Page 1 of 2 12 LastLast
Results 1 to 15 of 29

Math Help - arcwise connected set

  1. #1
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    arcwise connected set

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,642
    Thanks
    1594
    Awards
    1
    What is the question in all of that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Hartlw View Post
    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.

    This seems to be true. What's the "incomprehensible step" in Taylor's (Taylor who?) proof?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    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.

    From Taylor:

    "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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Hartlw View Post
    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.

    From Taylor:

    "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.
    Are you using "X" for intersection?? I know a proof of this theorem, but it's hard to comment on this one with the formatting. Is there any way you could tex it?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Hartlw View Post
    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.
    This ought to be a fairly straightforward result, quite reasonable to leave as a problem in a graduate text, and Taylor's proof seems unnecessarily complicated. (I find that surprising, because Taylor is usually an excellent author. His Introduction to functional analysis is still one of the best, 50+ years after its first publication.)

    Start as Taylor does, by fixing a point x\in S, 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 y\in A; 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 z\in B; there is a ball centred at z and contained in S. If that ball contains a point w\in A then there is a path in S consisting of a straight line from z to w and then from w to x. So z\in A – 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 x\in A. 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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Hartlw View Post
    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.
    The point is that B cannot contain accumulation points of A, because that would give a contradiction. The sets A and B are both open and closed in S. Therefore either A and B are completely separate, with no accumulation points in common (which would mean that S is not connected) or B is empty (and therefore S is arcwise-connected).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Opalg View Post
    The point is that B cannot contain accumulation points of A, because that would give a contradiction. The sets A and B are both open and closed in S. Therefore either A and B are completely separate, with no accumulation points in common (which would mean that S is not connected) or B is empty (and therefore S is arcwise-connected).
    From Buck, Advanced Calculus: "Definition 4. A set in E^n is said to be connected if it is impossible to split S into two disjoint sets A and B, neither one empty, without having one of the sets contain a boundary point of the other."

    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Hartlw View Post
    From Buck, Advanced Calculus: "Definition 4. A set in E^n is said to be connected if it is impossible to split S into two disjoint sets A and B, neither one empty, without having one of the sets contain a boundary point of the other."

    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.
    In this problem they are both open, and the conclusion is that B must therefore be empty.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Opalg View Post
    In this problem they are both open, and the conclusion is that B must therefore be empty.
    You haven't proved B is open because B contains a boundary point of A, with a ball which can have a point in A without creating a contradiction.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Hartlw View Post
    You haven't proved B is open because B contains a boundary point of A, with a ball which can have a point in A without creating a contradiction.
    I don't want to continue saying the same thing much more. But for the last time, B does not contain a boundary point of A. All the boundary points of A lie in A itself, and therefore do not belong to B.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Opalg View Post
    I don't want to continue saying the same thing much more. But for the last time, B does not contain a boundary point of A. All the boundary points of A lie in A itself, and therefore do not belong to B.
    But you proved A was open, so its boundary points have to be in B.

    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.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Hartlw View Post
    But you proved A was open, so its boundary points have to be in B.

    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.
    To get really nasty about this, how do you know that A isn't empty? Let x be a point in S with a spherical neighborhood. How do you know you can connect from x to any point in the spherical neighborhood without assuming what you are trying to prove?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98
    Quote Originally Posted by Hartlw View Post
    To get really nasty about this, how do you know that A isn't empty? Let x be a point in S with a spherical neighborhood. How do you know you can connect from x to any point in the spherical neighborhood without assuming what you are trying to prove?
    Taylor helps us out here:
    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.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Every path-connected metric space is connected
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 13th 2011, 06:31 AM
  2. arcwise connected open set
    Posted in the Differential Geometry Forum
    Replies: 10
    Last Post: January 21st 2011, 11:15 AM
  3. The unit sphere in R^3 is arcwise connected
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 15th 2010, 10:55 PM
  4. Proof that union of two connected non disjoint sets is connected
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 27th 2009, 08:22 AM
  5. Replies: 1
    Last Post: April 18th 2008, 07:19 AM

Search Tags


/mathhelpforum @mathhelpforum