Is the set ${ (x,y) in mathbb{R}^2 : xy=1 }$ open or closed in $mathbb{R}^2$












3















Determine whether the following sets are open or closed in
$mathbb{R}^2$ endowed with the eucledian metric



$1. { (x,y) in mathbb{R}^2 : xy=1 }$



$2. { (x,y) in mathbb{R}^2 : xyle1 }$



$3. { (x,y) in mathbb{R}^2 : xy<1 }$




The first two are not open because there is no ball that is contained in those sets and that doesn't contain an element from their complements. I'm having trouble expressing their complements as some type of union of open sets so I can prove that they are closed. The third one is a bit more interesting since the answer in my book says it's an open set. It's odd to visualize how there exists an open ball that is contained in $xy<1$ but doesn't contain any other points.



I would like to prove each of these a bit more rigorously and I have limited knowledge of topology/metric spaces. Hoping to prove it using only definition of open sets, (no sequences, continuity etc). Any hints or full answers are appreciated.










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  • 3




    Hint: $f(x,y)=xy$ is a continuous function, so use the criterion of open preimage by a continuous function.
    – GNUSupporter 8964民主女神 地下教會
    Nov 30 at 16:47










  • Try to visualize what the sets actually are (e.g. draw them on a piece of paper). That will make it somewhat easier to reason about their properties. For instance, on a simple drawing of set 2 one can see that your reasoning is flawed: the open ball centered at the origin with radius $1$ is contained in that set, among many other balls.
    – Arthur
    Nov 30 at 16:48












  • Well I know the first one is basically the function $frac{1}{x}$, the second one is the same function + bunch of other points but I can't see any patterns here. Also as I mentioned no continuity or sequences please.
    – DreaDk
    Nov 30 at 16:56
















3















Determine whether the following sets are open or closed in
$mathbb{R}^2$ endowed with the eucledian metric



$1. { (x,y) in mathbb{R}^2 : xy=1 }$



$2. { (x,y) in mathbb{R}^2 : xyle1 }$



$3. { (x,y) in mathbb{R}^2 : xy<1 }$




The first two are not open because there is no ball that is contained in those sets and that doesn't contain an element from their complements. I'm having trouble expressing their complements as some type of union of open sets so I can prove that they are closed. The third one is a bit more interesting since the answer in my book says it's an open set. It's odd to visualize how there exists an open ball that is contained in $xy<1$ but doesn't contain any other points.



I would like to prove each of these a bit more rigorously and I have limited knowledge of topology/metric spaces. Hoping to prove it using only definition of open sets, (no sequences, continuity etc). Any hints or full answers are appreciated.










share|cite|improve this question




















  • 3




    Hint: $f(x,y)=xy$ is a continuous function, so use the criterion of open preimage by a continuous function.
    – GNUSupporter 8964民主女神 地下教會
    Nov 30 at 16:47










  • Try to visualize what the sets actually are (e.g. draw them on a piece of paper). That will make it somewhat easier to reason about their properties. For instance, on a simple drawing of set 2 one can see that your reasoning is flawed: the open ball centered at the origin with radius $1$ is contained in that set, among many other balls.
    – Arthur
    Nov 30 at 16:48












  • Well I know the first one is basically the function $frac{1}{x}$, the second one is the same function + bunch of other points but I can't see any patterns here. Also as I mentioned no continuity or sequences please.
    – DreaDk
    Nov 30 at 16:56














3












3








3








Determine whether the following sets are open or closed in
$mathbb{R}^2$ endowed with the eucledian metric



$1. { (x,y) in mathbb{R}^2 : xy=1 }$



$2. { (x,y) in mathbb{R}^2 : xyle1 }$



$3. { (x,y) in mathbb{R}^2 : xy<1 }$




The first two are not open because there is no ball that is contained in those sets and that doesn't contain an element from their complements. I'm having trouble expressing their complements as some type of union of open sets so I can prove that they are closed. The third one is a bit more interesting since the answer in my book says it's an open set. It's odd to visualize how there exists an open ball that is contained in $xy<1$ but doesn't contain any other points.



I would like to prove each of these a bit more rigorously and I have limited knowledge of topology/metric spaces. Hoping to prove it using only definition of open sets, (no sequences, continuity etc). Any hints or full answers are appreciated.










share|cite|improve this question
















Determine whether the following sets are open or closed in
$mathbb{R}^2$ endowed with the eucledian metric



$1. { (x,y) in mathbb{R}^2 : xy=1 }$



$2. { (x,y) in mathbb{R}^2 : xyle1 }$



$3. { (x,y) in mathbb{R}^2 : xy<1 }$




The first two are not open because there is no ball that is contained in those sets and that doesn't contain an element from their complements. I'm having trouble expressing their complements as some type of union of open sets so I can prove that they are closed. The third one is a bit more interesting since the answer in my book says it's an open set. It's odd to visualize how there exists an open ball that is contained in $xy<1$ but doesn't contain any other points.



