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?