Topology/Connectedness

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[edit] Motivation

To best describe what is a connected space, we shall describe first what is a disconnected space. A disconnected space is a space that can be separated into two disjoint groups, or more formally:

A space $(X,\mathcal{T})$ is said to be disconnected iff a pair of disjoint, non-empty open subsets $X 1, X 2$ exists, such that $X = X_1 \cup X_2$ .

A space $X$ that is not disconnected is said to be a connected space.

[edit] Examples

A closed interval $[a, b]$ is connected. To show this, suppose that it was disconnected. Then there are two nonempty disjoint open sets $A$ and $B$ whose union is $[a, b]$ . Let $X$ be the the set equal to $A$ or $B$ and which does not contain $b$ . Let $s=\sup X$ . Since X does not contain b, s must be within the interval [a,b] and thus must be within either X or $[a,b]\setminus X$ . If $s$ is within $X$ , then there is an open set $(c-\varepsilon,c+\varepsilon)$ within $X$ . If $s$ is not within $X$ , then $s$ is within $[a,b]\setminus X$ , which is also open, and there is an open set $(c-\varepsilon,c+\varepsilon)$ within $[a,b]\setminus X$ . Either case implies that $s$ is not the supremum.
The topological space $X = (0,1)\setminus\{{\frac{1}{2} } \}$ is disconnected: $A = (0,\frac{1}{2} ), B = (\frac{1}{2}, 1)$
A picture to illustrate:

As you can see, the definition of a connected space is quite intuitive; when the space cannot be separated into (at least) two distinct subspaces.

[edit] Definitions

Definition 1.1

A subset $U$ of a topological space $X$ is said to be clopen if it is both closed and open.

Definition 1.2

A topological space X is said to be totally disconnected if every subset of X having more than one point is disconnected under the subspace topology

[edit] Theorems about connectedness

If $X$ and $Y$ are homeomorphic spaces and if $X$ is connected, then $Y$ is also connected.

Proof:
Let $X$ be connected, and let $f$ be a homeomorphism. Assume, if possible, $f (X)$ is disconnected. Then there exists two nonempty disjoint sets $A$ and $B$ whose union is $f (X)$ . As $f - 1$ is continuous, $f - 1 (A)$ and $f - 1 (B)$ are open. As $f - 1$ is a bijection, they are disjoint sets whose union is $X$ , contradicting the fact that $X$ is connected. Thus, $Y = f (X)$ is connected.
Note: this shows that connectedness is a topological property.

If two connected sets have a nonempty intersection, then their union is connected.

Proof:
Let $A$ and $B$ be two non-disjoint, connected sets. LEt $X$ and $Y$ be non-empty open sets such that $X\cup Y=A\cup B$ . Let $a_0\in A$ .
Without loss of generality, assume $a_0\in X$ .

As $A$ is connected, $a\in X\forall a\in A$ ...(1).

As $Y$ is non-empty, $\exists b\in B$ such that $b\in Y$ .

Hence, similarly, $b\in Y\forall b\in B$ ...(2)
Now, consider $c\in A\cap B$ . From (1) and (2), $c\in X\cap Y$ , and hence $X\cap Y\neq\emptyset$ . As $X,Y\in\mathcal{T}$ are arbitrary, $A\cup B$ is connected.

If two topological spaces are connected, then their product space is also connected.

Proof:
Let X₁ and X₂ be two connected spaces. Suppose that there are two nonempty open disjoint sets A and B whose union is X₁×X₂. If for every x∈X, {x}×X₂ is either completely within A or within B, then π₁(A) and π₁(B) are also open, and are thus disjoint and nonempty, whose union is X₁, contradicting the fact that X₁ is connected. Thus, there is an x∈X such that {x}×X₂ contains elements of both A and B. Then π₂(A∩{(x,y)}) and π₂(B∩{(x,y)}), where y is any element of X₂, are nonempty disjoint sets whose union is X₂, and which are are a union of open sets in {(x,y)} (by the definition of product topology), and are thus open. This implies that X₂ is disconnected, a contradiction. Thus, X₁×X₂ is connected.

[edit] Exercises

Show that a topological space $X$ is disconnected if and only if it has clopen sets other than $\emptyset$ and $X$ (Hint: Why is $X 1$ clopen?)
Prove that if $f:X\to Y$ is continuous (not homeomorphic), and if $X$ is connected, then $Y$ is connected.
Prove the Intermediate Value Theorem: if $f:[a,b]\to \mathbb{R}$ is continuous, then for any $y$ between $f (a)$ and $f (b)$ , there exists a $c\in [a,b]$ such that $f (c) = y$ .
Prove that $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^2$ (hint: removing a single point from $\mathbb{R}$ makes it disconnected).
Prove that an uncounable set given the countable complement topology is connected (this space is what mathematicians call 'hyperconnected')
a)Prove that the discrete topology on a set X is totally disconnected.

b) Does the converse of a) hold (Hint: Even if the subspace topology on a subset of X is the discrete topology, this need not imply that the set has the discrete topology)