I would like to prove each of these a bit more rigorously and I have limited knowledge of topology/metric spaces. Hoping to prove it using only definition of open sets, (no sequences, continuity etc). Any hints or full answers are appreciated.







general-topology metric-spaces






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edited Nov 30 at 17:05









mechanodroid

25.4k62245




25.4k62245










asked Nov 30 at 16:45









DreaDk

634318




634318








  • 3




    Hint: $f(x,y)=xy$ is a continuous function, so use the criterion of open preimage by a continuous function.
    – GNUSupporter 8964民主女神 地下教會
    Nov 30 at 16:47










  • Try to visualize what the sets actually are (e.g. draw them on a piece of paper). That will make it somewhat easier to reason about their properties. For instance, on a simple drawing of set 2 one can see that your reasoning is flawed: the open ball centered at the origin with radius $1$ is contained in that set, among many other balls.
    – Arthur
    Nov 30 at 16:48












  • Well I know the first one is basically the function $frac{1}{x}$, the second one is the same function + bunch of other points but I can't see any patterns here. Also as I mentioned no continuity or sequences please.
    – DreaDk
    Nov 30 at 16:56














  • 3




    Hint: $f(x,y)=xy$ is a continuous function, so use the criterion of open preimage by a continuous function.
    – GNUSupporter 8964民主女神 地下教會
    Nov 30 at 16:47










  • Try to visualize what the sets actually are (e.g. draw them on a piece of paper). That will make it somewhat easier to reason about their properties. For instance, on a simple drawing of set 2 one can see that your reasoning is flawed: the open ball centered at the origin with radius $1$ is contained in that set, among many other balls.
    – Arthur
    Nov 30 at 16:48












  • Well I know the first one is basically the function $frac{1}{x}$, the second one is the same function + bunch of other points but I can't see any patterns here. Also as I mentioned no continuity or sequences please.
    – DreaDk
    Nov 30 at 16:56








3




3




Hint: $f(x,y)=xy$ is a continuous function, so use the criterion of open preimage by a continuous function.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 16:47




Hint: $f(x,y)=xy$ is a continuous function, so use the criterion of open preimage by a continuous function.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 16:47












Try to visualize what the sets actually are (e.g. draw them on a piece of paper). That will make it somewhat easier to reason about their properties. For instance, on a simple drawing of set 2 one can see that your reasoning is flawed: the open ball centered at the origin with radius $1$ is contained in that set, among many other balls.
– Arthur
Nov 30 at 16:48






Try to visualize what the sets actually are (e.g. draw them on a piece of paper). That will make it somewhat easier to reason about their properties. For instance, on a simple drawing of set 2 one can see that your reasoning is flawed: the open ball centered at the origin with radius $1$ is contained in that set, among many other balls.
– Arthur
Nov 30 at 16:48














Well I know the first one is basically the function $frac{1}{x}$, the second one is the same function + bunch of other points but I can't see any patterns here. Also as I mentioned no continuity or sequences please.
– DreaDk
Nov 30 at 16:56




Well I know the first one is basically the function $frac{1}{x}$, the second one is the same function + bunch of other points but I can't see any patterns here. Also as I mentioned no continuity or sequences please.
– DreaDk
Nov 30 at 16:56










4 Answers
4






active

oldest

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2














Consider the complement of the second set ${(x,y) : xy > 1}$. We will show that ${(x,y) : xy > 1}$ is open.



Let $(x_0,y_0) in {(x,y) : xy > 1}$. Then $x_0y_0 > 1$ so we can pick $varepsilon > 0$ such that $x_0y_0 - varepsilon > 1$.



Now consider the open ball $B((x_0, y_0), delta)$ of radius $delta > 0$ and let $(x,y) in B((x_0, y_0), delta)$.



We have $$|x| le |(x,y)| le |(x,y) - (x_0,y_0)| + |(x_0,y_0)| < delta + |(x_0, y_0)|$$
so
begin{align}
|xy - x_0y_0| &= |x(y-y_0)+y_0(x - x_0)| \
&le |x||y-y_0|+|y_0||x - x_0| \
&le sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2}\
&< sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta
end{align}



Pick $delta$ small enough such that the last expression is $< varepsilon$ and we will get $$xy = (xy - x_0y_0) + x_0y_0 ge x_0y_0 - |xy - x_0y_0| ge x_0y_0 - varepsilon > 1$$



We conclude that $B((x_0, y_0), delta) subseteq {(x,y) : xy > 1}$ so ${(x,y) : xy > 1}$ is open.



An analogous argument shows that the third set ${(x,y) : xy < 1}$ is also open.



Now we have
$${(x,y) : xy = 1}^c = {(x,y) : xy > 1} cup {(x,y) : xy < 1}$$
so the complement of the first set is also open.






