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Math Help - closed and bounded sets

  1. #1
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    closed and bounded sets

    working in (Real)^2

    {(x, 0) : sin x ≥ 0}

    I Know this set is closed but is unbouned. But i am finding it different to show this.

    {(x^3y - z^2; xyz + y^3z) : 0 =< x, y,z =< 1).}

    I finding it difficult to show this is closed.

    Thanks for helping
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  2. #2
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    Quote Originally Posted by nerdo View Post
    working in (Real)^2

    {(x, 0) : sin x ≥ 0}

    I Know this set is closed but is unbouned. But i am finding it different to show this.


    Did you mean "difficult" instead of "different"? Let \displaystyle{X:=\{(x,0)\in\mathbb{R}^2\;;\;\sin x\geq 0\}=\bigcup\limits_{n\in\mathbb{Z}}[2n\pi,\,(2n+1)\pi]} .

    Now, let x_0\notin X\Longrightarrow \sin x_0<0\Longrightarrow \exists\epsilon > 0\,s.t.\,\,\sin x<0\,\,\forall\,x\in(x_0-\epsilon,\,x_0+\epsilon) (why??) , and

    thus the complement of X is an open set (why?)



    {(x^3y - z^2; xyz + y^3z) : 0 =< x, y,z =< 1).}


    Try to do something similar to the above

    Tonio



    I finding it difficult to show this is closed.

    Thanks for helping
    .
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  3. #3
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    Quote Originally Posted by tonio View Post
    .
    How does one show the second set is closed since it has two functions.

    The first one is closed since the complment being open means set is closed

    Thanks for helping me
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