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Thread: Closure, interior..

  1. #1
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    Closure, interior..

    1. Let $\displaystyle A=\{x\in R^3:x_2\geq 2, x_1^2+x_3^2=1\}$
    Find $\displaystyle A^\circ,\overline{A}$ and explain.
    id say that interior is empty set and closure is A, but if so, how to show that?
    2. Let $\displaystyle A=\{\|x\|_{max}\leq 2,x\in\mathbb{R}^2\}$ and $\displaystyle B=\{x^2-y^2>1,(x,y)\in\mathbb{R}^2\}$. Let $\displaystyle C=A\cap B$. i dont know how exactly its called but: is C partly opened with respect to A, or partly closed?

    definition: Y is partly opened with respect to X if and only if there exists opened set $\displaystyle G\in\mathbb{R}^d$ such that $\displaystyle Y=G\cap X$
    Last edited by Julius; Apr 30th 2010 at 01:02 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Julius View Post
    1. Let $\displaystyle A=\{x\in R^3:x_2\geq 2, x_1^2+x_3^2=1\}$
    Find $\displaystyle A^\circ,\overline{A}$ and explain.
    id say that interior is empty set and closure is A, but if so, how to show that?
    Hint:
    Spoiler:

    Call $\displaystyle \pi_2$ the canonical projection onto the second coordinate space. and $\displaystyle \gamma:\mathbb{R}^3\to\mathbb{R}^2x_1,x_2,x_3)\to (x_1,x_3)$. Clearly the first map is continuous and the second is because each of it's coordinate functions is continuous. So then $\displaystyle A=\pi_1^{-1}\left([2,\infty)\right)\cap\gamma^{-1}\left(\mathbb{S}^1\right)$ (where $\displaystyle \mathbb{S}^1$ is the unit circle). So what?


    2. Let $\displaystyle A=\{\|x\|_{max}\leq 2,x\in\mathbb{R^d}\}$ and $\displaystyle B=\{x^2-y^2>1,(x,y)\in\mathbb{R^d}\}$. Let $\displaystyle C=A\cap B$. i dont know how exactly its called but: is C partly opened with respect to A, or partly closed?

    definition: Y is partly opened with respect to X if and only if there exists opened set $\displaystyle G\in\mathbb{R}^d$ such that $\displaystyle Y=G\cap X$
    What do you think?
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  3. #3
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    So what?
    $\displaystyle A^\circ=X^\circ\cap Y^\circ$, $\displaystyle \gamma^{-1}(\mathbb{S}^{1})^\circ=\emptyset$ so $\displaystyle A^\circ=\emptyset$?

    What do you think?
    i dont know.. B isnt closed, so C wont be closed, $\displaystyle X=\{x^2-y^2\leq 1,(x,y)\in\mathbb{R}^2\}$, $\displaystyle R^2/ X$ is open so C is open?
    Last edited by Julius; Apr 30th 2010 at 01:52 AM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Julius View Post
    $\displaystyle A^\circ=X^\circ\cap Y^\circ$, $\displaystyle \gamma^{-1}(\mathbb{S}^{1})^\circ=\emptyset$ so $\displaystyle A^\circ=\emptyset$?
    Well, first-off it tells you that $\displaystyle A$ is the intersection of the preimages of two closed sets under a continuous map, in other words it's the intersection of two closed sets, in other words it's closed, in other words $\displaystyle \overline{A}=A$

    Why do you think $\displaystyle \left(\gamma^{-1}\left(\mathbb{S}^1\right)\right)^{\circ}=\varnot hing$?


    i dont know.. B isnt closed, so C wont be closed, $\displaystyle X=\{x^2-y^2\leq 1,(x,y)\in\mathbb{R}^2\}$, $\displaystyle R^2/ X$ is open so C is open?
    I don't understand I guess what's the problem. What you're calling "party open" just means that it's open in the subspace topology (or open as a metric subspace...whatever your prefer), in other words it's the intersection of $\displaystyle A$ with an open set. But, once again $\displaystyle \varphi:\mathbb{R}^2\to\mathbb{R}x,y)\mapsto x-y$ is continuous and $\displaystyle B=\varphi^{-1}\left((1,\infty)\right)$...soooo?
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