$(0,3)$ as a non-Hausdorff smooth manifold

Question

Consider the interval $(0,3)$ and let the neighborhoods of the points in $(0,3)-\{1,2\}$ be as those in the real line, while the neighborhoods of $i=1$ or $2$ are defined to be $(i-\epsilon,i] \cup (2,2+\epsilon)$ for $0<\epsilon <1$. This is clearly non-Hausdorff, but why is it a smooth manifold? I couldn’t find an open Euclidean set homeomorphic with a neighborhood of $1$ because this is impossible if we consider such neighborhoods as open in the reals.

Answer 1

For $i=2$ you obtain ordinary $\epsilon$-neighborhoods which are open in $\mathbb R$, thus this point is uninteresting. Let $U_\epsilon = (1-\epsilon,1] \cup (2,2+\epsilon)$. Now define $$h : (0,2) \to (0,3), h(x) = \begin{cases} x & x \le 1 \\ x+1 & x > 1 \end{cases} $$ This is an injection. Since $h^{-1}(U_\epsilon) = (1-\epsilon,1+\epsilon)$, it is continuos. Moreover, since $h((1-\epsilon,1+\epsilon)) = U_\epsilon$, it is open. Hence $h$ establishes a homeomorphism between $(0,2)$ and $h((0,2))$. We therefore get an atlas for $(0,3)$ by taking $h$ and all $h_x : (0,x) \hookrightarrow (0,3)$ for $x < 1$ and $h_x : (x,3) \hookrightarrow (0,3)$ for $x > 1$. It is now easy to verify to all transition maps are smooth.