S. D. V. Stephens

notes / Math 55b / chapter 2: Connectedness and Compactness / From my notebook

N11 Path-Connectedness

Lecture 6.

Definition N11.1 (Path). Let be a topological space and . A path from to is a continuous map (with a subspace of ) such that , .

An interval [a,b] mapped by f into a blob X, tracing a curve from the point x to the point y.

Definition N11.2 (Path-connected). is path-connected if any two points can be joined by a path.

Note. Define by: there exists a path from to . Claim: this is an equivalence relation.

  • Constant path: .
  • Backwards path.
  • Concatenated paths (“pasting lemma”):
A path f1 from x to y followed by a path f2 from y to z.

The equivalence classes are the path-connected components. (We’ll return to this in algebraic topology.)

Theorem N11.3. If is path-connected, then is connected.

Proof. Assume is not connected but is path-connected. So with . Let , , and let be a path from to .

Two disjoint blobs U and V with a path from x in U to y in V, red inside U and green inside V; below, the interval [a,b] split into the red preimage of U and the green preimage of V.

Well, we know intervals are connected, and with and , a separation of : contradiction. ∎

Note (Margin question). connected but not path-connected?

The converse is false in general, but true for nice enough spaces (e.g. CW spaces).

Example N11.4 (The “topologist’s sine curve”).

Graph of sin(1/x) for x > 0, oscillating faster and faster between -1 and 1 as x approaches 0, together with the origin marked in red.

is the image of (which is connected) under the continuous map , therefore is a connected subspace of .

Claim: is connected. If with disjoint and open, then both open in the subspace topology. is connected, so one of them is empty (they cannot both be). Without loss of generality and . Then which is not open in , so .

We claim that is not path-connected.

Proof. Take a path in from to , . By the IVT, takes every value in , so the image of the path contains the points

By compactness (pass to a convergent subsequence ) there is with , which is not a point of . Contradiction. ∎

Theorem N11.5. Let be open. Then is connected iff is path-connected.

Proof. Assume is open in . Then the path components of are open. Indeed, if then there is such that . Any two points of can be joined by a path inside (e.g. a straight line), so is contained in the path component of .

An open set A in the plane containing a point x and an open ball B_r(x) inside A, with a straight-line segment joining two points of the ball.

If is not path-connected, then is a separation, so is not connected. ∎

Remark. Similarly for manifolds and CW complexes (something with -cells?). For such spaces, the path components are also the connected components of , i.e. they partition into disjoint connected (nonempty) open subsets.