Separation and connectedness · Definition

Path-connected space

topologyconnectednesspaths

Before you start you need to know

It is the «with curves» version of connectedness: instead of asking that the space does not split, we ask to travel from any point to any other without leaving it.

Definition

A space \(X\) is path-connected if every pair of points \(a,b\in X\) is joined by a path: a continuous function \(\gamma:[0,1]\to X\) with \(\gamma(0)=a\) and \(\gamma(1)=b\).

\[\forall a,b\in X\ \exists\,\gamma:[0,1]\to X\text{ continuous},\ \gamma(0)=a,\ \gamma(1)=b\]

A path from \(a\) to \(b\) is a continuous function \(\gamma:[0,1]\to X\) with \(\gamma(0)=a\) and \(\gamma(1)=b\). In \(\mathbb R\) or \(\mathbb R^n\) the segment \(\gamma(t)=(1-t)a+tb\) is enough, so convex sets are path-connected. Other examples are the circle, spheres and \(\mathbb R^2\) minus a point.

Paths can be reversed with \(t\mapsto\gamma(1-t)\) and concatenated: if \(\alpha\) goes from \(a\) to \(b\) and \(\beta\) from \(b\) to \(c\), the path equal to \(\alpha(2t)\) on the first half and \(\beta(2t-1)\) on the second goes from \(a\) to \(c\). Hence «can be joined» is an equivalence relation, whose classes are the path components.

Every path-connected space is connected: the image of \([0,1]\) under a path is connected, so two points joined by a path never lie in disjoint open sets that split the space. Thus \(\mathbb Z\), \(\mathbb R\setminus\{0\}\) or a two-point discrete space admit no paths between their pieces.

The converse is false in general: the topologist's sine curve, \(\{(x,\sin(1/x)):x>0\}\) together with its closure on the axis, is connected but not path-connected. By contrast, in finite spaces both notions coincide, and so they do in open subsets of \(\mathbb R^n\). A typical mistake is to believe that a connected space can always be traversed by a path.

Think about it

If \(X\) is path-connected and \(f:X\to Y\) is continuous, is \(f(X)\) path-connected?

Yes. Given \(f(a)\) and \(f(b)\), if \(\gamma\) joins \(a\) to \(b\) in \(X\), then \(f\circ\gamma\) joins them in \(f(X)\).

Why is \(\mathbb R\) not homeomorphic to \(\mathbb R^2\) by path-connectedness arguments?

Removing a point disconnects \(\mathbb R\), but \(\mathbb R^2\) minus a point stays path-connected. A homeomorphism would preserve that property.

Exercises 10

References

  • James R. Munkres, Topology, 2nd edition, Pearson, 2000. §24
  • Stephen Willard, General Topology, Dover reprint, Dover Publications, 2004. cap. 6, 27