# Space which is connected but not path-connected

Consider the following two definitions:

Connected : A topologiocal space X is connected if it is not the disjoint union of two open subsets, i.e. if X is a disjoint union of two open sets A and B, then A or B is empty set.

Path Connected : A topological space X is path-connected if any two points in X can be joined by a continuous path.

So far I can picture, I think they should be equivalent. If X is connected iff it is path-connected. But, the Thm I got from lecture is just if X path connected then X is connected (The converse not necessarily true.)

Can anyone give me some picture of space which is connected but not path-connected? Just picture please, cuz I wanna grab the idea. I have already seen some examples like in connected but not path connected?, I cannot grab the idea if it (such space) consist of a single piece, then I couldn't make a continuous path from any two points.

Thank you :)

• Do a google search for the topologist's sine curve. Commented Apr 10, 2016 at 5:14
• @John, thanks for your suggestion. It seems quite similar to below answers :) Commented Apr 10, 2016 at 6:17
• If my example works for you could you upvote it plz. Commented Jul 21, 2022 at 1:07
• Path components of Topologist's Sine Curve. As a side note, it is possible to close this example to obtain the (path-connected) Warsaw circle, another interesting shape. Commented Sep 21, 2023 at 9:21

Others have mentioned the topologist's sine curve, that's the canonical example. But I like this one, though it is the same idea:

This is the image of the parametric curve $\gamma(t)=\langle (1+1/t)\cos t,(1+1/t)\sin t\rangle,$ along with the unit circle. There is no path joining a point on the curve with a point on the circle. Yet, the space is the closure of the connected set $\gamma((1,\infty))$, so it is connected.

• If that is the union of $\gamma (t)$ with the circle doesn't that mean that it is disjoint union of two sets. But, circle is not even open. But if it is not disjoint union, then I can always make a path, continue the path from the circle to the curve. *Confused Commented Apr 10, 2016 at 5:28
• @ChenMLing It is a disjoint union of the image of $\gamma$ and a circle, but as sets. The topology is vastly different than the disjoint union topology (namely, it's the subspace topology on $\gamma(1, \infty)\cup S^1$ where $S^1$ is the unit circle). Commented Apr 10, 2016 at 5:34
• This space is the union of two sets, but the circle is not open in the subspace topology. The issue is that the spiral gets arbitrarily close to, but does not touch the circle. If you start on the circle, you cannot leave the circle and go onto the spiral, because there is not place where the spiral meets the circle. Commented Apr 10, 2016 at 5:35
• [...] The intuition for why the resulting space is not path-connected is the following: if a path went round and round the spiral and hit a point on the limiting unit circle, then $\lim_{t \to \infty} \gamma(t)$ must exist. That doesn't, so no such continuous path exists. Intuitively, the path has to be "infinitely long", so it cannot be image of $[0, 1]$, which is a nice compact thing. Commented Apr 10, 2016 at 5:36
• @ChenMLing Suppose you look at $X = (0, 1) \cup \{1\}$ and give it the subspace topology from $\Bbb R$ where it lives. That's a disjoint union of sets, because $(0, 1)$ does not contain $\{1\}$. Yet it's connected. This is because due to the topology $1$ is not isolated: it's a limit point. The analogous thing here is that the unit circle is a limit circle of the spiral $\gamma$. Commented Apr 10, 2016 at 10:18

This picture of the comb space from the Wikipedia article on it may help:

This answer contains a pretty extensive discussion of why the deleted comb space, though connected, is not path-connected. (The deleted comb space has only two points on the $y$-axis, the origin, and $\langle 0,1\rangle$. There is no path from $\langle 0,1\rangle$ to any other point of the space.)

• I don't understand this picture. The blue lines seems not connected to each other. No? Commented Apr 10, 2016 at 5:31
• @ChenMLing: Did you look at the Wikipedia article? The space includes $[0,1]\times\{0\}$, the spine of the comb. Commented Apr 10, 2016 at 5:33
• Oh sorry, yeah it's connected. But then I can make a path from any point in the blue line, going down then go to other point in other blue line. Am I wrong? Commented Apr 10, 2016 at 5:38
• @ChenMLing: You’re not wrong: if you remove the point $\langle 0,1\rangle$ from the deleted comb space, what remains is path-connected. That’s one thing that makes it nice: there’s just one point that isn’t connected by a path to anything else, so one can see relatively clearly what is keeping it from being path-connected. Commented Apr 10, 2016 at 5:40
• Why I cannot make a path from $<0,1>$ to other point in the space? Is it because the $y$-axis only contain two points, 0 and $<0,1>$? So, it's not connected along the $y$-axis? Commented Apr 10, 2016 at 5:45

How about $$\mathbb{R}_K$$. That is the all open intervals together with intervals of the form

$$(a,b) \setminus K$$

where $$K=\{\frac{1}{n} : n \in \mathbb{Z}^+\}$$. This space is connected and Hausdorff but not path-connected.

• Why it us not parh connected?
– MANI
Commented Dec 10, 2021 at 18:21
• @MANI suppose there is a path $\gamma: [a,b] \rightarrow \mathbb{R}$ by continuity, $\gamma([a,b])$ is connected and compact in $\mathbb{R}_K$ but must also be connected as subspace of usual topology then the image contains $[a,b]$ thus $[a,b]$ is compact in the $K$ topology which is not true. contradiction. Commented Dec 10, 2021 at 22:07