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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?

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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 $A$?

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$U$ 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.

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