# If a topological space is a path-connected, it is also connected. The converse fails. Can this be remedied?

If a topological space is a path-connected, it is also connected. However, the converse of this theorem is false.

Can we generalize the notion of a path so that the converse also holds? I was thinking that maybe if the domain of a path was allowed to be an arbitrary connected totally ordered set, this might fix the problem.

EDIT. As Brian M. Scott explains in his answer, this doesn't fix the problem; so here's a last ditch-effort at salvaging the idea. Definition: A connected path in a topological space $X$ is a mapping $\gamma : T \rightarrow X,$ that preserves connectedness under direct images, where $T$ is a connected totally ordered set possessing both a least and a greatest element. Note in particular that $\gamma$ needn't be continuous.

Can anyone see whether this fixes the problem or not? We want to be able to prove that a topological space is path connected iff it is connected.

Anyway, here's the motivation.

The problem, as Brian explains, is that a connected linearly ordered space with endpoints is compact, so the continuous image of such a space is also compact and connected. This is a problem, because:

1. No compact, connected subset of the topologist’s sine curve contains the origin and at least one other point of the curve, and
2. There exist countably infinite connected Hausdorff spaces, and there are no paths between distinct points in such a space; because if such a path existed, its image would be a compact countably infinite connected Hausdorff space with at least two points, and no such space exist.

So the problem seems to be the compactness of the image of the path, which is implied by the compactness of the domain, because the path is defined as a continuous function. Thus, weakening the requirement that paths need to be continuous might fix the problem.

## 3 Answers

Your suggestion would take care of the long line, since it is a connected linearly ordered space, but it wouldn’t help with the topologist’s sine curve for instance. A connected linearly ordered space with endpoints is compact, so the continuous image of such a space is also compact and connected, but no compact, connected subset of the topologist’s sine curve contains the origin and at least one other point of the curve.

It also won’t help with any of the various countably infinite connected Hausdorff spaces that have been constructed. A generalized path in such a space would be a countable, connected, compact Hausdorff space, and no such space exists.

• You write: "No compact, connected subset of the topologist’s sine curve contains the origin and at least one other point of the curve." Are you sure about this? It seems to me that $T \cap [0,1]\times \mathbb{R}$ is both compact and connected. Which one fails? – goblin Apr 16 '13 at 21:54
• @user18921: $T\cap([0,1]\times\Bbb R=T$ is not a closed subset of $\Bbb R^2$, so it’s not compact. Its closure includes $\{0\}\times[0,1]$. – Brian M. Scott Apr 16 '13 at 23:19
• Okay, good point. What if instead of requiring that a path be continuous, we merely require that it preserves connectedness? This allows that the image of a compact set can fail to be compact. – goblin Apr 17 '13 at 1:24
• @user18921: I really have no intuition about that. Once you get away from continuous maps, you’re out of any purely topological setting. – Brian M. Scott Apr 17 '13 at 9:40

The following theorem might be the generalization you are looking for:

Theorem. A topological space $X$ is connected iff for every $x,y\in X$ and for every open cover $\mathcal U$ of $X$ there is a chain $U_1,\ldots,U_n\in\mathcal U$ that connects $x$ and $y$, more precisely: $x\in U_1$, $y\in U_n$ and $U_i\cap U_{i+1}\neq\emptyset$ for $i=1,2,\ldots,n-1$.

This is basically Theorem 8 on page 136 in Kuratowski - Topology, vol.2. (1968).

• Thanks, that's a cool theorem. – goblin Apr 17 '13 at 5:48

Connected plus Locally Path Connected Implies Path Connected. Here is the short proof if you are interested you may go through it. Let $C$ be a connected set that is also locally path connected. Pick any point $x$ in $C$, and let $U$ be the set of points in $C$ that are path connected to $x$. Thus $U$ is a subset of $C$. Let $y$ be a point in $U$. Enclose $y$ in an open set $H$ in $C$, such that $y$ is path connected to all of $H$. Since an arc can run from $x$ to $y$ to anything in $H$, $H$ is in $U$. Therefore $U$ is the union of open sets and is open, relative to $C$.

Let $y$ be a point in $C$ that is a limit point of $U$. Put an open set $H$ around $y$ such that $H$ is path connected. Let $z$ be common to $H$ and $U$. Now $x$ connects to $z$ connects to $y$, and $y$ is in $U$.

Since $U$ contains its limit points it is closed. thus $U$ is both open and closed in $C$. If $U$ is not all of $C$, separate $U$ and the rest of $C$ in open sets. This contradicts the fact that $C$ is connected. Therefore $U$ is all of $C$, and $C$ is path connected.