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Thread: Open set

  1. #1
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    Open set

    Hi! I have problems with this exercises

    Let $(\mathbb{R^2},d)$ where d is Euclidean metric. Let $A= \{ (x,y) | -1<x<1 , -1<y<1\} $. Show that $A$ is an open set

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  2. #2
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    Re: Open set

    Quote Originally Posted by cristianoceli View Post
    Let $(\mathbb{R^2},d)$ where d is Euclidean metric. Let $A= \{ (x,y) | -1<x<1 , -1<y<1\} $.
    Show that $A$ is an open set.
    To show that a set is open you must show that $\forall (x,y)\in A$ then $\exists\mathscr{B}_r(x,y)\subset A$
    So if $(x,y)\in A$ by definition $ -1<x<1~\&~-1<y<1$
    Now suppose that $\delta=\frac{1}{2}\min\{x+1,1-x,y+1,1-y\}$
    you must show that $\mathscr{B}_{\delta}(x,y)\subset A$
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  3. #3
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    Re: Open set

    Quote Originally Posted by Plato View Post
    To show that a set is open you must show that $\forall (x,y)\in A$ then $\exists\mathscr{B}_r(x,y)\subset A$
    So if $(x,y)\in A$ by definition $ -1<x<1~\&~-1<y<1$
    Now suppose that $\delta=\frac{1}{2}\min\{x+1,1-x,y+1,1-y\}$
    you must show that $\mathscr{B}_{\delta}(x,y)\subset A$
    in case it's not obvious, (and it's not)

    $\mathscr{B}_r(x,y) = \{(u,v) \ni d[(x,y),(u,v)]< r\}$

    i.e. a open disk centered at $(x,y)$ of radius $r$
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  4. #4
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    Re: Open set

    Quote Originally Posted by Plato View Post
    To show that a set is open you must show that $\forall (x,y)\in A$ then $\exists\mathscr{B}_r(x,y)\subset A$
    So if $(x,y)\in A$ by definition $ -1<x<1~\&~-1<y<1$
    Now suppose that $\delta=\frac{1}{2}\min\{x+1,1-x,y+1,1-y\}$
    you must show that $\mathscr{B}_{\delta}(x,y)\subset A$
    Why do you choose $\delta=\frac{1}{2}\min\{x+1,1-x,y+1,1-y\}$ ?
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  5. #5
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    Re: Open set

    x+1 is the distance from the point (x, y) to the boundary x= -1. 1- x is the distance from the point (x, y) to the line x= 1, 1- y is the distance from the point (x, y) to the line y= 1, and 1- y is the distance from the point (x, y) to the line y= -1. Take the smallest of those and divide by two to get a radius smaller than any of those distances so the circle around (x, y) is inside the set. Of course, you could multiply by any number less than 1, not just 1/2.
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  6. #6
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    Re: Open set

    Ok thanks!!!
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