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  • Could you explain this further? I'm having trouble understanding how you got this inequality. Mainly the terms in the square root. $sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta$
    – DreaDk
    Nov 30 at 20:31






  • 1




    @DreaDk I already showed that $|x| < delta + |(x_0, y_0)|$ so since the square root is strictly increasing we get $$sqrt{|x|^2 + |y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2}$$ The second term $sqrt{|x-x_0|^2 + |y-y_0|^2}$ is simply equal to $|(x-x_0,y-y_0)| = |(x,y) - (x_0, y_0)|$ which is $< delta$ because $(x,y) in B((x_0,y_0), delta)$.
    – mechanodroid
    Nov 30 at 20:46





















2














Let, $S={ (x,y) in R^2 : xy<1 }$
Suppose, $(x_0,y_0)in S$.
If $x_0y_0<0$ then it's easy to see that $S$ contains a nbd of $(x_0,y_0)$.
Suppose, $x_0y_0ge 0$
Let, $epsilon<min{1,frac{1-|x_0y_0|}{1+|x_0|+|y_0|}}$.
Then, for $|x-x_0|,|y-y_0|<epsilon$ we have,
$xy-1le |x||y|-1<(|x_0|+epsilon)(|y_0|+epsilon)-1=|x_0y_0|-1+epsilon(|x_0|+|y_0)+epsilon^2<|x_0y_0|-1+epsilon(|x_0|+|y_0|+1)<0$.
Hence, $S$ contains a nbd of $(x_0,y_0)$.



Hence $S$ is open.






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    2















    " It's odd to visualize how there exists an open ball that is contained in xy<1 but doesn't contain any other points."




    It's actually very easy.



    $x_0y_0 < 1$ so $(x_0,y_0)$ is not on the curve $xy = 1$. So there is some distance between the point $(x_0, y_0)$ and the curve $xy=1$. Take a ball whose diameter is smaller than that distance. All the points in that ball aren't far enough to be that distance. so they are all not on the curve.



    I guess you are thinking, if $x_0y_0 < 1$ it can get arbitrarily close to the curve and it can get close enough to the curve that nothing can get closer. Well, in that case get out a magnifying glass and zoom in so that that distance can appear large. Take a radius half that distance. Space (at least in the Euclidean $mathbb R^n$ universe) is continous and there is no such thing as "so close nothing can be closer". We just have to zoom in with a magnifying glass and a distance of $frac {1}{10^{100}}$ might as well be a million miles.



    ====



    Okay to do these formally.



    1) $A = {(x,y)| xy =1}$



    $A^c = {(x,y)|xy < 1} cup {(x,y)|xy > 1}$.



    that's the union of two sets call them $M = {(x,y)|xy < 1}$ and $N = {(x,y)|xy > 1}$. If we can prove they are both open we are done.



    Any my instinct as described above says that they are.



    2) $B = {(x,y)| xy le 1}$.



    $B^c = {(x,y)|xy > 1} = N$.



    So if we prove $N$ is open we are done.



    3) $M = {(x,y)| xy < 1}$.



    That's just $M$. If we prove it is open we are done.



    So...



    So we just have to prove $M$ and $N$ are open.






    share|cite|improve this answer























    • I was more worried about an intersection with $xy>1$. But I guess since they are also on some distance we would need to choose a minimum of the both and then we have the required ball.
      – DreaDk
      Nov 30 at 17:37










    • Cant intersect $xy > 1$ without also intersecting $xy = 1$.
      – fleablood
      Nov 30 at 17:49



















    0














    Lets start with $xy<1$ is open.



    for any $x_0y_0$ in the set. $x_0y_0 < 1$



    or there is a $epsilon >0$ such that $epsilon = 1 - x_0y_0$



    Let $delta = min (frac {epsilon}{4x_0},frac {epsilon}{4y_0})$



    Then all $x,y$ in the ball of radius $delta$



    Next $xy le 1$



    The compliment of this set is ${x,y|xy > 1}$
    By an identical argument, show that this set is open.



    and finally $xy = 1$



    The compliment is ${x,y|xy<1} cup {x,y|xy>1}$ which is the union of two open sets.






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      4 Answers
      4






      active

      oldest

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      4 Answers
      4






      active

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      active

      oldest

      votes






      active

      oldest

      votes









      2














      Consider the complement of the second set ${(x,y) : xy > 1}$. We will show that ${(x,y) : xy > 1}$ is open.



      Let $(x_0,y_0) in {(x,y) : xy > 1}$. Then $x_0y_0 > 1$ so we can pick $varepsilon > 0$ such that $x_0y_0 - varepsilon > 1$.



      Now consider the open ball $B((x_0, y_0), delta)$ of radius $delta > 0$ and let $(x,y) in B((x_0, y_0), delta)$.



