# Path-connected and locally connected space that is not locally path-connected

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?

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 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.