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 :)


Solution 1:

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

enter image description here

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.

Solution 2:

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

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.)

Solution 3:

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.