      We have $$|x| le |(x,y)| le |(x,y) - (x_0,y_0)| + |(x_0,y_0)| < delta + |(x_0, y_0)|$$
      so
      begin{align}
      |xy - x_0y_0| &= |x(y-y_0)+y_0(x - x_0)| \
      &le |x||y-y_0|+|y_0||x - x_0| \
      &le sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2}\
      &< sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta
      end{align}



      Pick $delta$ small enough such that the last expression is $< varepsilon$ and we will get $$xy = (xy - x_0y_0) + x_0y_0 ge x_0y_0 - |xy - x_0y_0| ge x_0y_0 - varepsilon > 1$$



      We conclude that $B((x_0, y_0), delta) subseteq {(x,y) : xy > 1}$ so ${(x,y) : xy > 1}$ is open.



      An analogous argument shows that the third set ${(x,y) : xy < 1}$ is also open.



      Now we have
      $${(x,y) : xy = 1}^c = {(x,y) : xy > 1} cup {(x,y) : xy < 1}$$
      so the complement of the first set is also open.






      share|cite|improve this answer























      • Could you explain this further? I'm having trouble understanding how you got this inequality. Mainly the terms in the square root. $sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta$
        – DreaDk
        Nov 30 at 20:31






      • 1




        @DreaDk I already showed that $|x| < delta + |(x_0, y_0)|$ so since the square root is strictly increasing we get $$sqrt{|x|^2 + |y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2}$$ The second term $sqrt{|x-x_0|^2 + |y-y_0|^2}$ is simply equal to $|(x-x_0,y-y_0)| = |(x,y) - (x_0, y_0)|$ which is $< delta$ because $(x,y) in B((x_0,y_0), delta)$.
        – mechanodroid
        Nov 30 at 20:46


















      2














      Consider the complement of the second set ${(x,y) : xy > 1}$. We will show that ${(x,y) : xy > 1}$ is open.



      Let $(x_0,y_0) in {(x,y) : xy > 1}$. Then $x_0y_0 > 1$ so we can pick $varepsilon > 0$ such that $x_0y_0 - varepsilon > 1$.



      Now consider the open ball $B((x_0, y_0), delta)$ of radius $delta > 0$ and let $(x,y) in B((x_0, y_0), delta)$.



      We have $$|x| le |(x,y)| le |(x,y) - (x_0,y_0)| + |(x_0,y_0)| < delta + |(x_0, y_0)|$$
      so
      begin{align}
      |xy - x_0y_0| &= |x(y-y_0)+y_0(x - x_0)| \
      &le |x||y-y_0|+|y_0||x - x_0| \
      &le sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2}\
      &< sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta
      end{align}



      Pick $delta$ small enough such that the last expression is $< varepsilon$ and we will get $$xy = (xy - x_0y_0) + x_0y_0 ge x_0y_0 - |xy - x_0y_0| ge x_0y_0 - varepsilon > 1$$



      We conclude that $B((x_0, y_0), delta) subseteq {(x,y) : xy > 1}$ so ${(x,y) : xy > 1}$ is open.



      An analogous argument shows that the third set ${(x,y) : xy < 1}$ is also open.



      Now we have
      $${(x,y) : xy = 1}^c = {(x,y) : xy > 1} cup {(x,y) : xy < 1}$$
      so the complement of the first set is also open.






      share|cite|improve this answer























      • Could you explain this further? I'm having trouble understanding how you got this inequality. Mainly the terms in the square root. $sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta$
        – DreaDk
        Nov 30 at 20:31






      • 1




        @DreaDk I already showed that $|x| < delta + |(x_0, y_0)|$ so since the square root is strictly increasing we get $$sqrt{|x|^2 + |y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2}$$ The second term $sqrt{|x-x_0|^2 + |y-y_0|^2}$ is simply equal to $|(x-x_0,y-y_0)| = |(x,y) - (x_0, y_0)|$ which is $< delta$ because $(x,y) in B((x_0,y_0), delta)$.
        – mechanodroid
        Nov 30 at 20:46
















      2












      2








      2






      Consider the complement of the second set ${(x,y) : xy > 1}$. We will show that ${(x,y) : xy > 1}$ is open.



      Let $(x_0,y_0) in {(x,y) : xy > 1}$. Then $x_0y_0 > 1$ so we can pick $varepsilon > 0$ such that $x_0y_0 - varepsilon > 1$.



      Now consider the open ball $B((x_0, y_0), delta)$ of radius $delta > 0$ and let $(x,y) in B((x_0, y_0), delta)$.



      We have $$|x| le |(x,y)| le |(x,y) - (x_0,y_0)| + |(x_0,y_0)| < delta + |(x_0, y_0)|$$
      so
      begin{align}
      |xy - x_0y_0| &= |x(y-y_0)+y_0(x - x_0)| \
      &le |x||y-y_0|+|y_0||x - x_0| \
      &le sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2}\
      &< sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta
      end{align}



      Pick $delta$ small enough such that the last expression is $< varepsilon$ and we will get $$xy = (xy - x_0y_0) + x_0y_0 ge x_0y_0 - |xy - x_0y_0| ge x_0y_0 - varepsilon > 1$$



      We conclude that $B((x_0, y_0), delta) subseteq {(x,y) : xy > 1}$ so ${(x,y) : xy > 1}$ is open.



