1433 words - 6 pages

MTH 333 Writing Project Abdullah Aurko

MTH 333 Writing Project Topic- Connectedness: A topological property I am working on concept of connectedness of a topological space as the topic for my writing project. I wish to proceed by giving a formal definition of the concept with proof and proving that connectedness is a topological property. This will be followed up with the proofs of several interesting and essential theorems involving connectedness of a topological space and how it enables us to distinguish between different topological spaces.

A topological space X is disconnected if X=A B, where A and B are disjoint, nonempty, open subsets of X. (Roseman, 1999) Definition A ...view middle of the document...

Proof: Let X be connected, with f:X→Y and let f be a homeomorphism. Assume Y= f(X) is disconnected. Then there exists two nonempty disjoint open sets A and B whose union is f(X). As f-1 is continuous, f-1 (A) and f-1 (B) are open in X. As f-1 is a bijection, f-1 (A) and f-1 (B) are disjoint nonempty open sets whose union is X, making X disconnected, a contradiction. Thus, Y = f(X) is connected if X is connected , thus also showing that connectedness is a topological property. (Sieradski,1992)

Path Connected Spaces One of the nice properties of the connected spaces Rn is that we can construct a continuous path between any two points. Specifically, given x, y in Rn the function f:[0,1] → Rn defined by f(t) = (1-t)x+ty is continuous with f(0) = x and f(1) = y. This can be generalized to more general spaces as follows: Definition A topological space X is path connected if for any x,y in X there exists a continuous map f:[0,1] → X such that f(0) = x and f(1) = y. We call such a map a path from x to y. Lemma Any path-connected space X is connected. Proof Suppose that X = U V, where U and V are disjoint open sets. We aim to show that either U or V is empty. We do this by showing that any two points of X are both in U or both in V. Let x,y in X be points with f a path from x to y. Then A = f([0, 1]), the image set of f, is a connected subspace of X. As A = (A U) (A V) is a decomposition of A into a disjoint union of open subsets. Thus one of these sets, say 2

A \ V must be empty. As x, y is in A, we have x, y in U. Thus given any two points x, y is in X, they are both in U or both in V. If z is any third point it must be in the same set as x and y by considering the pairs x,z and y,z. The converse of this above lemma is false. (Sieradski,1992) Example The following space is connected but not path connected.

(Wikipedia)

Figure 1

Define the function f : R R by f(x) = sin(1/x) [Figure 1] if x 0, and f(0) = 0. Let X be the graph of f with its topology as a subspace of R2. No continuous path can "get from the LH side to the RH side. However, any open neighborhood of (0, 0) meets both sides and the space is connected. Path connectedness gives an easy way of verifying that many of our standard examples are connected.

Major Lemma If f: X → Y is a homeomorphism and X is path-connected, then Y is path-connected. That is, the continuous...

Tap into the world’s largest open writing community