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My understanding of the definition of a manifold is essentially a topological space that locally looks like $\mathbb{R}^n$ for some $n$. That is, a topological space $M$ is a manifold (of dimension $n$) iff $M$ is Hausdorff and second-countable, and each $p\in M$ is contained in a chart $(U, \varphi)$ such that $\varphi$ is a homeomorphism from $U$ to an open subset of $\mathbb{R}^n$.

My question: Why do we need the Hausdorff condition? That is, isn't $M$ being Hausdorff implied from $M$ being "locally homeomorphic" to $\mathbb{R}^n$? (Note: I am aware this isn't precisely the definition of "locally homeomorphic").

For instance, we can take any two points $p, q\in M$ with $p\neq q$. If they both lie in some chart $(U, \varphi)$, then we may construct disjoint open balls $B_p, B_q$ around $\varphi(p)$ and $\varphi(q)$ respectively. Then, since $\varphi$ is a homeomorphism, $\varphi^{-1}(B_p)\cap \varphi^{-1}(B_q)=\emptyset$ with $\varphi^{-1}(B_p), \varphi^{-1}(B_q)$ disjoint open sets in $M$ containing $p, q$ respectively.

If they don't lie in the same chart, then pick a chart $(U, \varphi)$ containing $p$ and $(V, \psi)$ containing $q$. If $U\cap V=\emptyset$ then we are done. So, suppose $U\cap V\neq \emptyset$. Then, since $p\in U$ and $q\in V$, we have that $U\setminus V$ and $V\setminus U$ are nonempty. Now, consider an open ball $B_p\subseteq \varphi(U\setminus (U\cap V))$ containing $\varphi(p)$ and an open ball $B_q\subseteq \psi(V\setminus (U\cap V))$ containing $\psi(q)$. We see that, again, $\varphi^{-1}(B_p)$ and $\psi^{-1}(B_q)$ are disjoint open sets in $M$ containing $p, q$ respectively.

I'm also fairly sure I can prove that this condition implies that $M$ is second-countable, but I'll leave that for another question perhaps. Is there something wrong with this proof? If not, why is Hausdorff included in the definition?

Martin Brandenburg
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IAAW
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    The standard example of a non-Hausdorff manifold is the line with two origins, i.e. the quotient space of $\mathbb{R} \sqcup \mathbb{R}$ where $x$ in the first copy is identified with itself in the second copy except when $x = 0$, so that the space looks like the real line but has "two copies" of the origin. It might be instructive to trace through your proof and see where it fails (Hint: "Now, consider an open ball [...]"). – Ben Steffan Nov 18 '24 at 23:40
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    @BenSteffan Your comment should be an answer. – jjagmath Nov 18 '24 at 23:46
  • @BenSteffan ah, so if I'm following you, the problem with my proof is that there is no guarantee that such open balls exist? This was very helpful. – IAAW Nov 18 '24 at 23:46
  • @IAAW If you've figured this out you should answer this question yourself :) – Ben Steffan Nov 18 '24 at 23:47
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    @BenSteffan Will do! Just making sure my understanding is correct, so I'm gonna go over it a few times. Will probably answer soon. I don't want to write an answer I'm not 100% confident in lol. Thanks for your help! – IAAW Nov 18 '24 at 23:48
  • The assumption is that $B_p$ exists. How come $U\cap V$ can't be dense in $U?$ – Thomas Andrews Nov 19 '24 at 00:25
  • It is worth noting that non-Hausdorff manifolds are used in some areas of math, notably dynamical systems. – Deane Nov 19 '24 at 02:42

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(Thanks to Ben Steffan for the help)

In the proof, $U\setminus V$ and $V\setminus U$ being nonempty does not guarantee that they're open. We cannot be sure that such open balls around $\varphi(p)$ and $\psi(q)$ exist! For instance, what if $\varphi(p)$ is a boundary point of $\varphi(U\setminus (U\cap V))$? Or if $\varphi(U\setminus (U\cap V))$ is dense in $\varphi(U)$?

The line with two origins is an example of a "manifold" which is not Hausdorff: for the two origins, each origin is in every neighborhood of the other, but they are distinct points. However, each point in this set has a chart containing it.

IAAW
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