      An analogous argument shows that the third set ${(x,y) : xy < 1}$ is also open.



      Now we have
      $${(x,y) : xy = 1}^c = {(x,y) : xy > 1} cup {(x,y) : xy < 1}$$
      so the complement of the first set is also open.






      share|cite|improve this answer














      Consider the complement of the second set ${(x,y) : xy > 1}$. We will show that ${(x,y) : xy > 1}$ is open.



      Let $(x_0,y_0) in {(x,y) : xy > 1}$. Then $x_0y_0 > 1$ so we can pick $varepsilon > 0$ such that $x_0y_0 - varepsilon > 1$.



      Now consider the open ball $B((x_0, y_0), delta)$ of radius $delta > 0$ and let $(x,y) in B((x_0, y_0), delta)$.



      We have $$|x| le |(x,y)| le |(x,y) - (x_0,y_0)| + |(x_0,y_0)| < delta + |(x_0, y_0)|$$
      so
      begin{align}
      |xy - x_0y_0| &= |x(y-y_0)+y_0(x - x_0)| \
      &le |x||y-y_0|+|y_0||x - x_0| \
      &le sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2}\
      &< sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta
      end{align}



      Pick $delta$ small enough such that the last expression is $< varepsilon$ and we will get $$xy = (xy - x_0y_0) + x_0y_0 ge x_0y_0 - |xy - x_0y_0| ge x_0y_0 - varepsilon > 1$$



      We conclude that $B((x_0, y_0), delta) subseteq {(x,y) : xy > 1}$ so ${(x,y) : xy > 1}$ is open.



      An analogous argument shows that the third set ${(x,y) : xy < 1}$ is also open.



      Now we have
      $${(x,y) : xy = 1}^c = {(x,y) : xy > 1} cup {(x,y) : xy < 1}$$
      so the complement of the first set is also open.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 30 at 17:11

























      answered Nov 30 at 17:04









      mechanodroid

      25.4k62245




      25.4k62245












      • Could you explain this further? I'm having trouble understanding how you got this inequality. Mainly the terms in the square root. $sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta$
        – DreaDk
        Nov 30 at 20:31






      • 1




        @DreaDk I already showed that $|x| < delta + |(x_0, y_0)|$ so since the square root is strictly increasing we get $$sqrt{|x|^2 + |y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2}$$ The second term $sqrt{|x-x_0|^2 + |y-y_0|^2}$ is simply equal to $|(x-x_0,y-y_0)| = |(x,y) - (x_0, y_0)|$ which is $< delta$ because $(x,y) in B((x_0,y_0), delta)$.
        – mechanodroid
        Nov 30 at 20:46




















      • Could you explain this further? I'm having trouble understanding how you got this inequality. Mainly the terms in the square root. $sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta$
        – DreaDk
        Nov 30 at 20:31






      • 1




        @DreaDk I already showed that $|x| < delta + |(x_0, y_0)|$ so since the square root is strictly increasing we get $$sqrt{|x|^2 + |y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2}$$ The second term $sqrt{|x-x_0|^2 + |y-y_0|^2}$ is simply equal to $|(x-x_0,y-y_0)| = |(x,y) - (x_0, y_0)|$ which is $< delta$ because $(x,y) in B((x_0,y_0), delta)$.
        – mechanodroid
        Nov 30 at 20:46


















      Could you explain this further? I'm having trouble understanding how you got this inequality. Mainly the terms in the square root. $sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta$
      – DreaDk
      Nov 30 at 20:31




      Could you explain this further? I'm having trouble understanding how you got this inequality. Mainly the terms in the square root. $sqrt{|x|^2 + |y_0|^2}sqrt{|x-x_0|^2 + |y-y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2} cdot delta$
      – DreaDk
      Nov 30 at 20:31




      1




      1




      @DreaDk I already showed that $|x| < delta + |(x_0, y_0)|$ so since the square root is strictly increasing we get $$sqrt{|x|^2 + |y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2}$$ The second term $sqrt{|x-x_0|^2 + |y-y_0|^2}$ is simply equal to $|(x-x_0,y-y_0)| = |(x,y) - (x_0, y_0)|$ which is $< delta$ because $(x,y) in B((x_0,y_0), delta)$.
      – mechanodroid
      Nov 30 at 20:46






      @DreaDk I already showed that $|x| < delta + |(x_0, y_0)|$ so since the square root is strictly increasing we get $$sqrt{|x|^2 + |y_0|^2} < sqrt{(delta + |(x_0, y_0)|)^2 + |y_0|^2}$$ The second term $sqrt{|x-x_0|^2 + |y-y_0|^2}$ is simply equal to $|(x-x_0,y-y_0)| = |(x,y) - (x_0, y_0)|$ which is $< delta$ because $(x,y) in B((x_0,y_0), delta)$.
      – mechanodroid
      Nov 30 at 20:46













