The set $H={(x,y)in Bbb{R^2}:;3x+2y=5 },$ is a closed subset of $Bbb{R^2}$












6












$begingroup$


Kindly check if my proof is correct. Thanks for your time and efforts.



MY PROOF



For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $3x_n+2y_n=5,$ for all $nin Bbb{N}.$ As $nto infty, $



$$ 3x+2y=limlimits_{ntoinfty}(3x_n+2y_n)=limlimits_{ntoinfty}5=5 . $$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;3x+2y=5 },$ is a closed subset of $Bbb{R^2}$






linear-algebra analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Looks good. Maybe cite the algebra of limits theorem?
    $endgroup$
    – Theo Bendit
    Jan 11 at 15:08










  • $begingroup$
    I mean, sure I guess. But since it's a straight line you could just say that.
    $endgroup$
    – Ben W
    Jan 11 at 15:08






  • 2




    $begingroup$
    This seems a bit odd, you start assuming that $3x_n + 2y_n = 5$ for all $n$ and that $(x_n,y_n)to (x,y)$, of course $3 x + 2y = 5$ follows from that. But I don't think you proved anything, because you assumed everything
    $endgroup$
    – caverac
    Jan 11 at 15:23






  • 2




    $begingroup$
    Equivalent statement is: the sets $H_{+}={(x,y)in Bbb{R^2}:;3x+2y>5 }$ and $H_{-}={(x,y)in Bbb{R^2}:;3x+2y<5 }$ are open subsets of $ Bbb{R^2}$.
    $endgroup$
    – Aleksas Domarkas
    Jan 11 at 15:28








  • 3




    $begingroup$
    The statement $lim(3x_n + 2x_n = 5)$ does not make sense and is meaningless. What I believe you mean is: $3x + 2x = lim (3x_n + 2x_n) = lim 5 = 5$. But you should state explicitly that $lim (3x_n + 2x_n) = 3lim x_n + 2lim x_n$
    $endgroup$
    – fleablood
    Jan 11 at 16:17
















6












$begingroup$


Kindly check if my proof is correct. Thanks for your time and efforts.



MY PROOF



For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $3x_n+2y_n=5,$ for all $nin Bbb{N}.$ As $nto infty, $



$$ 3x+2y=limlimits_{ntoinfty}(3x_n+2y_n)=limlimits_{ntoinfty}5=5 . $$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;3x+2y=5 },$ is a closed subset of $Bbb{R^2}$






linear-algebra analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Looks good. Maybe cite the algebra of limits theorem?
    $endgroup$
    – Theo Bendit
    Jan 11 at 15:08










  • $begingroup$
    I mean, sure I guess. But since it's a straight line you could just say that.
    $endgroup$
    – Ben W
    Jan 11 at 15:08






  • 2




    $begingroup$
    This seems a bit odd, you start assuming that $3x_n + 2y_n = 5$ for all $n$ and that $(x_n,y_n)to (x,y)$, of course $3 x + 2y = 5$ follows from that. But I don't think you proved anything, because you assumed everything
    $endgroup$
    – caverac
    Jan 11 at 15:23






  • 2




    $begingroup$
    Equivalent statement is: the sets $H_{+}={(x,y)in Bbb{R^2}:;3x+2y>5 }$ and $H_{-}={(x,y)in Bbb{R^2}:;3x+2y<5 }$ are open subsets of $ Bbb{R^2}$.
    $endgroup$
    – Aleksas Domarkas
    Jan 11 at 15:28








  • 3




    $begingroup$
    The statement $lim(3x_n + 2x_n = 5)$ does not make sense and is meaningless. What I believe you mean is: $3x + 2x = lim (3x_n + 2x_n) = lim 5 = 5$. But you should state explicitly that $lim (3x_n + 2x_n) = 3lim x_n + 2lim x_n$
    $endgroup$
    – fleablood
    Jan 11 at 16:17














6












6








6


1



$begingroup$


Kindly check if my proof is correct. Thanks for your time and efforts.



