I tried to prove that with the set being subset of a space X with metric d,
" the interior of the closure of a set equal to the interior of that set".
I proved that the interior of,namely, $A$ is included in the interior of the closure of $A$. But I could not prove the reverse, in special because I think that there can be points that are limit points of A and is contained in the interior of the closure, am I wrong?
I am doing this to prove that the closure is equal to the union of the interior points of the closure with the set of all limit points of the set.
Is the aforementioned statement true?
Thank you.
Best Answer
The claim isn't true.
The set of rational numbers in the unit interval $[0,1]$ has empty interior, but its closure is the whole interval, so the interior of its closure is the open interval $(0,1)$.