# Uncountable Point exclusion topology: Arc connectedness

I am trying to prove some facts about the uncountable point exclusion topology: in particular I have some doubts about arc-connectedness.

Let $(\mathbb R,\tau)$ indicate the real line equipped with the topology $$\tau:=\{A\subset \mathbb R : A \cap \{0\} = \varnothing \}\cup \mathbb R$$ $\tau$ is discrete on $\mathbb R\setminus \{0\}$: every subset $A$ of $\mathbb R \setminus \{0\}$ is such that $A\cap \{0\}=\varnothing$, so $A$ is open.

$(\mathbb R, \tau)$ is clearly connected: suppose by contradiction that there exists a nontrivial clopen set $A$. If $A$ is an open set, then $A\cap \{0\}=\varnothing$. If $A$ is closed, $A^C\cap \{0\}=\varnothing \iff A\cup (\mathbb R \setminus \{0\})=\mathbb R$ and $0 \in A$, a contradiction.

Now, I'd like to believe that the space is not arc-connected: this is claimed here. On my topology course notes (in Italian, you can find them here) it is claimed on page $104$ that $\mathbb R$ with the point exclusion topology is arc-connected.

If we suppose that $(\mathbb R, \tau)$ is arc-connected, this would mean that for every $x_1, x_2 \in \mathbb R$ there exists a continuous function $$\begin{array}{r c c c }\gamma: & [0,1]& \longrightarrow & \mathbb R\\ & t & \mapsto &\gamma(t)\end{array}$$ such that $\gamma(0)=x_1$, $\gamma(1)=x_2$; taking $x_1>0$, $x_2>0$, this would imply that every fiber $\gamma^{-1}(x)$ with $x\in (x_1,x_2)$ would be an open set; since $[0,1]$ is $T_1$, points are closed, but it is also connected so fibers must be either empty or at least intervals of the form $(a,b)$. This is kind of far-fetched but I cannot find an explicit contradiction.

Would a function like $$\gamma(t)=\begin{cases} x_1 & t\in[0,\frac{1}{3}) \\ 0 & t\in[\frac{1}{3},\frac{2}{3}]\\x_2 & t \in(\frac{2}{3}, 1] \end{cases}$$ work? Let's say that $A$ is an open set in the codomain $(\mathbb R, \tau)$: then $0 \not \in A$. Then

• If $A$ does not contain $x_1,x_2$, $\gamma^{-1}(A)=\varnothing$, which is open.
• If $A$ contains $x_1, x_2$ or both, then its preimage can be either $[0,\frac{1}{3})$, $(\frac{2}{3},1]$ or $[0,\frac{1}{3})\cup (\frac{2}{3},1]$ which are open sets in $[0,1]$ with the subset topology.

Is $(\mathbb R, \tau)$ an arc connected space or not? Can somebody point to a proof?

• Add: I found out that different authors require invertibility for the path, so I guess this could be the issue. Choosing $x_1=0$, $x_2=x$: if $\gamma(0)=0$ and $\gamma(1)=x$, injectivity implies that $\gamma^{-1}(x)=\{1\}$; but $x$ is open because $\{ x\} \cap \{0\}=\varnothing$, while $\{1\}$ is closed in $[0,1]$, which is connected. QED – Lonidard Aug 27 '16 at 14:43

In one definition, "arc connected" requires the path to be a homeomorphism onto its image, in which case the topology $\tau$ is not arc connected: there are no subspaces whatsoever that are homeomorphic to $[0,1]$, because any $t \in [0,1]$ which does not map to $0 \in \mathbb{R}$ would have to form a singleton open set $\{t\} \subset [0,1]$, which is absurd.
In the other definition, "arc connected" is a synonym for "path connected", in which case the topology $\tau$ is arc connected, for instance if $x_1,x_2 \ne 0$ then they are connected by the path $$\gamma(t) = \begin{cases} x_1 & \quad\text{if t \in [0,1/2)} \\ 0 &\quad\text{if t=1/2} \\ x_2 &\quad\text{if t \in (1/2,1]} \end{cases}$$ whereas $x_1=0$, $x_2 \ne 0$ are connected by the the path $$\gamma(t) = \begin{cases} 0 &\quad\text{if t=0} \\ x_2 &\quad\text{if 0 < t \le 1} \end{cases}$$ and similarly for $x_1 \ne 0$, $x_2 = 0$.