I'm trying to classify the various topological concepts about connectedness. According to 3 assertions ((Locally) path-connectedness implies (locally) connectedness. Connectedness together with locally path-connectedness implies path-connectedness.), we can draw this diagram:

|Connected                 |
|             1      +-----+----------------------------+
|                    |  3  |           Locally connected|
|   +----------------+-----+     6                      |
|   |Path-connected  |  4  |                            |
|   |                +-----+------------------------+   |
|   |    2           |  5  |  Locally path-connected|   |
+---+----------------+-----+                        |   |
                  8  |           7                  |   |

So, I want to find examples of all these 8 categories, but I can't find an example for 4.

  1. The topologist's sine curve
  2. The comb space
  3. The ordered square
  4. See below
  5. The real line
  6. The disjoint union of two spaces of the 3rd type
  7. $[0,1] \cup [2,3]$
  8. The rationals

Actually there is an answer that gives an example of type 4, but there isn't any explanation. Can anyone please explain it (why it is not locally path-connected, to be specific) or give another example?

  • $\begingroup$ You won't find explanations, but Spacebook is a good source of examples. They are mostly drawn from Steen and Seebach's excellent text "Counterexamples in Topology". $\endgroup$ – Austin Mohr Jun 22 '14 at 19:32
  • $\begingroup$ Thanks. But that site doesn't have an example having these properties. $\endgroup$ – TheoYou Jun 23 '14 at 1:44
  • $\begingroup$ Just a stray remark, but for those who find the counterexamples below unsatisfyingly pathological, unfortunately the counterexamples are always going to be pretty bad. Any connected and locally connected, compact metric space (Peano continuum) is locally path-connected. It follows easily that any open subset of a connected and locally connected sigma-compact metric space is locally path-connected. I'm not sure if weakening from "metric" to "Hausdorff" changes the answer, it's probably somewhere in Kuratowski's Topology II. $\endgroup$ – John Samples Jun 19 '17 at 4:03

Let $\mathbb{R}_c$ be $\mathbb{R}$ with the cocountable topology. We'll show that its cone is not locally path-connected, where the cone is $$X=(\mathbb{R}_c \times [0,1])/\sim$$ with $(x,t)\sim (x',t')$ if $t=t'=1$.

Lemma. Let $A$ be an uncountable set with the cocountable topology. Then compact subsets of $A$ are finite and connected subsets are either singletons or uncountable.

Proof. Suppose $B\subset A$ is compact and (countably or uncountably) infinite. Given any countably infinite subset $B_0\subset B$, we can cover $A$ with sets of the form $U_b=b \cup (A \setminus B_0)$ for $b \in B_0$. A finite subcollection of these sets $U_b$ can contain only finitely many elements of $B$, hence $B$ is not compact. Now suppose that $C \subset A$ is connected and contains more than one element. Since finite/countable subspaces in the cocountable topology have the discrete (subspace) topology, $C$ is uncountable. $\square$

Claim 1. Every path into $\mathbb{R}_c$ is constant. Thus no open subsets of $\mathbb{R}_c$ are path-connected; in particular, $\mathbb{R}_c$ is not locally path-connected.

Proof. Let $f:[0,1] \to \mathbb{R}_c$ be continuous. The image $f([0,1])$ is compact and connected, so it must be a singleton set by the lemma. $\square$

Claim 2. The cone on $\mathbb{R}_c$, denoted $X$, is not locally path-connected.

Proof. Fix a point $(x,t)$ in the open subset $\mathbb{R}_c \times [0,1) \subset X$. Any path $f:[0,1] \to\mathbb{R}_c \times [0,1)$ projects to a path $p \circ f:[0,1] \to \mathbb{R}_c$ under the projection $p: \mathbb{R}_c \times [0,1) \to \mathbb{R}_c$. By Claim 1, $p \circ f$ is constant, so all paths into $\mathbb{R}_c \times [0,1)$ have a fixed first coordinate. Because every open neighborhood of $(x,t)$ includes points $(x',t')$ with $x'\neq x$, it follows that no neighborhood of $(x,t)$ is path-connected. Thus $X$ is not locally path-connected. $\square$

Remark. Any uncountable set with the cocountable topology is connected because any two (nonempty) open sets intersect. It follows that open subsets of such a space are themselves uncountable sets with the cocountable (subspace) topology, hence all open subsets are connected. Then certainly the original space is locally connected. Since a finite product of locally connected spaces is locally connected, $\mathbb{R}_c \times [0,1]$ is locally connected. Quotient maps also preserve local connectedness, so this implies that the cone $X$ is locally connected.

  • $\begingroup$ Ingenious! Except there's one little problem with your proof of the lemma. The set $U_b = b \cup (A \setminus B)$ may not be open, since it's possible that $B$ is not countable. So we should choose a countable subset $B_0$ of $B$ which has more than 1 element, and let $U_b = b \cup (A \setminus B_0)$. $\endgroup$ – TheoYou Jun 23 '14 at 1:49
  • $\begingroup$ @TheoYou: Thanks, and excellent catch and salvage! I'll make the edit now. $\endgroup$ – Kyle Jun 23 '14 at 3:29
  • $\begingroup$ As an update, can you find one on $\Bbb{R}^2$ with the normal topology? math.stackexchange.com/questions/1492772/… $\endgroup$ – user223391 Oct 26 '15 at 17:20
  • $\begingroup$ Sorry to necro this, but isn't it possible that $A\setminus B_0$ contain infinitely many points of $B$? As in, isn't is possible that $B\setminus B_0$ be infinite? EDIT: I think it's enough to say that 'A finite subcollection of these sets $U_b$ can contain only finitely many elements of $B_0$', rather than 'finitely many elements of $B$'. $\endgroup$ – Fimpellizieri Jul 4 '17 at 22:26

A very nice example for (4) is a "long circle". That is, let $L$ be the closed long line, with endpoints $0$ and $\omega_1$, and adjoin to $L$ a path from $0$ to $\omega_1$ to get a space $S$. Then $S$ is path-connected and locally connected, but it is not locally path-connected at $\omega_1$.


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