2.04 Connectedness, path-connectedness
Below the video you will find accompanying notes and some pre-class questions.
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Notes
Connectedness
(00.23) Let X be a topological space. We say that X is
disconnected if there exist open sets U,V⊂X such that
U∩V=∅, U≠∅, V≠∅ and
U∪V=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]=U∪V. We want to prove that one
of them is empty.
(4.06) Define a function F:[0,1]→R by F(x)={0 if x∈U1 if x∈V.
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,y∈X
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,V⊂X be disjoint open sets such that U∪V=X. If they are both
nonempty then we can pick a point x∈U and y∈V. By
path-connectedness, there is a continuous path γ from x
to y. Now γ−1(U) and γ−1(V) are disjoint
open sets in [0,1] whose union is [0,1] and they are both
nonempty because 0∈γ−1(U) and
1∈γ−1(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,y∈X, 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:R→R2 such that F−1 is also
continuous (for more about homeomorphisms, see this section).
(16.51) If there is a homeomorphism
F:R→R2 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
- True or false? Every indiscrete space is connected.
- True or false? Every indiscrete space is path-connected.
- True or false? A subspace of a connected space is connected.
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