      2














      Let, $S={ (x,y) in R^2 : xy<1 }$
      Suppose, $(x_0,y_0)in S$.
      If $x_0y_0<0$ then it's easy to see that $S$ contains a nbd of $(x_0,y_0)$.
      Suppose, $x_0y_0ge 0$
      Let, $epsilon<min{1,frac{1-|x_0y_0|}{1+|x_0|+|y_0|}}$.
      Then, for $|x-x_0|,|y-y_0|<epsilon$ we have,
      $xy-1le |x||y|-1<(|x_0|+epsilon)(|y_0|+epsilon)-1=|x_0y_0|-1+epsilon(|x_0|+|y_0)+epsilon^2<|x_0y_0|-1+epsilon(|x_0|+|y_0|+1)<0$.
      Hence, $S$ contains a nbd of $(x_0,y_0)$.



      Hence $S$ is open.






      share|cite|improve this answer


























        2














        Let, $S={ (x,y) in R^2 : xy<1 }$
        Suppose, $(x_0,y_0)in S$.
        If $x_0y_0<0$ then it's easy to see that $S$ contains a nbd of $(x_0,y_0)$.
        Suppose, $x_0y_0ge 0$
        Let, $epsilon<min{1,frac{1-|x_0y_0|}{1+|x_0|+|y_0|}}$.
        Then, for $|x-x_0|,|y-y_0|<epsilon$ we have,
        $xy-1le |x||y|-1<(|x_0|+epsilon)(|y_0|+epsilon)-1=|x_0y_0|-1+epsilon(|x_0|+|y_0)+epsilon^2<|x_0y_0|-1+epsilon(|x_0|+|y_0|+1)<0$.
        Hence, $S$ contains a nbd of $(x_0,y_0)$.



        Hence $S$ is open.






        share|cite|improve this answer
























          2












          2








          2






          Let, $S={ (x,y) in R^2 : xy<1 }$
          Suppose, $(x_0,y_0)in S$.
          If $x_0y_0<0$ then it's easy to see that $S$ contains a nbd of $(x_0,y_0)$.
          Suppose, $x_0y_0ge 0$
          Let, $epsilon<min{1,frac{1-|x_0y_0|}{1+|x_0|+|y_0|}}$.
          Then, for $|x-x_0|,|y-y_0|<epsilon$ we have,
          $xy-1le |x||y|-1<(|x_0|+epsilon)(|y_0|+epsilon)-1=|x_0y_0|-1+epsilon(|x_0|+|y_0)+epsilon^2<|x_0y_0|-1+epsilon(|x_0|+|y_0|+1)<0$.
          Hence, $S$ contains a nbd of $(x_0,y_0)$.



          Hence $S$ is open.






          share|cite|improve this answer












          Let, $S={ (x,y) in R^2 : xy<1 }$
          Suppose, $(x_0,y_0)in S$.
          If $x_0y_0<0$ then it's easy to see that $S$ contains a nbd of $(x_0,y_0)$.
          Suppose, $x_0y_0ge 0$
          Let, $epsilon<min{1,frac{1-|x_0y_0|}{1+|x_0|+|y_0|}}$.
          Then, for $|x-x_0|,|y-y_0|<epsilon$ we have,
          $xy-1le |x||y|-1<(|x_0|+epsilon)(|y_0|+epsilon)-1=|x_0y_0|-1+epsilon(|x_0|+|y_0)+epsilon^2<|x_0y_0|-1+epsilon(|x_0|+|y_0|+1)<0$.
          Hence, $S$ contains a nbd of $(x_0,y_0)$.



          Hence $S$ is open.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 at 17:14









          Tom.

          14118




          14118























              2















              " It's odd to visualize how there exists an open ball that is contained in xy<1 but doesn't contain any other points."




              It's actually very easy.



              $x_0y_0 < 1$ so $(x_0,y_0)$ is not on the curve $xy = 1$. So there is some distance between the point $(x_0, y_0)$ and the curve $xy=1$. Take a ball whose diameter is smaller than that distance. All the points in that ball aren't far enough to be that distance. so they are all not on the curve.



              I guess you are thinking, if $x_0y_0 < 1$ it can get arbitrarily close to the curve and it can get close enough to the curve that nothing can get closer. Well, in that case get out a magnifying glass and zoom in so that that distance can appear large. Take a radius half that distance. Space (at least in the Euclidean $mathbb R^n$ universe) is continous and there is no such thing as "so close nothing can be closer". We just have to zoom in with a magnifying glass and a distance of $frac {1}{10^{100}}$ might as well be a million miles.



              ====



              Okay to do these formally.



              1) $A = {(x,y)| xy =1}$



              $A^c = {(x,y)|xy < 1} cup {(x,y)|xy > 1}$.



              that's the union of two sets call them $M = {(x,y)|xy < 1}$ and $N = {(x,y)|xy > 1}$. If we can prove they are both open we are done.



              Any my instinct as described above says that they are.



              2) $B = {(x,y)| xy le 1}$.



              $B^c = {(x,y)|xy > 1} = N$.



