First time here, so sorry for the rookie mistakes. I'm a $4^{th}$ year physics student taking Riemannian Geometry so my background on the subject is very small.
I'm trying to solve Boothby's exercise $I.3.1$ on showing that a subset of $\mathbb{R}^{2}$ is not a manifold. The subset is $X = A_{+} \cup A_{-} \cup B$ with
$A_{+} = \{(x,y): x \geq 0, y=+1 \}$
$A_{-} = \{(x,y): x \geq 0, y=-1 \}$
$B = \{(x,y): x < 0, y=0 \}$
EDIT sugested by Travis: Boothby indeed tells the reader how to define a topology. So, what Boothby does is to use the subspace topology on $A_{+}-\{(0,1)\}$, $A_{-}-\{(0,-1)\}$ and B, then for $\epsilon>0$ he lets $N_{\epsilon}^{\pm} = \{(x,\pm 1): 0 \leq x < \epsilon\} \cup \{(x,0): −\epsilon \leq x<0\}$, finally he uses this as a basis of neighborhoods of $(0,1)$ and $(0,−1)$, respectively for $+$ and for $−$.
I've shown that the space X is locally Euclidean but I don't know what to do from here. Obviously it must have something to do with the remaining conditions for a manifold, but I'm unfamiliarized with the concepts of Hausdorff spaces and the second axiom of numerability.
If someone could point me in the right direction I would be very grateful!
Thanks in advance!
This is the so-called branching line (counter)example, which is related to a perhaps better-known example known as the line with two origins.
To show that a space $X$ is not Hausdorff, it suffices to pick two points $x, y$ such that for every neighborhood $U$ of $x$ and $V$ of $y$, $U \cap V \neq 0$. We know that away from the points on the $y$-axis the topology looks like the usual (subspace) topology, which is Hausdorff, which suggests the problem points should be on the $y$-axis, and there are only two of these, namely $(0, \pm 1)$, and we know what the neighborhoods of these spaces are.
Once you've sorted this problem, you might like to try to prove that another space is not a manifold, namely, the union of three distinct rays (in, say, $\mathbb{R}^2$) with a common vertex, which is what I initially thought your problem might have referred to.