# An explanation as to why U equals the indiscrete topology on A.

Let $$A\subset X$$, and assume that $$X$$ has the indiscrete topology. If $$U$$ is the induced subspace topology on $$A$$, explain why $$U$$ is equal to the indiscrete topology on $$A$$.

So far I understand that the indiscrete topology has the fewest open sets: the empty set and the set itself. So i am assuming that i have to show that $$U$$ has these properties.

Also, considering the definition of subspace topology, wouldnt the set $$U$$ have the same topological properties as the set $$X$$, since $$A\subset X$$ and $$U$$ is induced on $$A$$.

Am I on the right track?

By definition, a set $$V\subset A$$ is open in the induced topology if there exists a set $$W$$ which is open in $$X$$ such that $$V=A\cap W$$. Now, the only open sets in $$X$$ in your case are $$X$$ itself and the empty set. So what are the open sets in $$A$$?
$$U$$ just contains the subsets of $$A$$ of the form $$A \cap V$$ where $$V$$ is an open set of $$X$$. Now the proof is easy.