MY PROOF



For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $3x_n+2y_n=5,$ for all $nin Bbb{N}.$ As $nto infty, $



$$ 3x+2y=limlimits_{ntoinfty}(3x_n+2y_n)=limlimits_{ntoinfty}5=5 . $$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;3x+2y=5 },$ is a closed subset of $Bbb{R^2}$






linear-algebra analysis







share|cite|improve this question











$endgroup$




Kindly check if my proof is correct. Thanks for your time and efforts.



MY PROOF



For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $3x_n+2y_n=5,$ for all $nin Bbb{N}.$ As $nto infty, $



$$ 3x+2y=limlimits_{ntoinfty}(3x_n+2y_n)=limlimits_{ntoinfty}5=5 . $$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;3x+2y=5 },$ is a closed subset of $Bbb{R^2}$






linear-algebra analysis




linear-algebra analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 16:45







Omojola Micheal

















asked Jan 11 at 15:06









Omojola MichealOmojola Micheal

1,831324




1,831324








  • 2




    $begingroup$
    Looks good. Maybe cite the algebra of limits theorem?
    $endgroup$
    – Theo Bendit
    Jan 11 at 15:08










  • $begingroup$
    I mean, sure I guess. But since it's a straight line you could just say that.
    $endgroup$
    – Ben W
    Jan 11 at 15:08






  • 2




    $begingroup$
    This seems a bit odd, you start assuming that $3x_n + 2y_n = 5$ for all $n$ and that $(x_n,y_n)to (x,y)$, of course $3 x + 2y = 5$ follows from that. But I don't think you proved anything, because you assumed everything
    $endgroup$
    – caverac
    Jan 11 at 15:23






  • 2




    $begingroup$
    Equivalent statement is: the sets $H_{+}={(x,y)in Bbb{R^2}:;3x+2y>5 }$ and $H_{-}={(x,y)in Bbb{R^2}:;3x+2y<5 }$ are open subsets of $ Bbb{R^2}$.
    $endgroup$
    – Aleksas Domarkas
    Jan 11 at 15:28








  • 3




    $begingroup$
    The statement $lim(3x_n + 2x_n = 5)$ does not make sense and is meaningless. What I believe you mean is: $3x + 2x = lim (3x_n + 2x_n) = lim 5 = 5$. But you should state explicitly that $lim (3x_n + 2x_n) = 3lim x_n + 2lim x_n$
    $endgroup$
    – fleablood
    Jan 11 at 16:17














  • 2




    $begingroup$
    Looks good. Maybe cite the algebra of limits theorem?
    $endgroup$
    – Theo Bendit
    Jan 11 at 15:08










  • $begingroup$
    I mean, sure I guess. But since it's a straight line you could just say that.
    $endgroup$
    – Ben W
    Jan 11 at 15:08






  • 2




    $begingroup$
    This seems a bit odd, you start assuming that $3x_n + 2y_n = 5$ for all $n$ and that $(x_n,y_n)to (x,y)$, of course $3 x + 2y = 5$ follows from that. But I don't think you proved anything, because you assumed everything
    $endgroup$
    – caverac
    Jan 11 at 15:23






  • 2




    $begingroup$
    Equivalent statement is: the sets $H_{+}={(x,y)in Bbb{R^2}:;3x+2y>5 }$ and $H_{-}={(x,y)in Bbb{R^2}:;3x+2y<5 }$ are open subsets of $ Bbb{R^2}$.
    $endgroup$
    – Aleksas Domarkas
    Jan 11 at 15:28








  • 3




    $begingroup$
    The statement $lim(3x_n + 2x_n = 5)$ does not make sense and is meaningless. What I believe you mean is: $3x + 2x = lim (3x_n + 2x_n) = lim 5 = 5$. But you should state explicitly that $lim (3x_n + 2x_n) = 3lim x_n + 2lim x_n$
    $endgroup$
    – fleablood
    Jan 11 at 16:17








2




2




$begingroup$
Looks good. Maybe cite the algebra of limits theorem?
$endgroup$
– Theo Bendit
Jan 11 at 15:08




$begingroup$
Looks good. Maybe cite the algebra of limits theorem?
$endgroup$
– Theo Bendit
Jan 11 at 15:08












$begingroup$
I mean, sure I guess. But since it's a straight line you could just say that.
$endgroup$
– Ben W
Jan 11 at 15:08




