Is $[0,1) \cup \{2\}$ a manifold with boundary? My issue is the $2$.

The subset $[0,1[ \cup \{2\}$ of the real line is a manifold with boundary having two connected components of different dimensions. The component $[0,1[$ is a 1-dimensional manifold with boundary, and the single point {2} is a 0-dimensional manifold.

We need to refer to a definition of manifold with boundary.

Tu says

A topological $n$-manifold with boundary is a second countable, Hausdorff topological space that is locally ${\cal H}^n$.

where ${\cal H}^1 = [0,\infty)$ with the usual topology, and

We say that a topological space $M$ is locally ${\cal H}^n$ if every point $p \in M$ has a neighborhood $U$ homeomorphic to an open subset of ${\cal H}^n$.

Now consider $[0,1) \cup \{2\}$. Because of the line segment, it can only be a manifold with boundary if it is $n=1$ dimensional.

Now consider $p = 2$. Then every neighbourhood $U$ of $p$ contains an open set containing only one point. But no open set in ${\cal H}^1$ contains only one point. Hence $U$ cannot be homeomorphic to any subset of ${\cal H}^1$. Therefore, this is not a manifold with boundary.

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As far as I can see, this isn't really open for debate given Tu's definitions (though perhaps I'm missing something!).

Edit: I should emphasize that it's fairly arbitrary whether one defines only "topological $n$-manifold" and "topological $n$-manifold with boundary", or if one generalizes to "topological manifold" and "topological manifold with boundary" where the $n$ can vary between different $p \in M$. As far as I can see, Tu does not actually define the latter notions.

If one does define these notions, then clearly this is a topological manifold which is a union of an $n=0$ manifold $\{2\}$ and a $n=1$ dimensional manifold with boundary $[0,1)$.