How can I show that the interior of the set of interior points is an interior set? $(X^\circ)^\circ = X^\circ$ I am trying to show that the interior of a set of interior points is an interior point, that is, if $X$ is a subset of a metric space $M$, that $(X^\circ)^\circ = X^\circ$ for $X \subset M$. My approach is to show that each is a subset of the other. It is quite obvious that $(X^\circ)^\circ \subset X^\circ$. However, I am bogged down when trying to have an intuitive understanding how to show the other direction. Can anyone give me help? It would be greatly appreciated!

One way to see this for a metric space, is to use that every point of an open ball $B(x,r) = \\{y: d(x,y) < r \\}$ is indeed an interior point of that open ball (justifying the name "open ball", one could say).

(Proof: if $y \in B(x,r)$, define $s = r - d(x,y) > 0$, and for any $z$, if $d(y,z) < s$, we know that $d(x,z) <= d(x,y) + d(y,z) < d(x,y) + s = d(x,y) + (r - d(x,y)) = r$, so indeed $B(y, s) \subset B(x,r)$. This $s$ shows that $y$ is an interior point of $B(x,r)$.)

Now, if $x \in X^\circ$, this means that there is some $r>0$ such that $B(x,r) \subset X$. For any $y \in B(x,r)$, there exists $s>0$ such that $B(y,s) \subset B(x,r) \subset X$, see above. But this just says that every point of $B(x,r)$ is itself an interior point of $X$, i.e. $B(x,r) \subset X^\circ$. And this by definition means $x \in (X^\circ)^\circ$, as required.