$begingroup$
I mean, sure I guess. But since it's a straight line you could just say that.
$endgroup$
– Ben W
Jan 11 at 15:08




2




2




$begingroup$
This seems a bit odd, you start assuming that $3x_n + 2y_n = 5$ for all $n$ and that $(x_n,y_n)to (x,y)$, of course $3 x + 2y = 5$ follows from that. But I don't think you proved anything, because you assumed everything
$endgroup$
– caverac
Jan 11 at 15:23




$begingroup$
This seems a bit odd, you start assuming that $3x_n + 2y_n = 5$ for all $n$ and that $(x_n,y_n)to (x,y)$, of course $3 x + 2y = 5$ follows from that. But I don't think you proved anything, because you assumed everything
$endgroup$
– caverac
Jan 11 at 15:23




2




2




$begingroup$
Equivalent statement is: the sets $H_{+}={(x,y)in Bbb{R^2}:;3x+2y>5 }$ and $H_{-}={(x,y)in Bbb{R^2}:;3x+2y<5 }$ are open subsets of $ Bbb{R^2}$.
$endgroup$
– Aleksas Domarkas
Jan 11 at 15:28






$begingroup$
Equivalent statement is: the sets $H_{+}={(x,y)in Bbb{R^2}:;3x+2y>5 }$ and $H_{-}={(x,y)in Bbb{R^2}:;3x+2y<5 }$ are open subsets of $ Bbb{R^2}$.
$endgroup$
– Aleksas Domarkas
Jan 11 at 15:28






3




3




$begingroup$
The statement $lim(3x_n + 2x_n = 5)$ does not make sense and is meaningless. What I believe you mean is: $3x + 2x = lim (3x_n + 2x_n) = lim 5 = 5$. But you should state explicitly that $lim (3x_n + 2x_n) = 3lim x_n + 2lim x_n$
$endgroup$
– fleablood
Jan 11 at 16:17




$begingroup$
The statement $lim(3x_n + 2x_n = 5)$ does not make sense and is meaningless. What I believe you mean is: $3x + 2x = lim (3x_n + 2x_n) = lim 5 = 5$. But you should state explicitly that $lim (3x_n + 2x_n) = 3lim x_n + 2lim x_n$
$endgroup$
– fleablood
Jan 11 at 16:17










4 Answers
4






active

oldest

votes


















3












$begingroup$

Looks Good! Alternatively, take $alpha=(x,y) in Bbb R^2 setminus H$. Take $r=d(alpha,H)>0$, the distance from the point $alpha$ to the line $H$. Then $$B(alpha,r) subset Bbb R^2 setminus H$$ showing $Bbb R^2 setminus H$ is open.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I love this one! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:46



















1












$begingroup$

Maybe I'm overcomplicating this, but here's where I think your proof fails



Imagine that $H = {(x, y)in Bbb{R}^2: x^2 + y^2 < 1}$. And now I am going to apply your same argument




For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $x_n^2+y_n^2<1,$ for all $nin Bbb{N}.$ As $nto infty, $



$$limlimits_{ntoinfty} x_n^2 + y_n^2 = x^2 + y^2 < 1$$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;x^2+y^2<1 },$ is a closed subset of $Bbb{R}^2$




Do you see the problem?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hmm... You are right because, $x^2+y^2leq 1$ which will imply that $(x,y)notin H.$ However, it's all about the approach used. In that scenerio, I believe mine is correct, right?
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:04










  • $begingroup$
    This has the same general form as the OP's proof, but the error stems from something not present there. The OP is tacitly using the theorem that $lim_{ntoinfty}(u_n+v_n)=lim_{ntoinfty}u_n+lim_{ntoinfty}v_n$ (provided the limits on the right hand side both exist), while the "proof" here tacitly uses the incorrect (non-)theorem that $u_nlt cimplieslim_{ntoinfty}u_nlt c$. The correct theorem is $u_nlt cimplieslim_{ntoinfty}u_nle c$.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:10












  • $begingroup$
    @BarryCipra Oh yeah! That's the missing part here $u_n < c implies lim_{ntoinfty} color{red}{leq} c$. I'm going to delete this, don't want to contribute to the clutter. Thanks for the clarification
    $endgroup$
    – caverac
    Jan 11 at 16:26












