What is the question in all of that?
Taylor proves the following (with one incomprehensible step), and Rosenlicht gives it as a problem, a problem!! Is the answer given? Do bears poop in the woods?
Any connected open subset of E^n is arcwise connected.
Definitions from Rosenlicht:
"A metric space E is connected if the only subsets of E which are both open and closed are E and 0. A subset S of a metric space is a connected subset if the subspace S is connected."
"A metric space E is said to be arcwise connected if, given any p,q eE, there is a continuous function f:[0,1] -> E such that f(0)=p , f(1)=q."
Rudin doesn't touch this.
The question is how do you prove the statement about arcwise connectedness.
Taylor, "General Theory of Functions and Integration"
Taylor gives a long and involved proof, most of which can be followed.
"Theorem: Let S be an open and connected set in R^k. Then, given any two distinct points x,y in S, there exists a polygonal arc lying in S and having x and y as its end points."
"Let A consist of x and all points a in s such that there exists a polygonal arc lying in S with x and a as its end points. We have to prove that y is in A. We shall assume the contrary and deduce a contradiction. Let B=S-A. Evidentally A unequal 0, B unequal 0, S = AUB, AXB = 0."
To be disconnected, AXB must equal zero (just proved) and AXB' = 0 and A'XB = 0. (Prime means the set of all accumulation points). Then
He proves A is open so that AXB'=0 (no accumulation points in A).
That leaves A'XB =0. That's hard. OK, I'll copy the damn thing:
"If we now prove that A'XB ==0, S will be expressed as the union of the separated sets A, B, contrary to the fact that S is connected, and so we shall have completed our proof. Suppose then that A'XB unequal zero, say z eA'XB. Then zeS, and there exists a spherical neighborhood of z lying entirely in S.This neighborhood must contain a point w in A because zeA'. There is then a polygonal arc L in S joining x to w. Let M be the line segment from w to z, (ends included). Now z is not on L because L is a subset of A and zeB."
Now I get stuck. Continuing with Taylor:
"The set LXM is compact, and hence there is a point y of L at minimum distance from z. We then obtain a polygonal arc joining x to z by proceeding from x along L until we come to y, and then following the straight line segment from y to z."
The rest then is easy:
"But this means that zeA, which is a contradiction. Hence we must conclude that A'XB=0, and the proof is complete."
comment: Not only is this an excercise in Rosenlicht, it is also an excercise in Buck, "Advanced Calculus," with of course, no answer. Are these people serious?
What I was really looking for is an understandable proof, or a reference to one. If someone could explain the underlined sentence, that would work and be greatly appreciated.
Start as Taylor does, by fixing a point , and defining A to be the set of points in S that can be reached by a path from x. Let B be the complement of A in S. Then A is open. Reason: suppose that ; since S is open, there is a ball B(y,e) centred at y and contained in S. Then every point in that ball belongs to A, because you can connect it to y by a straight line which lies within the ball, and then from y there is a path to x.
A very similar argument shows that B is open: suppose that ; there is a ball centred at z and contained in S. If that ball contains a point then there is a path in S consisting of a straight line from z to w and then from w to x. So – contradiction. Therefore the ball contains no points in A and hence lies entirely in B, showing that B is open.
Thus S is the disjoint union of the open sets A and B. But if S is connected then one of those sets must be empty. It can't be A, because . Therefore B is empty, which means that S is arcwise-connected.
The basic idea in all of this is that the composition of two paths is a path. If there is a path from p to q, and a path from q to r, then you can stitch them together to get a path from p to r.
Drexel28: X is intersection and e is epsilon (belongs to).
Opalg: Thanks so much for reply and very impeccable and transparent proof. The problem with clarity is that it enables someone to see a possible fault, a problem that dosn't occur with Taylor's proof.
If B is the complement of A in S, then B must contain accumulation points of A. In that case, a ball in B centered on an accumulation point can contain points of A without a contradiction.
So A and B cannot both be open and the argument for B being open fails because a ball exists in B (the boundary point) which can contain a point of A without a contradiction.
From Wolfram Math World:
A connected set is a set which cannot be partitioned into two nonempty subsets such that each subset has no points in common with the set closure of the other.
I suspect that's why Taylor is so obstruse: You have to get past the boundary points.
Two distinct points in R^k determine a unique straight line- the line joining these two points. If x1 and x2 are the points, the parametric equation of the line is: x=(1-t)x1+x2, t in [0,1]. So if x is any point in S and w is a point in its spherical neighborhood, there is a line from x to w and A is not empty.