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Math Help - Analysis help

  1. #1
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    Analysis help

    Hello,
    I need help with the following problem:
    Let S = {(x,y) in R2| xy>1}. Show that S is open.

    This is what I've done so far:
    1. Let u be a point in S.
    2. So, u = (u1,u2) for some u1, u2 in R and u1*u2>1
    3. I chose my epsilon to be equal to (u1*u2 - 1) / ||u||, which is > 0.
    4. Now I need to show that the open ball B(u, epsilon) is contained in S.
    So, let v = (v1,v2) be a point in the ball B(u, epsilon).
    That means dist(u,v) < epsilon.
    In order for v to be in S, we need to show v1*v2 > 1. However, I'm having difficulty showing this. I tried to use the inequality dist(u,v) < epsilon, but I didn't get anything about v1*v2. Any suggestions on what should I do? Appreciate anyone's help and thanks for reading my post
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  2. #2
    Junior Member Dark Sun's Avatar
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    Could you use \widetilde S=\{(x,y)|xy\leq 1\} is closed, since clearly the open ball centered at (1,1) contains elements outside of the complement of S for all epsilon. Since the complement of S is closed, then S itself must be open. (Specifically, (1+\epsilon ,1+\epsilon )\notin \widetilde S)
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  3. #3
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    Quote Originally Posted by anlys View Post
    Hello,
    I need help with the following problem:
    Let S = {(x,y) in R2| xy>1}. Show that S is open.

    This is what I've done so far:
    1. Let u be a point in S.
    2. So, u = (u1,u2) for some u1, u2 in R and u1*u2>1
    3. I chose my epsilon to be equal to (u1*u2 - 1) / ||u||, which is > 0.
    4. Now I need to show that the open ball B(u, epsilon) is contained in S.
    So, let v = (v1,v2) be a point in the ball B(u, epsilon).
    That means dist(u,v) < epsilon.
    In order for v to be in S, we need to show v1*v2 > 1. However, I'm having difficulty showing this. I tried to use the inequality dist(u,v) < epsilon, but I didn't get anything about v1*v2. Any suggestions on what should I do? Appreciate anyone's help and thanks for reading my post
    The function f(x,y) = xy is a continuous map from R^2 to R. So

    \{(x,y): xy > 1\} = f^{-1}((1, \infty))

    is the inverse image of an open set, hence open.
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  4. #4
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    Quote Originally Posted by awkward View Post
    The function f(x,y) = xy is a continuous map from R^2 to R. So

    \{(x,y): xy > 1\} = f^{-1}((1, \infty))

    is the inverse image of an open set, hence open.
    Thanks Dark Sun and Awkward for replying,
    Your explanation makes sense to me. However, the problem is, my prof hasn't cover closed sets yet. It will on the next section, so I'm not allowed to use other parts that has not been covered yet. What I know is the definition of open sets, and its property. Nothing about the inverse function or closed sets was mentioned in the section. That's why I'm trying to follow the procedure directly from the definition. Is there any way we can do this by direct proof from the definition?

    Thanks...
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