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 topological space X is connected if it is not disconnected. Examples 1. A closed interval [a,b] is connected. To show this, suppose that it was disconnected. Then there are two nonempty disjoint open sets and whose union is [a,b]. Let X be the set equal to A or B and which does not contain the point b. Let s= sup X, the supremum of 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]\X. If s is within X, then by the definition of supremum there is an open set (s-ε, s+ε) within X. If s is not within X, then is within [a,b]\X, which is also open, and there is an open set (s-ε, s+ε) within [a,b]\X. Either case implies that s is not the supremum. 2. The topological space X=(0,1)\{1/2) is disconnected: A=(0,1/2), B=(1/2,1) (MathWorld) Lemma The following are equivalent for a topological space X: 1) X is connected 2) The only subsets of X that are both open and closed are Φ and X, where Φ denotes the empty set 3) Every continuous function f:X→2disc is constant (Adams and Franzosa, 2008). Definition A function f: X → Y between two topological spaces X and Y is called a homeomorphism if it has the following properties:

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f is a bijection (one-to-one and onto), f is continuous, the inverse function f −1 is continuous

Theorem The continuous image of a connected space is connected Corollary Connectedness is a topological property This also means that if X and Y are homeomorphic spaces and if X is connected, then Y is also connected. 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 image of a path-connected space is path-connected. Proof Let x, y ∈ Y. Then there exists a continuous path g: [a, b] → X such that g(a) = f−1 (x) and g(b) = f−1b). Define h:= f ◦ g. Then h: [a, b] → Y is continuous and h(a) = (f ◦ g)(a) = f(g(a)) = f(f−1(x)) = x and h(b) = (f ◦ g)(b) = f(g(b)) = f(f−1(y)) = y since f is bijective. Hence, h is a path from x to y, so, since our choice of x and y was arbitrary, Y is path connected.

Major Theorem The interval (0, 1) R with its usual topology is connected. 3

Proof Suppose that (0, 1) = A B with A, B disjoint non-empty closed and open subsets. Choose a A and b B with (say) a < b. Then let be the least upper bound of the set C = { [a, b] A }. This least upper bound exists by the standard properties of R. Since C is a closed subset it contains its limit points and so C and hence is in A. Since A is open an -neighborhood lying inside A and so unless = b it would not be an upper bound of C. But = b contradicts the fact that b B = (0, 1) - A. (Sieradski,1992) Theorem Let T be a topological space. Let T′⊆T be a subspace of T. Then V⊆T′ is closed in T′ iff V=T′∩W for some W closed in T. Corollary Let subspace T′ be closed in T. Then V⊆T′ is closed in T′ iff V is closed in T. has

Proof Suppose V⊆T′ is closed in T′. Then T′∖V is open in T′ by definition. So, by definition of subspace topology, T′∖V=T′∩U for some U open in T. Then: V = T′∖(T′∖V) = T′∖(T′∩U) = T′∖U = T′∩(T∖U) Thus T∖U is closed in T. (MathWorld) Relative Complement of Relative Complement from above Set Difference with Intersection is Difference

Definition A and B are separated in a topological space X if each is disjoint from the other's closure.

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The closures themselves do not have to be disjoint from each other; for example, the [0,1) and (1,2] are separated in the R, even though the point 1 belongs to both of their closures. (Sutherland 1975) Theorem Let the real number line R be considered as a topological space. Let S be a subspace of R. Then S is connected if and only if S is an interval of R. That is, the only subspaces of R that are connected are intervals. Proof Suppose S is an interval of R. Suppose further that A∣B is a separation of S. Let a∈A, b∈B, and suppose without loss of generality that aa. Now let A′′=A′∩[a..b′]. Using the same argument as for B′, we have that a′′=sup(A′′) exists, that a′′∈A′′ and also a′′