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.

- The topologist’s sine curve
- The comb space
- The ordered square
- See below
- The real line
- The disjoint union of two spaces of the 3rd type
- [0,1]∪[2,3]
- 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?

**Answer**

Let Rc be R with the cocountable topology. We’ll show that its cone is not locally path-connected, where the cone is X=(Rc×[0,1])/∼ with (x,t)∼(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⊂A is compact and (countably *or* uncountably) infinite. Given any countably infinite subset B0⊂B, we can cover A with sets of the form Ub=b∪(A∖B0) for b∈B0. A finite subcollection of these sets Ub can contain only finitely many elements of B, hence B is not compact. Now suppose that C⊂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. ◻

**Claim 1.** Every path into Rc is constant. Thus no open subsets of Rc are path-connected; in particular, Rc is not locally path-connected.

*Proof.* Let f:[0,1]→Rc be continuous. The image f([0,1]) is compact and connected, so it must be a singleton set by the lemma. ◻

**Claim 2.** The cone on Rc, denoted X, is not locally path-connected.

*Proof.* Fix a point (x,t) in the open subset Rc×[0,1)⊂X. Any path f:[0,1]→Rc×[0,1) projects to a path p∘f:[0,1]→Rc under the projection p:Rc×[0,1)→Rc. By Claim 1, p∘f is constant, so all paths into Rc×[0,1) have a fixed first coordinate. Because every open neighborhood of (x,t) includes points (x′,t′) with x′≠x, it follows that no neighborhood of (x,t) is path-connected. Thus X is not locally path-connected. ◻

**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, Rc×[0,1] is locally connected. Quotient maps also preserve local connectedness, so this implies that the cone X is locally connected.

**Attribution***Source : Link , Question Author : Qian , Answer Author : Kyle*