  • $begingroup$
    @caverac, you're welcome for the clarification. The flaw is somewhat subtle (it took me, at least, a bit of thought to see what was wrong in it) and possibly worth preserving on that basis.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:34





















1












$begingroup$

I would modify your proof to make one tacit step explicit in the display: If $(x_n,y_n)to(x,y)$, then



$$3x+2y=3lim_{ntoinfty}x_n+2lim_{ntoinfty}y_n=lim_{ntoinfty}(3x_n+2y_n)=lim_{ntoinfty}5=5$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45












  • $begingroup$
    If your comment here was intended for dmtri, you can delete it (here) by hovering your cursor to the right of the timestamp until an X appears.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:56










  • $begingroup$
    It was meant for you as well. As you know, I made typos and you did well to make corrections. Thanks, once again!
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:57










  • $begingroup$
    OK. But I didn't even notice the typo!
    $endgroup$
    – Barry Cipra
    Jan 11 at 17:00



















1












$begingroup$

I think you should have $3x+2y$ in the 3 line from the bottom....






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45













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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Looks Good! Alternatively, take $alpha=(x,y) in Bbb R^2 setminus H$. Take $r=d(alpha,H)>0$, the distance from the point $alpha$ to the line $H$. Then $$B(alpha,r) subset Bbb R^2 setminus H$$ showing $Bbb R^2 setminus H$ is open.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I love this one! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:46
















3












$begingroup$

Looks Good! Alternatively, take $alpha=(x,y) in Bbb R^2 setminus H$. Take $r=d(alpha,H)>0$, the distance from the point $alpha$ to the line $H$. Then $$B(alpha,r) subset Bbb R^2 setminus H$$ showing $Bbb R^2 setminus H$ is open.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I love this one! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:46














3












3








3





$begingroup$

Looks Good! Alternatively, take $alpha=(x,y) in Bbb R^2 setminus H$. Take $r=d(alpha,H)>0$, the distance from the point $alpha$ to the line $H$. Then $$B(alpha,r) subset Bbb R^2 setminus H$$ showing $Bbb R^2 setminus H$ is open.






share|cite|improve this answer









$endgroup$



Looks Good! Alternatively, take $alpha=(x,y) in Bbb R^2 setminus H$. Take $r=d(alpha,H)>0$, the distance from the point $alpha$ to the line $H$. Then $$B(alpha,r) subset Bbb R^2 setminus H$$ showing $Bbb R^2 setminus H$ is open.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 15:33









Chinnapparaj RChinnapparaj R

5,3641828




5,3641828












  • $begingroup$
    I love this one! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:46


















  • $begingroup$
    I love this one! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:46
















$begingroup$
I love this one! (+1)
$endgroup$
– Omojola Micheal
Jan 11 at 16:46




$begingroup$
I love this one! (+1)
$endgroup$
– Omojola Micheal
Jan 11 at 16:46











1












$begingroup$

Maybe I'm overcomplicating this, but here's where I think your proof fails



Imagine that $H = {(x, y)in Bbb{R}^2: x^2 + y^2 < 1}$. And now I am going to apply your same argument




For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $x_n^2+y_n^2<1,$ for all $nin Bbb{N}.$ As $nto infty, $



$$limlimits_{ntoinfty} x_n^2 + y_n^2 = x^2 + y^2 < 1$$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;x^2+y^2<1 },$ is a closed subset of $Bbb{R}^2$




Do you see the problem?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hmm... You are right because, $x^2+y^2leq 1$ which will imply that $(x,y)notin H.$ However, it's all about the approach used. In that scenerio, I believe mine is correct, right?
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:04










  • $begingroup$
    This has the same general form as the OP's proof, but the error stems from something not present there. The OP is tacitly using the theorem that $lim_{ntoinfty}(u_n+v_n)=lim_{ntoinfty}u_n+lim_{ntoinfty}v_n$ (provided the limits on the right hand side both exist), while the "proof" here tacitly uses the incorrect (non-)theorem that $u_nlt cimplieslim_{ntoinfty}u_nlt c$. The correct theorem is $u_nlt cimplieslim_{ntoinfty}u_nle c$.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:10












