Example of a space that is locally homeomorhpic to $\mathbb{R}^n$ but not Hausdorff

Question

I'm reading Brocker and Janich and am stumped already on the first page! I am very weak in topology so please bear with me.

The authors define a manifold by saying that it is locally homeomorphic to $\mathbb{R}^n$ everywhere (as well as being Hausdorff and second countable). That is, for every $x \in M^n$ there exists an open set $U \subset M^n$ and a homeomorphism $h:U \to U' \subset \mathbb{R}^n$. Now they say that this condition does not imply Hausdorff and then proceed to give a counter example.

Take the set $\mathbb{R} \cup \{ p \}$, where $p$ is an isolated point. Let us "define our topology" by letting $\mathbb{R}$ be open and that neighborhoods of $p$ are the sets $\{ U -\{0\} \} \cup \{ p \}$ where $U$ is a neighborhood of $\mathbb{R}$ at $0$.

${\bf Question \ 1}$: What is the dimension of this "set" (I suppose it is not a manifold as it is apparently not Hausdorff)? If it is locally homeomorphic to $\mathbb{R}^n$ then I would think $n=2$ due to the presence of the point $\{ p \}$. But it's weird to me because I think of homeomorphisms as deformations, and this set doesn't nicely deform into $\mathbb{R}^2$.

${\bf Question \ 2}$: What is the topology of this set? We "took $\mathbb{R}$ to be open" so I guess that is included in the topology. The topology also has to include the whole set and the null set. So is $T = \{M, \mathbb{R}, \emptyset, U_{\alpha}\}$ and how is it possibly second countable?

${\bf Question \ 3}$: How do we even define a "neighborhood" of $p$ without invoking a metric? What is the definition of a neighborhood?

${\bf Question \ 4}$: What is it about this space that fails to make it Hausdorff?

Thank you.

Answer 1

1. There's no such thing as the "dimension" of a topological space. There are various notions of dimension that make sense in various contexts, but you'd have to specify which one you mean. However, each point of the space has an open neighborhood homeomorphic to $\mathbb{R}^1$, so many notions of dimension would say that it's one-dimensional.
2. You're meant to regard $\mathbb{R}$ as a topological space here, so the open sets are open subsets of $\mathbb{R}$ together with the sets $(U \setminus \{0\}) \cup \{p\}$.
3. A "neighborhood" of a point, by definition, is an open set containing the point. [EDIT: See comments]
4. Try to separate $0$ and $p$ by open sets.

Answer 2

Question $1$: it is locally Euclidean for $n=1$. It is trivial everywhere except at $p$. You can picture neighbourhoods of $p$ as being neighbourhoods of $0$, replacing $0$ by $p$.

Question $2$: To define a topology, we must be able to take an arbitrary subset of $\Bbb R \cup \{p\}$ and tell if it is open or not. Let $V\subseteq \Bbb R \cup \{p\}$ be such a set. If $p \not\in V$, then $V \subseteq \Bbb R$ and we can decide. If $p \in V$, then $V$ is of the form $(U\setminus\{0\})\cup \{p\}$ for some "usual" neighbourhood of $0$, or it is not, so we can also decide. It is second-countable because we can take ${\cal B}$ a countable basis for the topology of $\Bbb R$, ${\cal B}_0$ a countable local basis at $0$, define $${\cal B}_p = \{ (U \setminus \{0\})\cup \{p\} \mid U \in {\cal B}_0\}$$and check that ${\cal B}\cup {\cal B}_p$ is a countable basis for the topology of $\Bbb R \cup \{p\}$ (and in fact ${\cal B}_p$ is a countable local basis at $p$).

Question $3$: we're dealing with topological spaces, not metric spaces. We need some collection of subsets, which we'll call open, that enables us to study continuity (this is possible, since continuity is characterized by the property "inverse image of any open set is again open"). Meaning that if we have sets to call "open", we don't need a metric anymore. We abstract it. An open neighbourhood of a point is an open subset containing it. It need not be a ball.

Question $4$: The points $p$ and $0$ screw up Hausdorffness. Every neighbourhood of $p$ intersects some neighbourhood of $0$. If $V$ is a neighbourhood of $0$, then $V = (U\setminus\{0\})\cup \{p\}$, for $U$ some "usual" neighbourhood of $0$, then $V$ intercepts $U$ at points distinct from $0$. But every $U$ gives arise to such a neighbourhood of $p$.

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Note that everywhere we had $\Bbb R$, we could've written $\Bbb R^n$ instead and the reasoning would be exactly the same.