I'm trying to characterize the path-connected subsets of the compact complement topology over $\mathbb{R}$ (let's call $X$ this topological space).

I have already proved the connected subsets are the intervals (so, in particular, the whole space is connected) and the unbounded ones. As the compact complement topology is coarser than the usual topology on $\mathbb{R}$, I already know every path-connected subset of $\mathbb{R}$ is a path-connected subset of $X$, so every interval is a path-connected subset of $X$. If there were others, they would be connected,so they would be unbounded. However, as the unit interval cannot be decomposed in a denumerable quantity of disjoint, nonempty closed sets, we know a denumerable subset of $X$ is never path-connected, even if it is connected.

So, my question is, are all uncountable unbounded sets of $X$ path connected? It doesn't seem so, but I don't know the correct way of reasoning about these things. For example, $(-\infty,-1)\cup[0,1]\cup(2,\infty)$ doesn't seem path connected, while $(-\infty,-1)\cup(2,\infty)$ does. But this is just an intuition, as I don't really know hwo to work properly with functions that end in this space.


1 Answer 1


Any set containing an unbounded interval is path-connected. The intuitiv idea is that you can walk to infinity and jump from there to any point you like.

E.g. if it contains $(y_0,\infty)$ for some $y_0\in \mathbb{R}_{>0}$. Let $x_0\in \mathbb{R}$. If $x_0\geq y_0$, then they are clearly connected by a continuous path ($\gamma:[0,1]\rightarrow X, \gamma(t)= x_0+ t(y_0-x_0$)). On the other hand, if $x_0< y_0$, then we have the path $$ \gamma: [0,1] \rightarrow X, \gamma(t)=\begin{cases} \frac{y_0}{t},& t\neq 0, \\ x_0,& t=0. \end{cases} $$ Let me show that is continuous in the case $x_0\geq 0$ (the case $x_0<0$ is similar). Note that any open nbhd $U$ of $x_0$ in $X$ can be written as $$ U= V \setminus \{x_0\} \cup (-\infty, -m) \cup (\{x_0\} \cup (n, \infty)) $$ where $m,n\in \mathbb{R}_{>0}$ and $V\subseteq \mathbb{R}$ open and bounded. Then we have $$ \gamma^{-1}(U) = \gamma^{-1}(V\setminus \{x_0\}) \cup \gamma^{-1}((-\infty, -m)) \cup \gamma^{-1}(\{x_0\} \cup (n,\infty)).$$ We show that all of those sets are open. First we note $\gamma^{-1}((-\infty, -m))=\emptyset$, which is open in $[0,1]$. Next we have $$ \gamma^{-1}(\{x_0\} \cup (n,\infty)) = \begin{cases} [0, \frac{y_0}{n}),& n>y_0, \\ [0,1],& n\leq y_0. \end{cases} $$ In both cases the sets are open in $[0,1]$.

Finally, as $V$ is bounded, there exists $R>0$ such that $V\subseteq (-R,R)$. Then we define $$ \tau : [0,1] \rightarrow \mathbb{R},\tau(t):= \min \{ R, \gamma(t) \}. $$ As $\tau$ is continuous and $\tau^{-1}(V\setminus \{x_0\} ) = \gamma^{-1}(V\setminus \{x_0\})$, we get that also $\gamma^{-1}(V\setminus \{x_0\})$ is open in $[0,1]$.

Right now I do not have time to work it out, but I guess that that if a set does not contain an unbounded interval, then it is path-connected in $X$ iff it is path-connected in $\mathbb{R}$. The idea is that the space is first-countable and thus continuity and sequential continuity coinced (see here Sequentially continuous implies continuous). Then we should be able to use that bounded sequences in $X$ converge iff they converge in $\mathbb{R}$. This means you cannot jump unless you are at infinity.

  • $\begingroup$ thank you, I love your explanation $\endgroup$
    – Seven
    Apr 10, 2020 at 16:27
  • $\begingroup$ @Seven Thank you for your kind words. I would suggest that you try to make the argument in my last paragraph rigorous (or show that it does not work). This way you gain deeper understanding and the community can give you feedback when you post it here :) I do not see myself doing it in the next couple of days (I'm busy with a lot of other things). $\endgroup$ Apr 10, 2020 at 18:09
  • $\begingroup$ why is $\tau$ continuous? $\endgroup$
    – Ben
    Jul 29, 2021 at 18:42
  • $\begingroup$ $\tau(V\backslash\{x_0\})$ should be $\tau^{-1}(V\backslash\{x_0\})$ $\endgroup$
    – Ben
    Jul 29, 2021 at 18:45
  • $\begingroup$ @Ben: We have that $\tau$ is continuous away from the origin (it is a rational function and we are working in the euclidean topology). However, the cut-off (i.e. taking the minimum) ensures that $\tau$ is constant in a nbhd of the origin and thus is also continuous there. You are right about the typo, thanks, I corrected it. $\endgroup$ Jul 29, 2021 at 21:53

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