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Math Help - is the set A={ (x,y)∈R : x+ y =1} connected?

  1. #1
    Junior Member sorv1986's Avatar
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    Cool is the set A={ (x,y)∈R : x+ y =1} connected?

    is the set A={ (x,y)∈R : x+ y =1} connected? how do we prove it?
    is there any general proof? or some quick method by which we can tell any set is connected or not
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  2. #2
    Senior Member Tinyboss's Avatar
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    It's a subset of the plane, so you can visualize it or draw a picture. What does it look like?
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  3. #3
    Junior Member sorv1986's Avatar
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    Cool

    Quote Originally Posted by Tinyboss View Post
    It's a subset of the plane, so you can visualize it or draw a picture. What does it look like?
    looks like a unit circle centred at origin
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  4. #4
    Senior Member Tinyboss's Avatar
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    And is that connected, under the subspace topology inherited from R^2? Here's a hint: a continuous image of a connected set is connected. Can you express the unit circle as the continuous image of some set you know is connected?
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  5. #5
    Junior Member sorv1986's Avatar
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    Quote Originally Posted by Tinyboss View Post
    And is that connected, under the subspace topology inherited from R^2? Here's a hint: a continuous image of a connected set is connected. Can you express the unit circle as the continuous image of some set you know is connected?
    If f : [0,1] -> R2 defined by f(t) = (cost, sint) then the map is continuous and the closed interval [0,1] is connected. So the image of f should be connected.and i think the image of f is exactly the unit circle centerd at origin. is it right?


    but is there any general method for proving or disproving any subset of R2 connected?
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  6. #6
    Senior Member Tinyboss's Avatar
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    I don't know of a general way.
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    Quote Originally Posted by sorv1986 View Post
    If f : [0,1] -> R2 defined by f(t) = (cost, sint) then the map is continuous and the closed interval [0,1] is connected. So the image of f should be connected.and i think the image of f is exactly the unit circle centerd at origin. is it right?


    but is there any general method for proving or disproving any subset of R2 connected?
    Well, yes, but it isn't that easy to implement: a space X is NOT connected iff there is a

    continuous surjection X\rightarrow \{0,1\} , with the second set with the inherited

    topology from the reals (is thus a discrete space).

    Tonio
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