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Math Help - closed and convex sets

  1. #1
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    closed and convex sets

    How does one proof the closeness in this case?

    the questions is show that (C)_{\mathbf{v}} is closed and convex if C is.

    (C)_{\mathbf{v}} = {x : x = \mathbf{v} + y for some y in C }

    To prove that (C)_{\mathbf{v}} is convex i did this:

    Let x_{1}, x_{2} be in (C)_{\mathbf{v}}, then

    x_{1} = \mathbf{v} + y_{1} and
    x_{2} = \mathbf{v} + y_{2} , y_{1}, y_{2} in C.

    thus,
    x = \mu x_{1} + (1 - \mu)x_{2}
    x   =\mathbf{v} +  \mu y_{1} + (1 - \mu)y_{2}

    but we know that C is convex so
    \mu y_{1} + (1 - \mu)y_{2} is in C

    thus
    x   =\mathbf{v} +  \mu y_{1} + (1 - \mu)y_{2} is in (C)_{\mathbf{v}}.

    but how does one prove that C_{\mathbf{v} is closed?
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  2. #2
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    Quote Originally Posted by mtmath View Post
    How does one proof the closeness in this case?

    the questions is show that (C)_{\mathbf{v}} is closed and convex if C is.

    (C)_{\mathbf{v}} = {x : x = \mathbf{v} + y for some y in C }

    To prove that (C)_{\mathbf{v}} is convex i did this:

    Let x_{1}, x_{2} be in (C)_{\mathbf{v}}, then

    x_{1} = \mathbf{v} + y_{1} and
    x_{2} = \mathbf{v} + y_{2} , y_{1}, y_{2} in C.

    thus,
    x = \mu x_{1} + (1 - \mu)x_{2}
    x   =\mathbf{v} +  \mu y_{1} + (1 - \mu)y_{2}

    but we know that C is convex so
    \mu y_{1} + (1 - \mu)y_{2} is in C

    thus
    x   =\mathbf{v} +  \mu y_{1} + (1 - \mu)y_{2} is in (C)_{\mathbf{v}}.

    but how does one prove that C_{\mathbf{v} is closed?


    It'd be great if you'd tell us what is C: is this a vector/metric/norm/in general, topological space or what?

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    It'd be great if you'd tell us what is C: is this a vector/metric/norm/in general, topological space or what?

    Tonio
    Actually the whole question reads as:

    Let (C)_{\mathbf{v}} denote the translation of a set C by vector \mathbf{v}, i.e.,

    (C)_{\mathbf{v}} \triangleq \{x: x = \mathbf{v} + y for some y  \in C\}.

    Show that (C)_{\mathbf{v}} is also closed and convex if C is.
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  4. #4
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    Quote Originally Posted by mtmath View Post
    Actually the whole question reads as:

    Let (C)_{\mathbf{v}} denote the translation of a set C by vector \mathbf{v}, i.e.,

    (C)_{\mathbf{v}} \triangleq \{x: x = \mathbf{v} + y for some y  \in C\}.

    Show that (C)_{\mathbf{v}} is also closed and convex if C is.


    I understood the question from the beginning: where are C and v taken from?! What's the question's context?

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    I understood the question from the beginning: where are C and v taken from?! What's the question's context?

    Tonio
    I meant to say this is an exercise after a chapter on projections Onto convex sets. C and \mathbf{v} are not defined on the question.
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  6. #6
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    Quote Originally Posted by mtmath View Post
    I meant to say this is an exercise after a chapter on projections Onto convex sets. C and \mathbf{v} are not defined on the question.

    C must be defined as some kind of topological structure, otherwise I can't see how "closedness" can

    be defined...

    Tonio
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  7. #7
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    Quote Originally Posted by tonio View Post
    C must be defined as some kind of topological structure, otherwise I can't see how "closedness" can

    be defined...

    Tonio
    C is not defined anywhere in the question. So should i take my answer as a complete answer to the question?
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