              So if we prove $N$ is open we are done.



              3) $M = {(x,y)| xy < 1}$.



              That's just $M$. If we prove it is open we are done.



              So...



              So we just have to prove $M$ and $N$ are open.






              share|cite|improve this answer























              • I was more worried about an intersection with $xy>1$. But I guess since they are also on some distance we would need to choose a minimum of the both and then we have the required ball.
                – DreaDk
                Nov 30 at 17:37










              • Cant intersect $xy > 1$ without also intersecting $xy = 1$.
                – fleablood
                Nov 30 at 17:49
















              2















              " It's odd to visualize how there exists an open ball that is contained in xy<1 but doesn't contain any other points."




              It's actually very easy.



              $x_0y_0 < 1$ so $(x_0,y_0)$ is not on the curve $xy = 1$. So there is some distance between the point $(x_0, y_0)$ and the curve $xy=1$. Take a ball whose diameter is smaller than that distance. All the points in that ball aren't far enough to be that distance. so they are all not on the curve.



              I guess you are thinking, if $x_0y_0 < 1$ it can get arbitrarily close to the curve and it can get close enough to the curve that nothing can get closer. Well, in that case get out a magnifying glass and zoom in so that that distance can appear large. Take a radius half that distance. Space (at least in the Euclidean $mathbb R^n$ universe) is continous and there is no such thing as "so close nothing can be closer". We just have to zoom in with a magnifying glass and a distance of $frac {1}{10^{100}}$ might as well be a million miles.



              ====



              Okay to do these formally.



              1) $A = {(x,y)| xy =1}$



              $A^c = {(x,y)|xy < 1} cup {(x,y)|xy > 1}$.



              that's the union of two sets call them $M = {(x,y)|xy < 1}$ and $N = {(x,y)|xy > 1}$. If we can prove they are both open we are done.



              Any my instinct as described above says that they are.



              2) $B = {(x,y)| xy le 1}$.



              $B^c = {(x,y)|xy > 1} = N$.



              So if we prove $N$ is open we are done.



              3) $M = {(x,y)| xy < 1}$.



              That's just $M$. If we prove it is open we are done.



              So...



              So we just have to prove $M$ and $N$ are open.






              share|cite|improve this answer























              • I was more worried about an intersection with $xy>1$. But I guess since they are also on some distance we would need to choose a minimum of the both and then we have the required ball.
                – DreaDk
                Nov 30 at 17:37










              • Cant intersect $xy > 1$ without also intersecting $xy = 1$.
                – fleablood
                Nov 30 at 17:49














              2












              2








              2







              " It's odd to visualize how there exists an open ball that is contained in xy<1 but doesn't contain any other points."




              It's actually very easy.



              $x_0y_0 < 1$ so $(x_0,y_0)$ is not on the curve $xy = 1$. So there is some distance between the point $(x_0, y_0)$ and the curve $xy=1$. Take a ball whose diameter is smaller than that distance. All the points in that ball aren't far enough to be that distance. so they are all not on the curve.



              I guess you are thinking, if $x_0y_0 < 1$ it can get arbitrarily close to the curve and it can get close enough to the curve that nothing can get closer. Well, in that case get out a magnifying glass and zoom in so that that distance can appear large. Take a radius half that distance. Space (at least in the Euclidean $mathbb R^n$ universe) is continous and there is no such thing as "so close nothing can be closer". We just have to zoom in with a magnifying glass and a distance of $frac {1}{10^{100}}$ might as well be a million miles.



              ====



              Okay to do these formally.



              1) $A = {(x,y)| xy =1}$



              $A^c = {(x,y)|xy < 1} cup {(x,y)|xy > 1}$.



              that's the union of two sets call them $M = {(x,y)|xy < 1}$ and $N = {(x,y)|xy > 1}$. If we can prove they are both open we are done.



              Any my instinct as described above says that they are.



              2) $B = {(x,y)| xy le 1}$.



              $B^c = {(x,y)|xy > 1} = N$.



              So if we prove $N$ is open we are done.



              3) $M = {(x,y)| xy < 1}$.



              That's just $M$. If we prove it is open we are done.



              So...



              So we just have to prove $M$ and $N$ are open.






              share|cite|improve this answer















              " It's odd to visualize how there exists an open ball that is contained in xy<1 but doesn't contain any other points."




              It's actually very easy.



              $x_0y_0 < 1$ so $(x_0,y_0)$ is not on the curve $xy = 1$. So there is some distance between the point $(x_0, y_0)$ and the curve $xy=1$. Take a ball whose diameter is smaller than that distance. All the points in that ball aren't far enough to be that distance. so they are all not on the curve.



              I guess you are thinking, if $x_0y_0 < 1$ it can get arbitrarily close to the curve and it can get close enough to the curve that nothing can get closer. Well, in that case get out a magnifying glass and zoom in so that that distance can appear large. Take a radius half that distance. Space (at least in the Euclidean $mathbb R^n$ universe) is continous and there is no such thing as "so close nothing can be closer". We just have to zoom in with a magnifying glass and a distance of $frac {1}{10^{100}}$ might as well be a million miles.