  • $begingroup$
    @BarryCipra Oh yeah! That's the missing part here $u_n < c implies lim_{ntoinfty} color{red}{leq} c$. I'm going to delete this, don't want to contribute to the clutter. Thanks for the clarification
    $endgroup$
    – caverac
    Jan 11 at 16:26












  • $begingroup$
    @caverac, you're welcome for the clarification. The flaw is somewhat subtle (it took me, at least, a bit of thought to see what was wrong in it) and possibly worth preserving on that basis.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:34


















1












$begingroup$

Maybe I'm overcomplicating this, but here's where I think your proof fails



Imagine that $H = {(x, y)in Bbb{R}^2: x^2 + y^2 < 1}$. And now I am going to apply your same argument




For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $x_n^2+y_n^2<1,$ for all $nin Bbb{N}.$ As $nto infty, $



$$limlimits_{ntoinfty} x_n^2 + y_n^2 = x^2 + y^2 < 1$$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;x^2+y^2<1 },$ is a closed subset of $Bbb{R}^2$




Do you see the problem?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hmm... You are right because, $x^2+y^2leq 1$ which will imply that $(x,y)notin H.$ However, it's all about the approach used. In that scenerio, I believe mine is correct, right?
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:04










  • $begingroup$
    This has the same general form as the OP's proof, but the error stems from something not present there. The OP is tacitly using the theorem that $lim_{ntoinfty}(u_n+v_n)=lim_{ntoinfty}u_n+lim_{ntoinfty}v_n$ (provided the limits on the right hand side both exist), while the "proof" here tacitly uses the incorrect (non-)theorem that $u_nlt cimplieslim_{ntoinfty}u_nlt c$. The correct theorem is $u_nlt cimplieslim_{ntoinfty}u_nle c$.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:10












  • $begingroup$
    @BarryCipra Oh yeah! That's the missing part here $u_n < c implies lim_{ntoinfty} color{red}{leq} c$. I'm going to delete this, don't want to contribute to the clutter. Thanks for the clarification
    $endgroup$
    – caverac
    Jan 11 at 16:26












  • $begingroup$
    @caverac, you're welcome for the clarification. The flaw is somewhat subtle (it took me, at least, a bit of thought to see what was wrong in it) and possibly worth preserving on that basis.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:34
















1












1








1





$begingroup$

Maybe I'm overcomplicating this, but here's where I think your proof fails



Imagine that $H = {(x, y)in Bbb{R}^2: x^2 + y^2 < 1}$. And now I am going to apply your same argument




For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $x_n^2+y_n^2<1,$ for all $nin Bbb{N}.$ As $nto infty, $



$$limlimits_{ntoinfty} x_n^2 + y_n^2 = x^2 + y^2 < 1$$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;x^2+y^2<1 },$ is a closed subset of $Bbb{R}^2$




Do you see the problem?






share|cite|improve this answer









$endgroup$



Maybe I'm overcomplicating this, but here's where I think your proof fails



Imagine that $H = {(x, y)in Bbb{R}^2: x^2 + y^2 < 1}$. And now I am going to apply your same argument




For all $nin Bbb{N},$ let $(x_n,y_n)in H$ such that $(x_n,y_n)to (x,y),$ as $nto infty.$ We show that $(x,y)in H.$



However, $(x_n,y_n)in H$ implies $x_n^2+y_n^2<1,$ for all $nin Bbb{N}.$ As $nto infty, $



$$limlimits_{ntoinfty} x_n^2 + y_n^2 = x^2 + y^2 < 1$$
Hence, $(x,y)in H$ which implies that the set, $H={(x,y)in Bbb{R^2}:;x^2+y^2<1 },$ is a closed subset of $Bbb{R}^2$




Do you see the problem?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 15:37









caveraccaverac

14.5k31130




14.5k31130












  • $begingroup$
    Hmm... You are right because, $x^2+y^2leq 1$ which will imply that $(x,y)notin H.$ However, it's all about the approach used. In that scenerio, I believe mine is correct, right?
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:04










