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

Personally I'd say $$M$$ wasn't a valid manifold with boundary because the $$\{2\}$$ doesn't have a neighborhood with any structure like an open ball/half-ball.

• This is actually an exercise from An Introduction to Manifolds by Loring W. Tu and is not mentioned in an errata.

• I have spent almost 2 hours thinking about an exercise that looked like it would take only 15 minutes and even tried pasting lemma (it's a good thing Professor Tu has solutions unlike Professor Lee): The result of all that thinking is that I don't think $$[0,1) \cup \{2\}$$, $$(\varepsilon,1) \cup \{2\}$$ or $$\{2\}$$ is homeomorphic to any open subset of $$\mathscr H^1$$ or $$\mathscr L^1$$. I was able to show $$\{0\} \subseteq \partial M$$ and $$(0,1)\subseteq M^0$$, but I don't quite know where $$2$$ belongs. I believe $$M$$ is not locally $$\mathscr H^1$$.

• Also, I have double checked: I believe "manifold boundary" was defined for manifolds with boundary, so this isn't some trick where "manifold boundary" is actually defined for a Hausdorff and second countable space that need not be locally $$\mathscr H^n$$.

To generalize,

Is a half-open interval and a point not in the interval's closure a manifold with boundary?

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.

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 $$[0,1)$$.

• Does Tu define topological manifold or just topological n-manifold? I agree that $[0,1) \cup \{2\}$ is not a topological $n$-manifold with boundary for any fixed $n$, but I would definitely call it a topological manifold with boundary. – Jason DeVito Mar 13 '19 at 13:56
• I don't know about Tu, but the standard definition is that $M$ is a topological manifold if there exists $n \in \{0,1,2,\ldots\}$ such that $M$ is a topological $n$-manifold. So by this definition, $[0,1) \cup \{2\}$ is not a topological manifold (with boundary or otherwise). @JasonDeVito – Lee Mosher Mar 13 '19 at 14:28
• @Lee: The standard definition I am used to allows $M$ to have components of different dimensions. I agree that by your "standard" definition, $[0,1)\cup\{2\}$ is not a manifold with boundary. But I also think that with my "standard" definitition, it is. – Jason DeVito Mar 13 '19 at 14:58
• @JasonDeVito The given definition is the only definition of a manifold with boundary. For what it's worth, the without-boundary definitions technically also restrict to topological $n$-manifold: "A topological space $M$ is locally Euclidean of dimension $n$...", and "A topological manifold is a Hausdorff, second countable, locally Euclidean space. It is said to be of dimension $n$ if it is locally Euclidean of dimension $n$." It's purely a semantic issue though. I'll edit the answer to emphasize this a matter of convention only. – Sharkos Mar 13 '19 at 21:07
• Thanks Sharkos! – user636532 Apr 13 '19 at 8:28

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.

• Prof Tu, does the idea in Eric Wofsey's answer on question for your definition of manifolds apply to your definition of manifolds with boundary? I mean, for example you say that manifolds with boundary have dimension at least 1, but what is meant is that manifolds with boundary have dimension at least 1 if they have dimensions? Also, I think there's another mistake: You meant manifolds with boundary have dimension at least 1 provided they have dimensions in the first place and have nonempty boundary? – user636532 May 10 '19 at 11:55
• I asked about the latter here – user636532 May 10 '19 at 13:18