There is a second sense in which a space can be one piece: any two of its points can be joined by a path inside it. A path is a continuous map from a closed interval into the space, so this is a statement about journeys rather than about splittings. Path connectedness implies connectedness, and the two are not the same — but for the spaces met in this course they usually agree, and paths are far easier to produce.

A country is one piece on the map if it cannot be split in two; it is one piece to a traveller if you can drive from any town to any other. Driving is the stronger claim, and the easier to demonstrate.

A path from one point to another: a continuous $\gamma:[0,1]\to X$ with $\gamma(0)$ at the start and $\gamma(1)$ at the finish. Its image is a continuous image of an interval, so it is connected — and a space in which every pair of points is joined this way is a union of connected sets all sharing one point, which is the argument that path connected implies connected.

A path from $x$ to $y$ in $X$ is a continuous map $\gamma:[0,1]\to X$ with $\gamma(0)=x$ and $\gamma(1)=y$, where $[0,1]$ carries its topology from $\mathbb{R}$. The space is path connected when every pair of its points is joined by some path. Nothing asks the path to be injective: it may double back, and it may visit the same point many times.

• $\mathbb{R}^n$ is path connected, and so is any convex subset of it: the straight path $\gamma(t)=(1-t)x+ty$ stays inside and is continuous.
• The plane with the origin removed is path connected — walk round the hole — although the line with a point removed is not. Punctures cost more in one dimension than in two.
• Path connected implies connected, which is the theorem below. The converse fails: the topologist's sine curve, the graph of $\sin(1/x)$ for $x>0$ together with the segment it presses against, is connected and not path connected. It is named here and not verified: the argument belongs to a later course.
• Paths join end to end. If a path runs from $x$ to $y$ and another from $y$ to $z$, running the first at double speed and then the second gives a path from $x$ to $z$, which is why path connectedness behaves like a relation between points.
• A continuous image of a path connected space is path connected: send each path forward through the map, and continuity of a composition does the rest.

Fix one point of the space. For every other point there is a path from the fixed one to it, and the image of that path is a continuous image of an interval. An interval is connected, and continuous images of connected spaces are connected, so each such image is a connected subset. Every one of them holds the fixed point, and between them they hold every point of the space, since each point is the far end of its own path. So the space is a union of connected sets sharing a point, and the union theorem of the previous lesson makes it connected.

Fix a point and take, for each point, a path to it from the fixed one.

Each image is a continuous image of a connected interval, so it is connected.

Every one of these sets holds the fixed point, since every path starts there.

Every point lies in the image of its own path, so the images cover the space.

A union of connected sets sharing a point is connected.

A path is a continuous map from $[0,1]$ into the space, with named ends.

• A path is a continuous map from $[0,1]$ with named ends.
• Path connected implies connected, by the union theorem applied to path images.
• $\mathbb{R}^n$, convex sets and the punctured plane are path connected; the punctured line is not.
• The converse fails, and the topologist's sine curve is the standard witness.

Great work! You now know how to:

• write down a path and check its two ends;
• prove that path connected spaces are connected;
• produce paths in convex sets and round a puncture;
• keep the two notions apart, and name where they differ.