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?

  • $\begingroup$ 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 $\endgroup$ – Lonidard Aug 27 '16 at 14:43

The concept of arc connected seems to be inconsistently defined in the literature: that link refers to two definitions.

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 $x_1 \ne 0$, $x_2 = 0$.

  • $\begingroup$ This clears it up. Thank you very much. $\endgroup$ – Lonidard Aug 27 '16 at 16:36
  • $\begingroup$ It is my pleasure. $\endgroup$ – Lee Mosher Aug 27 '16 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.