  • $begingroup$
    This has the same general form as the OP's proof, but the error stems from something not present there. The OP is tacitly using the theorem that $lim_{ntoinfty}(u_n+v_n)=lim_{ntoinfty}u_n+lim_{ntoinfty}v_n$ (provided the limits on the right hand side both exist), while the "proof" here tacitly uses the incorrect (non-)theorem that $u_nlt cimplieslim_{ntoinfty}u_nlt c$. The correct theorem is $u_nlt cimplieslim_{ntoinfty}u_nle c$.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:10












  • $begingroup$
    @BarryCipra Oh yeah! That's the missing part here $u_n < c implies lim_{ntoinfty} color{red}{leq} c$. I'm going to delete this, don't want to contribute to the clutter. Thanks for the clarification
    $endgroup$
    – caverac
    Jan 11 at 16:26












  • $begingroup$
    @caverac, you're welcome for the clarification. The flaw is somewhat subtle (it took me, at least, a bit of thought to see what was wrong in it) and possibly worth preserving on that basis.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:34




















  • $begingroup$
    Hmm... You are right because, $x^2+y^2leq 1$ which will imply that $(x,y)notin H.$ However, it's all about the approach used. In that scenerio, I believe mine is correct, right?
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:04










  • $begingroup$
    This has the same general form as the OP's proof, but the error stems from something not present there. The OP is tacitly using the theorem that $lim_{ntoinfty}(u_n+v_n)=lim_{ntoinfty}u_n+lim_{ntoinfty}v_n$ (provided the limits on the right hand side both exist), while the "proof" here tacitly uses the incorrect (non-)theorem that $u_nlt cimplieslim_{ntoinfty}u_nlt c$. The correct theorem is $u_nlt cimplieslim_{ntoinfty}u_nle c$.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:10












  • $begingroup$
    @BarryCipra Oh yeah! That's the missing part here $u_n < c implies lim_{ntoinfty} color{red}{leq} c$. I'm going to delete this, don't want to contribute to the clutter. Thanks for the clarification
    $endgroup$
    – caverac
    Jan 11 at 16:26












  • $begingroup$
    @caverac, you're welcome for the clarification. The flaw is somewhat subtle (it took me, at least, a bit of thought to see what was wrong in it) and possibly worth preserving on that basis.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:34


















$begingroup$
Hmm... You are right because, $x^2+y^2leq 1$ which will imply that $(x,y)notin H.$ However, it's all about the approach used. In that scenerio, I believe mine is correct, right?
$endgroup$
– Omojola Micheal
Jan 11 at 16:04




$begingroup$
Hmm... You are right because, $x^2+y^2leq 1$ which will imply that $(x,y)notin H.$ However, it's all about the approach used. In that scenerio, I believe mine is correct, right?
$endgroup$
– Omojola Micheal
Jan 11 at 16:04












$begingroup$
This has the same general form as the OP's proof, but the error stems from something not present there. The OP is tacitly using the theorem that $lim_{ntoinfty}(u_n+v_n)=lim_{ntoinfty}u_n+lim_{ntoinfty}v_n$ (provided the limits on the right hand side both exist), while the "proof" here tacitly uses the incorrect (non-)theorem that $u_nlt cimplieslim_{ntoinfty}u_nlt c$. The correct theorem is $u_nlt cimplieslim_{ntoinfty}u_nle c$.
$endgroup$
– Barry Cipra
Jan 11 at 16:10






$begingroup$
This has the same general form as the OP's proof, but the error stems from something not present there. The OP is tacitly using the theorem that $lim_{ntoinfty}(u_n+v_n)=lim_{ntoinfty}u_n+lim_{ntoinfty}v_n$ (provided the limits on the right hand side both exist), while the "proof" here tacitly uses the incorrect (non-)theorem that $u_nlt cimplieslim_{ntoinfty}u_nlt c$. The correct theorem is $u_nlt cimplieslim_{ntoinfty}u_nle c$.
$endgroup$
– Barry Cipra
Jan 11 at 16:10














$begingroup$
@BarryCipra Oh yeah! That's the missing part here $u_n < c implies lim_{ntoinfty} color{red}{leq} c$. I'm going to delete this, don't want to contribute to the clutter. Thanks for the clarification
$endgroup$
– caverac
Jan 11 at 16:26






$begingroup$
@BarryCipra Oh yeah! That's the missing part here $u_n < c implies lim_{ntoinfty} color{red}{leq} c$. I'm going to delete this, don't want to contribute to the clutter. Thanks for the clarification
$endgroup$
– caverac
Jan 11 at 16:26














