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Thread: Open Set

  1. #1
    Super Member Showcase_22's Avatar
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    Open Set

    Show that $\displaystyle U=\{ (x,y,z) \in \mathbb{R}^3 \ : \ \frac{e^{x+y^2-z}-1}{x^2+y^2-z^3}>7 \}$ is open in $\displaystyle \mathbb{R}^3$.
    $\displaystyle U$ can be rearranged to get:

    $\displaystyle U=\{(x,y,z) \in \mathbb{R}^3 \ : \ e^{x^2+y^2-z}-1-7x^2-7y^2+7z^3>0 \}$

    The definition of an open ball is $\displaystyle (\forall u \in U)(\exists \varepsilon_w>0)[B_{\varepsilon_w}(u,d) \subset U]$ where $\displaystyle w \in U$.

    Define $\displaystyle \varepsilon_w:=\frac{1}{2} \left(e^{x+y^2-z}-1-7x^2-7y^2+7z^3 \right)$

    The fact that it is halved means that any point in any ball around a point $\displaystyle u$ will be greater than 0 and will therefore be in $\displaystyle U$. This ball is open.

    Since this is valid $\displaystyle \forall u \in U$, $\displaystyle U$ is expressed as a union of open balls ie. $\displaystyle \cup_{w \in U} B_{\varepsilon_w}(u,d)=U $

    Hence $\displaystyle U$ is open.

    Is this right? I'm wondering if I have made any syntax errors!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    $\displaystyle U$ can be rearranged to get:

    $\displaystyle U=\{(x,y,z) \in \mathbb{R}^3 \ : \ e^{x^2+y^2-z}-1-7x^2-7y^2+7z^3>0 \}$

    The definition of an open ball is $\displaystyle (\forall u \in U)(\exists \varepsilon_w>0)[B_{\varepsilon_w}(u,d) \subset U]$ where $\displaystyle w \in U$.

    Define $\displaystyle \varepsilon_w:=\frac{1}{2} \left(e^{x+y^2-z}-1-7x^2-7y^2+7z^3 \right)$

    The fact that it is halved means that any point in any ball around a point $\displaystyle u$ will be greater than 0 and will therefore be in $\displaystyle U$. This ball is open.

    Since this is valid $\displaystyle \forall u \in U$, $\displaystyle U$ is expressed as a union of open balls ie. $\displaystyle \cup_{w \in U} B_{\varepsilon_w}(u,d)=U $

    Hence $\displaystyle U$ is open.

    Is this right? I'm wondering if I have made any syntax errors!
    In a metric space $\displaystyle \left(X,d\right)$ as et $\displaystyle O$ is open if and only if for every $\displaystyle x\in O$ there exists a $\displaystyle \delta>0$ such that $\displaystyle B_{\delta}(x)\subseteq O$. Now, have you done that?

    I am honestly not even sure what you are doing here. Are we working with the usual metric?

    For your definition of an open ball, what is $\displaystyle \varepsilon_w$, I don't understand the $\displaystyle w$ part?
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  3. #3
    Super Member Showcase_22's Avatar
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    I'm trying to show that my set is a union of open balls so it's open. I'm not sure if it's the easiest way of doing it, but I think it might work.

    I'm afraid that my question does not say. Is it possible to prove it for every metric? I not, it probably means Euclidean.

    I was thinking that for every point (x,y,z) $\displaystyle e^{x+y^2-z}-1-7x^2-7y^2+7z^3
    $ will produce a single number. If I create a ball of radius $\displaystyle

    \delta=\frac{1}{2} \left(e^{x+y^2-z}-1-7x^2-7y^2+7z^3 \right)
    $ and centre (x,y,z), (the $\displaystyle B_{\delta}(x,y,z)$ that you wrote) then this will be an open ball that is always a subset of U.

    Is there something wrong with my idea?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I'm trying to show that my set is a union of open balls so it's open. I'm not sure if it's the easiest way of doing it, but I think it might work.

    I'm afraid that my question does not say. Is it possible to prove it for every metric? I not, it probably means Euclidean.

    I was thinking that for every point (x,y,z) $\displaystyle e^{x+y^2-z}-1-7x^2-7y^2+7z^3
    $ will produce a single number. If I create a ball of radius $\displaystyle

    \delta=\frac{1}{2} \left(e^{x+y^2-z}-1-7x^2-7y^2+7z^3 \right)
    $ and centre (x,y,z), (the $\displaystyle B_{\delta}(x,y,z)$ that you wrote) then this will be an open ball that is always a subset of U.

    Is there something wrong with my idea?
    What you are saying is analogous to this. Let $\displaystyle O=\left\{x\in\mathbb{R}:x^3>0\right\}$. To see that this is open we merely note that choosing $\displaystyle \frac{x^3}{2}=\delta$ guarantees that $\displaystyle B_{\delta}(x)\subseteq O$. Take $\displaystyle x=2\implies \delta=4$. Then, $\displaystyle B_\delta\left(2\right)\nsubseteq O$
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  5. #5
    Super Member Showcase_22's Avatar
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    ah, so I was wrong!

    In your example of $\displaystyle

    O=\left\{x\in\mathbb{R}:x^3>0\right\}
    $, does $\displaystyle \delta=\frac{x}{2}$ work in the ball $\displaystyle B_{\delta}(x)$?

    It works in the case $\displaystyle x=2$. The set is bounded below by 0 so if $\displaystyle a \in O$, then $\displaystyle \frac{a}{2}$ is always in the set since it's the same distance from a as it is 0.

    So extending this idea to the set $\displaystyle U$ gives $\displaystyle \delta= \frac{\frac{e^{x+y^2-z}-1}{x^2+y^2-z^3}-7}{2}$.

    I think i'm starting to get it!
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