2.04 Connectedness, path-connectedness

Below the video you will find accompanying notes and some pre-class questions.

Notes

Connectedness

(00.23) Let X be a topological space. We say that X is disconnected if there exist open sets U,VX such that UV=, U, V and UV=X. We say that X is connected if it is not disconnected.

(2.09) The interval [0,1], equipped with the subspace topology inherited from R, is connected.
(2.54) This is essentially the intermediate value theorem. Suppose there are open sets U,V[0,1] which are disjoint and such that [0,1]=UV. We want to prove that one of them is empty.

(4.06) Define a function F:[0,1]R by F(x)={0 if xU1 if xV.

This function is well-defined because every point is in either U or V but not both. It is continuous because the preimage of any open set is one of U, V, [0,1] or (depending on whether it contains 0, 1, both or neither) which are all open.

(6.18) If F takes on both values 0 and 1 then, by the intermediate value theorem, it takes on all the intermediate values. This function doesn't: it only takes on the values 0 and 1. That means F does not take on both values, hence either U= or V=.

Path-connectedness

(8.08) We can use the fact that [0,1] is connected to prove that lots of other spaces are connected:

A space X is path-connected if for all points x,yX there exists a path from x to y, that is a continuous map γ:[0,1]X such that γ(0)=x and γ(1)=y.

(9.16) A path-connected space is connected. (The converse fails.)
(9.57) Let X be a path-connected space and let U,VX be disjoint open sets such that UV=X. If they are both nonempty then we can pick a point xU and yV. By path-connectedness, there is a continuous path γ from x to y. Now γ1(U) and γ1(V) are disjoint open sets in [0,1] whose union is [0,1] and they are both nonempty because 0γ1(U) and 1γ1(V). This contradicts the previous theorem, so either U= or V=, and we deduce that X is connected.

Examples

(13.00) The space X=R2{(0,0)} is path-connected. Given x,yX, if the straight line ¯xy misses (0,0) then it gives a path in X connecting x and y. Otherwise, first follow a semicircle centred at (0,0) passing through x, then follow a straight path to y. Therefore any two points are connected by a path.

(14.56) By contrast, R{0}=(,0)(0,) is a disjoint union of nonempty open sets, so is disconnected.

The spaces R and R2 are not homeomorphic, in other words there is no continuous bijection F:RR2 such that F1 is also continuous (for more about homeomorphisms, see this section).
(16.51) If there is a homeomorphism F:RR2 then F|R{0}:R{0}R2{F(0)} is also a homomorphism. We may as well take F(0)=(0,0) (by composing F with a translation, for example). But we know that R{0} and R2{(0,0)} are not homeomorphic: one is connected, one is disconnected.

Pre-class questions

  1. True or false? Every indiscrete space is connected.
  2. True or false? Every indiscrete space is path-connected.
  3. True or false? A subspace of a connected space is connected.

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