Last modified on 17 April 2014, at 23:07

Topology/Connectedness

Topology
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MotivationEdit

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.

ExamplesEdit

  1. 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 (s-\varepsilon,s+\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 (s-\varepsilon,s+\varepsilon) within [a,b]\setminus X. Either case implies that s is not the supremum.
  2. 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:
    The logo for this Wiki

    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.

DefinitionsEdit

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

Theorems about connectednessEdit

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.

ExercisesEdit

  1. 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?)
  2. Prove that if f:X\to Y is continuous and surjective (not necessarily homeomorphic), and if X is connected, then Y is connected.
  3. 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.
  4. Prove that \mathbb{R} is not homeomorphic to \mathbb{R}^2 (hint: removing a single point from \mathbb{R} makes it disconnected).
  5. Prove that an uncountable set given the countable complement topology is connected (this space is what mathematicians call 'hyperconnected')
  6. 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)


Topology
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