S. D. V. Stephens

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

N10 Connected Spaces

Definition N10.1 (Connected). A topological space is connected if it cannot be separated into a disjoint union with disjoint nonempty open sets.

Two separate hatched blobs U and V with dashed boundaries, illustrating a space that is not connected.

Connected the only subsets that are open and closed at the same time are and .

A separation: , , with “super disjoint”. This means , so is closed; and , so is closed. (Additionally , .)

Proposition N10.2. A subset , with the subspace topology of the usual topology of , is connected iff it has “no gaps”: So the connected subspaces of are the intervals of all kinds (open/closed/half-open, finite/semi-infinite/all of ).

Proof. If there exist with and , then where is open in and is open in . Thus is not connected (find two opens and split it).

Conversely, assume with implies , and assume is not connected, so , , , , with open in . We know .

A real line with an interval U containing x and an adjacent interval V containing y, and the segment [x,y] drawn inside E.

Assume . We want the largest thing in the interval still in : let . Then , so or .

First assume . Since is open in , there is such that , the -ball around in . But then there is no element of between and , so , a contradiction.

So , hence . is open in , so there is such that . This contains elements (any , small), which lie in . This contradicts . ∎

Theorem N10.3. If is a continuous map and is connected, then the subspace is connected.

Proof. Consider as a map . Assume otherwise. If is a separation of , then is a separation of . (Here , with open, and .) ∎

IVT (the intermediate value theorem).

Note (Fact). connected for the subspace topology connected.

An annulus split into a left piece A and a right piece B that overlap in two separate red patches at top and bottom: the union is connected but the intersection is not.

Here is connected.

Theorem N10.4. Let , , be connected subspaces which all contain some given point , so . Then is connected.

Three overlapping ellipses, all of which contain a common point p.

Proof. Look at the lecture notes for the proof. ∎

Theorem N10.5. The product of two connected spaces is itself connected.

Proof. Fix . For all , let is homeomorphic to , therefore connected; is homeomorphic to , therefore connected. Both contain , so is connected.

The square X x Y with a horizontal line X x {y0} and a vertical line {x} x Y crossing at (x, y0); their union A_x passes through (x0, y0).

Then where the are connected and all contain . So is connected. ∎

Theorem N10.6. If all are connected, then with the product topology is connected.