Line with two origins is a manifold but not Hausdorff

Question

The line with two origins is $(\mathbb{R} \times \{0,1\})/\sim$ where $(x,0)\sim(x,1)$ for $x\neq 0$. I can see that it is not Hausdorff, since we cannot separate the points $(0,0)$ and $(0,1)$.

However, I'm not quite clear on why it is a manifold, in particular why it is locally Euclidean. Please help me understand why this is so.

Often (in the majority of texts I have read), a manifold is by definition Hausdorff [and also often, additionally second countable]. Just mentioning it so you'll be aware of the different usages of the word. — Daniel Fischer, Nov 25 '14 at 16:00
Now, a space is locally Euclidean if every point has an open neighbourhood that is homeomorphic to an open set in $\mathbb{R}^n$. How do the natural neighbourhood bases of points in the line with two origins look? — Daniel Fischer, Nov 25 '14 at 16:03
Every point on the line with two origins has an interval nbhd right? — iwriteonbananas, Nov 25 '14 at 16:30
yeah each of those intervals would be homeomorphic to $\mathbb{R}$ — iwriteonbananas, Nov 25 '14 at 16:44

score 6 · Accepted Answer · answered Oct 21 '15 at 16:28

The key here is what is the topology on this space?

A possible base for this topology is $\mathcal B$ which contains all open balls not containing zero, and at zero, we either have $\{(-\varepsilon, 0) \cup \{\text{origin } A\} \cup (0, \varepsilon)\}$ and $\{(-\varepsilon, 0) \cup \{\text{origin } B\} \cup (0, \varepsilon)\}$.

Note: If you restrict the basis to rational endpoints, we get a countable basis, making the space second countable, which might also be in your definition of manifold.

Consider for instance $\{(-\varepsilon, 0) \cup \{\text{origin } A\} \cup (0, \varepsilon)\}$. There is an obvious bijection to $(-\varepsilon, \varepsilon)$ which is continuous and whose inverse is continuous.

Line with two origins is a manifold but not Hausdorff

1 Answers1

Linked