Interior of Topological n-Manifold with Boundary is Topological n-Manifold without Boundary

Theorem

Let $M$ be a topological $n$-manifold with boundary.

The interior of $M$, denoted $\Int M$, is a topological $n$-manifold (without boundary).

Proof

Recall the definition of locally Euclidean space:

Let $M$ be a topological space.

Let $d \in \N$ be a natural number.

Then $M$ is a locally Euclidean space of dimension $d$ if and only if each point in $M$ has an open neighbourhood which is homeomorphic to an open subset of Euclidean space $\R^d$.

Recall the definition of manifold (without boundary):

Let $M$ be a Hausdorff second-countable locally Euclidean space of dimension $d$.

Then $M$ is a topological manifold of dimension $d$.

Recall the definition of manifold with boundary and its interior:

Let $\R^n$ denote the $n$-dimensional Euclidean space.

Let $\H^n$ denote the $n$-dimensional closed upper half-space of $\R^n$.

A $d$-dimensional topological manifold with boundary is a second-countable Hausdorff space $M$ in which every point has a neighborhood homeomorphic either to:

a) an open subset of $\R^n$

or:

b) an open subset of $\H^n$.

The interior of $M$, denoted $\Int M$, is the set of all its interior points.

A point $p \in M$ is an interior point of $M$ if and only if $p$ is in the domain of some interior chart of $M$.

The chart $\struct {U, \varphi}$ is an interior chart if and only if $\map \varphi U$ is an open subset of $\R^n$.

Hence, $\Int M$ is:

• second-countable,
• Hausdorff, and
• homeomorphic to an open subset of Euclidean space $\R^d$.

Therefore, $\Int M$ is a topological $n$-manifold (without boundary).

$\blacksquare$

Sources

• 2013: John M. Lee: Introduction to Smooth Manifolds (2nd ed.): Chapter $1$: Smooth Manifolds: $\S$ Manifolds with Boundary