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Math Help - Connected Subset

  1. #1
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    Connected Subset

    Let A be a connected subset of R^3. Suppose that the points (0,0,1) and (4,3,0) are in A.

    a. Prove that there is a point in A whose second component is 2.
    b. Prove that there is a point in A whose norm is 4.

    a. Let u = (0,0,1) v = (4,3,0) be points in A. Next, I find the symmetric equation of the line l that contains the points u,v.




    Thus,.


    We use the first point u to find the symmetric equation of l.
    Thus, the equation yields:





    It is sufficient to show that there exists a y such that y/3 = 2 that disconnects A into two disjoint sets?


    b. My question is could I just take the norm of the points u = (0,0,1) and v = (4,3,0) which is and that there must exist a a point whose norm is 4 because it is less than ?


    Thank you for your time.
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  2. #2
    Super Member Matt Westwood's Avatar
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    a) My immediate take on this (long time since I've done topology) is as follows.

    Since A is connected, there exists a continuous path from u to v.

    Thus (I expect there's a theorem somewhere that proves this), the projection of this path onto each of the 3 co-ordinates is also continuous.

    That is, the x-coordinate of that path must be continuous from 0 to 4, the y-coordinate from 0 to 3 and the z-coordinate from 1 to 0.

    As the y-coordinate is continuous from 0 to 3, it must pass through 2 somewhere along it.

    b) I'll have to pass on that one as I can't remember what a norm is.
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hello

    Quote Originally Posted by Matt Westwood View Post
    Since A is connected, there exists a continuous path from u to v.
    This is not necessarily true. (it'd be true if A were path-connected)

    Quote Originally Posted by Paperwings View Post
    Let A be a connected subset of R^3. Suppose that the points (0,0,1) and (4,3,0) are in A.

    a. Prove that there is a point in A whose second component is 2. (...)

    a. Let u = (0,0,1) v = (4,3,0) be points in A. Next, I find the symmetric equation of the line l that contains the points u,v.




    Thus,.


    We use the first point u to find the symmetric equation of l.
    Thus, the equation yields:





    It is sufficient to show that there exists a y such that y/3 = 2 that disconnects A into two disjoint sets?
    If you find two disjoint, proper subsets of A which are open in A, yes, A is disconnected... but how can you find two such sets given that A is, by definition, connected ?

    This idea of disconnecting A is however useful if you use contradiction :

    Assume that there is no point in A whose second coordinate is 2. This means that the intersection of A with the plane whose equation is y=2 is the empty set. Now, can you find two disjoint, proper subsets of A which are open in A ? (this would give us a contradiction : A would have to be both connected and disconnected)

    Hint (highlight to read) : *Consider the intersection of A with the half space { (x,y,z) in R^3 | y<2 } and the intersection of A with the half space { (x,y,z) in R^3 | y>2 }.*

    b. Prove that there is a point in A whose norm is 4. (...)

    b. My question is could I just take the norm of the points u = (0,0,1) and v = (4,3,0) which is and that there must exist a a point whose norm is 4 because it is less than ?
    No, the word "because" hides too many things in this sentence. Once again, you can try using contradiction : assume that there is no vector in A whose norm is 4. (the idea is the same as for the first question)
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  4. #4
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    Assume that there is no point in A whose second coordinate is 2. This means that the intersection of A with the plane whose equation is y=2 is the empty set. Now, can you find two disjoint, proper subsets of A which are open in A ? (this would give us a contradiction : A would have to be both connected and disconnected)
    Thanks flyingsquirrel.

    So, assume that for all x,z, . I need to show that the plane
    into two open disjoint, proper sets.

    The first set is (0,0,1) since the y component is 0 < 2 which belongs in the plane { (x,y,z) in R^3 | y<2 }
    The second set is (4,3,0) since the y component is 3 > 2 which belongs in the plane { (x,y,z) in R^3 | y>2 }

    Similarly, this can be done with the norm. There is point in A such that {a, ||a|| = 4}

    The first set is the norm of (0,0,1), which is 1.
    The second set is the norm of (4,3,0) which is 5.

    Which contradicts A being connected.

    I have a good hunch this is correct.
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Paperwings View Post
    (...) I have a good hunch this is correct.
    No, it's not correct.

    So, assume that for all x,z, . I need to show that the plane
    into two open disjoint, proper sets.
    My mistake : I forgot to say that the union of these two subsets has to be A. (the hint I gave remains correct though)

    The first set is (0,0,1) since the y component is 0 < 2 which belongs in the plane { (x,y,z) in R^3 | y<2 }
    No, there can be more than one point in the intersection of A with {(x,y,z) in R^3 such that y>2}. Plus you didn't check that this set is open*, that it is a proper subset of A... I wrote the answer for this question (see attached picture) hoping it'll help you. As the proof for the second question is similar, I let you write it on your own. (and let me know if something is not clear in the answer to the first question)

    (*) which it is not since it's a singleton of R^3 with the Euclidean topology...
    Attached Thumbnails Attached Thumbnails Connected Subset-sans-titre.png  
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by JaneBennet View Post
    It is true in \mathbb{R}^3. For topological spaces in general, connectedness does not imply path-connectedness – but in the case of \mathbb{R}^3 the two concepts happen to be equivalent.
    I searched the web for this and the only thing I found is that "In n dimensional space (...) every open connected set is path connected." (comes from the paragraph titled " Connected plus Locally Path Connected Implies Path Connected " of this page) If this is what you meant, what happens if the set A is not open ? If that's not what you meant, may I ask you to give me a link to a webpage where I could read a proof of the result you're talking about?

    Since A is (path-)connected (see point above) every point on the straight line joining two points in A lies in A.
    Either I'm missing something either you're saying that if a subset of \mathbb{R}^3 is path-connected then it is convex. That seems wrong : for example the union of the planes P=\{ (x,y,z)\in\mathbb{R}^3\,|\, x=0 \} and Q=\{ (x,y,z)\in\mathbb{R}^3\,|\, y=0 \} is path-connected since P and Q are both path connected and P \cap Q \neq \emptyset. However, P\cup Q isn't convex : for A=(0,1,0)\in P and B=(1,0,0)\in Q, the point C=\left( \frac12, \frac12,0 \right) lies in the line segment [AB] but is neither in P nor in Q.
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