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Thread: Set A open, but f(A) not open

  1. #1
    Member thaopanda's Avatar
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    Set A open, but f(A) not open

    Let $\displaystyle f : R^2 \rightarrow R$ given by $\displaystyle f(x,y) := x^2$. Find a set A open in $\displaystyle R^2$ such that f(A) in not open in R.

    I don't get how to do this...
    help..?
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  2. #2
    Moo
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    Hello,

    Is it really $\displaystyle x^2$ ?

    If so, you can just take $\displaystyle A=(-a,a) \times \mathcal{O}$, where $\displaystyle \mathcal{O}$ is any open set in $\displaystyle \mathbb{R}$ and $\displaystyle a>0$

    Then $\displaystyle f(A)=[0,a^2)$ which is not an open set.
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  3. #3
    Member thaopanda's Avatar
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    is that really all there is to it? I just assume all my answers are supposed to be super long, 'cause so far, most of them have been taken up half a page... hehe now I feel silly

    Thank you very much!
    Nicole
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Is it really $\displaystyle x^2$ ?

    If so, you can just take $\displaystyle A=(-a,a) \times \mathcal{O}$, where $\displaystyle \mathcal{O}$ is any open set in $\displaystyle \mathbb{R}$ and $\displaystyle a>0$

    Then $\displaystyle f(A)=[0,a^2)$ which is not an open set.
    Hey Moo I'm curious. Why did you chose $\displaystyle A=(-a,a) \times \mathcal{O}$ and not $\displaystyle A=(-a,a)$? I don't really know what your "$\displaystyle \times$" product means. I understand it as "multiplying" two sets to get another one... but I don't know if it makes sense.
    Wouldn't $\displaystyle A=(-a,a)$ be sufficient?
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  5. #5
    Moo
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    Quote Originally Posted by thaopanda View Post
    is that really all there is to it? I just assume all my answers are supposed to be super long, 'cause so far, most of them have been taken up half a page... hehe now I feel silly

    Thank you very much!
    Nicole
    I guess that's all. Unless I made a mistake in the definition of an open set in $\displaystyle \mathbb{R}^2$ ^^'

    Quote Originally Posted by arbolis View Post
    Hey Moo I'm curious. Why did you chose $\displaystyle A=(-a,a) \times \mathcal{O}$ and not $\displaystyle A=(-a,a)$? I don't really know what your "$\displaystyle \times$" product means. I understand it as "multiplying" two sets to get another one... but I don't know if it makes sense.
    Wouldn't $\displaystyle A=(-a,a)$ be sufficient?
    Because A has to be an open set in $\displaystyle \mathbb{R}^{\color{red}2}$.
    $\displaystyle \times$ stands for the cartesian product. Because we're having two values, one in $\displaystyle \mathbb{R}$, the other one in $\displaystyle \mathbb{R}$
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Moo View Post

    Because A has to be an open set in $\displaystyle \mathbb{R}^{\color{red}2}$.
    $\displaystyle \times$ stands for the cartesian product. Because we're having two values, one in $\displaystyle \mathbb{R}$, the other one in $\displaystyle \mathbb{R}$
    Thanks, I missed this part!
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