Name for topology generated by the standard open sets of reals and every point that isn't zero.

Question

I'm interested in topological spaces that are easy to construct. In particular, I'd like to know what this one is called so I can look it up in pi-base, assuming of course that it's interesting enough to be included there.

Let $(\mathbb{R}, \tau^R)$ be the standard topology on the reals.

Let $\sigma$ be the topology generated by $\tau^R \cup \{ \{x\} : x \in \mathbb{R} \setminus 0 \}$.

$(\mathbb{R}, \sigma)$ is a topological space. It is almost the discrete topology, except that any open set containing zero must have a standardly open subset.

Some preliminary observations.

• All of the open sets containing $0$ are in fact clopen.
• All the sets that are open but not closed get arbitrarily close to $0$, intuitively.

Answer 1

Your space is (subtlely) already in $\pi$-Base, under the name S133: Post office metric on the plane.

I'll prove something a bit more general. Let $X$ be a metric space containing a point $0$, and let the sets $A_0=X\setminus B_1(0)$ and $A_{n+1}=B_{1/2^n}(0)\setminus B_{1/2^{n+1}}(0)$. Note $X\setminus\{0\}=\bigcup\{A_n:n<\omega\}$ and suppose $|A_n|=|\mathbb R|$ for all $n<\omega$. Give $X$ the post-office metric $$ d(x,y)= \begin{cases} 0 &\text{if }x=y\\ d(x,0)+d(0,y)&\text{otherwise} \end{cases} $$

I claim that $X$ is homeomorphic to your space $\mathbb R$. Let $\theta:X\to\mathbb R$ be defined as follows. First, choose $\theta\upharpoonright A_0$ to be an arbitrary bijection from $A_0$ to $\mathbb R\setminus(-1,1)$. Then, choose $\theta\upharpoonright A_{n+1}$to be an arbitrary bijection from $A_{n+1}$ to $(-1/2^{n},1/2^n)\setminus(-1/2^{n+1},1/2^{n+1})$. Finally, let $\theta(0)=0$.

We have a bijection, so we now show it is a continuous open map.

To see that it is open, let $x\in X\setminus\{0\}$. Then $\{x\}$ is open, and $\theta[\{x\}]\subseteq\mathbb R\setminus\{0\}$ is open. For $0\in X$, choose the basic open set $B_{1/2^n}(0)$. Then $\theta[B_{1/2^n}(0)]=(-1/2^n,1/2^n)$ is open.

To see that it is continuous, let $y\in \mathbb R\setminus\{0\}$. Then $\{y\}$ is open, and $\theta^\leftarrow[\{y\}]\subseteq X\setminus\{0\}$ is open. For $0\in\mathbb R$, choose the basic open set $(-1/2^n,1/2^n)$. Then $\theta^\leftarrow[(-1/2^n,1/2^n)]=B_{1/2^n}(0)$ is open.

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I'd very much appreciate it if you'd like to contribute this new perspective on S133 to the $\pi$-Base. I'll create an issue and link it here.