\in T$$, $$\,\bigcap_{n=1}^{k+1} T_n \in T$$ by definition. Thus intersection of any finite number of sets in $$T$$ belongs to $$T$$.

I want to know every mistake of this proof or ways to improve it.

• I'm not sure what other proofs you've seen, but I think this is the standard proof to show that a finite intersection of elements in a topology $\tau$ is again in $\tau$. – Clayton Mar 14 at 19:30
• I see no mistakes. Your proof is fine. – Mark Mar 14 at 19:30
• The proof is perfect. The reason you don't see this exact argument elsewhere is because everyone has a different style in writing and not everyone has the same base of facts at their disposal. – Alberto Takase Mar 14 at 19:38
• If $0$ counts as finite (which I think it should), then you should mention that the intersection of $0$ subsets of $X$ is $X$, which is also in $T$. – Andreas Blass Mar 15 at 1:42
• “I hate to ask questions but ...” Why? Asking questions and communicating with other mathematicians is what makes math what it is! – Santana Afton Mar 15 at 1:53

The proof by induction is fine. Renaming to $$M$$ is not really necessary of course (I don't usually introduce "unnecessary" notation), just writing

$$\bigcap_{i=1}^{n+1} T_i = \left(\bigcap_{i=1}^{n} T_i\right) \cap T_{n+1}$$

(so using braces) is enough to make your point, I'd say.

It's actually a very common pattern: e.g. groups are closed under products of two members, but writing finite powers etc. is routinely done: having a set closed under an operation allows all finite "products' (or sums/intersections, etc. etc.) You do the proof once, and henceforth it's implicit.

It sometimes makes for slightly easier to read proofs, only having to check the $$n=2$$ case.

A minor nitpick: you don't mention the base case (which is $$n=1$$), which is almost too trivial to mention: $$T_1$$ in the topology gives us the $$1$$-ary intersection $$\bigcap_{i=1}^1 T_i = T_1$$ in the topology too, tautologically. The $$n=2$$ case could also serve as the base case, depending on the formulation of the statement to be proved. But even the $$0$$-ary intersection (i.e. $$X$$) is in the topology....