Topological Space, Finite Intersections

My book gives the following as an exercise:

Let (X,T) be any topological space. Verify that the intersection of any finite number of members of T is a member of T. [Hint: to prove this result use "mathematical induction."]

Doesn't this follow directly from the definition of a topology? If this weren't true, then T wouldn't be a topology and (X,T) wouldn't be a topological space. Or so I think. If I'm right, why do we need to prove this at all?

And if I do need to prove it, I'm confused about how to use induction. Let me see ... if u belongs to X, then {u} belongs to T; if u is the only member of X, then {u}, {X} and the null set are the only members of T and it's true. Now if there's a second element v belonging to X, then {v} also belongs to T, so {u} and {v} both belong to T and their intersection (if any) also belongs to T by definition of a topology ... except then I'm assuming what I'm trying to prove, am I not?

I hope one of you guys can straighten me out on this.

Re: Topological Space, Finite Intersections

Depends on the definition of topology you are using. Typically, the definition only requires that the intersection of two elements of the topology is in the topology. It typically does not expressly state that the intersection of any finite number of elements in the topology is still in the topology.

To use induction, you know by definition that the intersection of two sets is in the set. For the induction step, suppose it is true for the intersection of n elements and prove it for n+1.

Also, if $\displaystyle \{u\}\in \mathcal{T}$ for all $\displaystyle u\in X$, then you have the discrete topology. Through unions, you get $\displaystyle \mathcal{T}$ is the power set of $\displaystyle X$. That is rarely the case. In the Euclidean case, we look at topologies where no single point sets are in the topology.

Re: Topological Space, Finite Intersections

Thanks! Very clear now. Apparently I mixed up what the book said about unions with what it said about intersections. It did actually say "the intersection of any two