Are topological manifolds with boundary metrizable?

Question

It is standard that topological manifolds (without boundary) are metrizable. Is the same true for manifolds with boundary?. I'm using the following definition: Let $\mathbb{R}^n_{x_n\ge 0}=\{x\in \mathbb{R^n}:x_n\ge 0\}$. A topological manifold with boundary is a paracompact hausdorff topological space $M$ such that every point $p\in M$ is contained in some open set $U_p$ that is homeomorphic to an open subset of $\mathbb{R}^n_{x_n\ge 0}$.

It'd be nice to have some reference. The only reference about this I've found is John Lee's Introduction to Smooth manifolds but this deals with smooth manifolds.

Answer 1

Michael's answer is the right one in that it works directly for any manifold. But since you say you know the answer for manifolds:

Every manifold with boundary is a subspace of its double, which is a manifold (i.e., without boundary). Subspaces of metrizable spaces are metrizable.

Answer 2

They are metrisable. One way to see this is to use the following theorem.

Urysohn's Metrisation Theorem: Every Hausdorff, second countable, regular space is metrisable.

Let $M$ be a connected topological manifold with boundary. By definition, $M$ is Hausdorff.

As $M$ is paracompact and connected, $M$ is second countable - see this note by Hiro Lee Tanaka.

To see that $M$ is regular, let $C \subset M$ be closed and $p \in M\setminus C$. As $M\setminus C$ is open, there is an open neighbourhood $V$ of $p$ with $V \subseteq M\setminus C$ which we can take to be the domain of a coordinate chart $(V, \varphi)$ with $\varphi(p) = 0$ where $\varphi$ is a homeomorphism $V \to \mathbb{R}^n$ if $p \not\in \partial M$ or $V \to \mathbb{H}^n$ if $p \in \partial M$. If $p \not\in \partial M$, let $A = M\setminus\varphi^{-1}(\overline{B(0, 2)})$ and $B = \varphi^{-1}(B(0, 1))$; if $p \in \partial M$, instead let $A = M\setminus\varphi^{-1}(\overline{B(0, 2)}\cap\mathbb{H}^n)$ and $B = \varphi^{-1}(B(0, 1)\cap\mathbb{H}^n)$. In either case, $A$ and $B$ are open sets in $M$ with $C\subset A$, $p \in B$ and $A\cap B = \emptyset$ so $M$ is regular.

Therefore, by Urysohn's Metrisation Theorem, every connected topological manifold (with or without boundary) is metrisable. As an arbitrary disjoint union of metrisable spaces is metrisable, the conclusion also holds for disconnected topological manifolds (with or with boundary).

Answer 3

Although Michael Albanese and Peter L. Clark already gave good answers, I will post the solution I came up with.

In appendix A of Michael Spivak's book A Comprehensive Introduction to Differential Geometry he defines a manifold to be a topological space $M$ such that

1. $M$ is hausdorff
2. For each $x\in M$ there is a neighborhood $U$ of $x$ and an integer $n\ge 0$ such that $U$ is homeomorphic to $R^n$

Then he proves this theorem: The following properties are equivalent for any manifold $M$:

1. Each component of $M$ is $\sigma$-compact.
2. Each component of $M$ is second countable (has a countable base for the topology).
3. $M$ is metrizable.
4. $M$ is paracompact.

The first proof of this theorem also works even if the manifold has boundary. More explicitly, the key implication 4.$\implies$1. (which is put into a lemma) works because each connected component is locally compact and paracompact because it is a closed subset of $M$ (all components are closed) and $M$ itself is locally compact and paracompact.