              ====



              Okay to do these formally.



              1) $A = {(x,y)| xy =1}$



              $A^c = {(x,y)|xy < 1} cup {(x,y)|xy > 1}$.



              that's the union of two sets call them $M = {(x,y)|xy < 1}$ and $N = {(x,y)|xy > 1}$. If we can prove they are both open we are done.



              Any my instinct as described above says that they are.



              2) $B = {(x,y)| xy le 1}$.



              $B^c = {(x,y)|xy > 1} = N$.



              So if we prove $N$ is open we are done.



              3) $M = {(x,y)| xy < 1}$.



              That's just $M$. If we prove it is open we are done.



              So...



              So we just have to prove $M$ and $N$ are open.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 30 at 17:49

























              answered Nov 30 at 17:25









              fleablood

              68.1k22684




              68.1k22684












              • I was more worried about an intersection with $xy>1$. But I guess since they are also on some distance we would need to choose a minimum of the both and then we have the required ball.
                – DreaDk
                Nov 30 at 17:37










              • Cant intersect $xy > 1$ without also intersecting $xy = 1$.
                – fleablood
                Nov 30 at 17:49


















              • I was more worried about an intersection with $xy>1$. But I guess since they are also on some distance we would need to choose a minimum of the both and then we have the required ball.
                – DreaDk
                Nov 30 at 17:37










              • Cant intersect $xy > 1$ without also intersecting $xy = 1$.
                – fleablood
                Nov 30 at 17:49
















              I was more worried about an intersection with $xy>1$. But I guess since they are also on some distance we would need to choose a minimum of the both and then we have the required ball.
              – DreaDk
              Nov 30 at 17:37




              I was more worried about an intersection with $xy>1$. But I guess since they are also on some distance we would need to choose a minimum of the both and then we have the required ball.
              – DreaDk
              Nov 30 at 17:37












              Cant intersect $xy > 1$ without also intersecting $xy = 1$.
              – fleablood
              Nov 30 at 17:49




              Cant intersect $xy > 1$ without also intersecting $xy = 1$.
              – fleablood
              Nov 30 at 17:49











              0














              Lets start with $xy<1$ is open.



              for any $x_0y_0$ in the set. $x_0y_0 < 1$



              or there is a $epsilon >0$ such that $epsilon = 1 - x_0y_0$



              Let $delta = min (frac {epsilon}{4x_0},frac {epsilon}{4y_0})$



              Then all $x,y$ in the ball of radius $delta$



              Next $xy le 1$



              The compliment of this set is ${x,y|xy > 1}$
              By an identical argument, show that this set is open.



              and finally $xy = 1$



              The compliment is ${x,y|xy<1} cup {x,y|xy>1}$ which is the union of two open sets.






              share|cite|improve this answer


























                0














                Lets start with $xy<1$ is open.



                for any $x_0y_0$ in the set. $x_0y_0 < 1$



                or there is a $epsilon >0$ such that $epsilon = 1 - x_0y_0$



                Let $delta = min (frac {epsilon}{4x_0},frac {epsilon}{4y_0})$



                Then all $x,y$ in the ball of radius $delta$



                Next $xy le 1$



                The compliment of this set is ${x,y|xy > 1}$
                By an identical argument, show that this set is open.



                and finally $xy = 1$



                The compliment is ${x,y|xy<1} cup {x,y|xy>1}$ which is the union of two open sets.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Lets start with $xy<1$ is open.



                  for any $x_0y_0$ in the set. $x_0y_0 < 1$



                  or there is a $epsilon >0$ such that $epsilon = 1 - x_0y_0$



                  Let $delta = min (frac {epsilon}{4x_0},frac {epsilon}{4y_0})$



                  Then all $x,y$ in the ball of radius $delta$



                  Next $xy le 1$



                  The compliment of this set is ${x,y|xy > 1}$
                  By an identical argument, show that this set is open.



                  and finally $xy = 1$



                  The compliment is ${x,y|xy<1} cup {x,y|xy>1}$ which is the union of two open sets.






                  share|cite|improve this answer












                  Lets start with $xy<1$ is open.



                  for any $x_0y_0$ in the set. $x_0y_0 < 1$



                  or there is a $epsilon >0$ such that $epsilon = 1 - x_0y_0$



                  Let $delta = min (frac {epsilon}{4x_0},frac {epsilon}{4y_0})$



                  Then all $x,y$ in the ball of radius $delta$



                  Next $xy le 1$



                  The compliment of this set is ${x,y|xy > 1}$
                  By an identical argument, show that this set is open.



                  and finally $xy = 1$



                  The compliment is ${x,y|xy<1} cup {x,y|xy>1}$ which is the union of two open sets.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 at 17:21









                  Doug M

                  43.9k31854




                  43.9k31854






























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