If $M$ is an $n$-dimensional manifold with boundary, then $\Int{M}$ is an open subset of $M$, which is itself an $n$-dimensional manifold without boundary.
by Hidenori
Proposition
If $M$ is an $n$-dimensional manifold with boundary, then $\Int{M}$ is an open subset of $M$, which is itself an $n$-dimensional manifold without boundary.
Solution
The manifold interior is defined differently from the topological interior.
Let $x \in \Int{M}$. Then $x$ has a neighborhood $U$ that homeomorphic to an open subset $V$ of $\mathbb{R}^n$. Then for any $y \in U$, $U$ is a neighborhood of $x$ that is homeomorphic to $V$. Thus $U \subset \Int{M}$. This implies that $\Int{M}$ is an open subset of $M$.
$\Int{M}$ is a subspace of $M$, so it is Hausdorff and second countable because $M$ is an $n$-dimensional manifold with boundary. Let $x \in \Int{M}$. Then there exists an open set $U \subset M$ that is homeomorphic to an open subset $V \subset \mathbb{R}^n$. As shown above, $U \subset \Int{M}$. Therefore, $x$ has a neighborhood in $\Int{M}$ that is homeomorphic to an open subset in $\mathbb{R}^n$. Therefore, $\Int{M}$ is indeed an $n$-dimensional manifold without boundary.
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