$begingroup$
@caverac, you're welcome for the clarification. The flaw is somewhat subtle (it took me, at least, a bit of thought to see what was wrong in it) and possibly worth preserving on that basis.
$endgroup$
– Barry Cipra
Jan 11 at 16:34






$begingroup$
@caverac, you're welcome for the clarification. The flaw is somewhat subtle (it took me, at least, a bit of thought to see what was wrong in it) and possibly worth preserving on that basis.
$endgroup$
– Barry Cipra
Jan 11 at 16:34













1












$begingroup$

I would modify your proof to make one tacit step explicit in the display: If $(x_n,y_n)to(x,y)$, then



$$3x+2y=3lim_{ntoinfty}x_n+2lim_{ntoinfty}y_n=lim_{ntoinfty}(3x_n+2y_n)=lim_{ntoinfty}5=5$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45












  • $begingroup$
    If your comment here was intended for dmtri, you can delete it (here) by hovering your cursor to the right of the timestamp until an X appears.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:56










  • $begingroup$
    It was meant for you as well. As you know, I made typos and you did well to make corrections. Thanks, once again!
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:57










  • $begingroup$
    OK. But I didn't even notice the typo!
    $endgroup$
    – Barry Cipra
    Jan 11 at 17:00
















1












$begingroup$

I would modify your proof to make one tacit step explicit in the display: If $(x_n,y_n)to(x,y)$, then



$$3x+2y=3lim_{ntoinfty}x_n+2lim_{ntoinfty}y_n=lim_{ntoinfty}(3x_n+2y_n)=lim_{ntoinfty}5=5$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45












  • $begingroup$
    If your comment here was intended for dmtri, you can delete it (here) by hovering your cursor to the right of the timestamp until an X appears.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:56










  • $begingroup$
    It was meant for you as well. As you know, I made typos and you did well to make corrections. Thanks, once again!
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:57










  • $begingroup$
    OK. But I didn't even notice the typo!
    $endgroup$
    – Barry Cipra
    Jan 11 at 17:00














1












1








1





$begingroup$

I would modify your proof to make one tacit step explicit in the display: If $(x_n,y_n)to(x,y)$, then



$$3x+2y=3lim_{ntoinfty}x_n+2lim_{ntoinfty}y_n=lim_{ntoinfty}(3x_n+2y_n)=lim_{ntoinfty}5=5$$






share|cite|improve this answer









$endgroup$



I would modify your proof to make one tacit step explicit in the display: If $(x_n,y_n)to(x,y)$, then



$$3x+2y=3lim_{ntoinfty}x_n+2lim_{ntoinfty}y_n=lim_{ntoinfty}(3x_n+2y_n)=lim_{ntoinfty}5=5$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 16:27









Barry CipraBarry Cipra

59.4k653126




59.4k653126












  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45












  • $begingroup$
    If your comment here was intended for dmtri, you can delete it (here) by hovering your cursor to the right of the timestamp until an X appears.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:56










  • $begingroup$
    It was meant for you as well. As you know, I made typos and you did well to make corrections. Thanks, once again!
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:57










  • $begingroup$
    OK. But I didn't even notice the typo!
    $endgroup$
    – Barry Cipra
    Jan 11 at 17:00


















  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45












  • $begingroup$
    If your comment here was intended for dmtri, you can delete it (here) by hovering your cursor to the right of the timestamp until an X appears.
    $endgroup$
    – Barry Cipra
    Jan 11 at 16:56










  • $begingroup$
    It was meant for you as well. As you know, I made typos and you did well to make corrections. Thanks, once again!
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:57










  • $begingroup$
    OK. But I didn't even notice the typo!
    $endgroup$
    – Barry Cipra
    Jan 11 at 17:00
















$begingroup$
Thanks a lot! It was a typo! (+1)
$endgroup$
– Omojola Micheal
Jan 11 at 16:45






$begingroup$
Thanks a lot! It was a typo! (+1)
$endgroup$
– Omojola Micheal
Jan 11 at 16:45














$begingroup$
If your comment here was intended for dmtri, you can delete it (here) by hovering your cursor to the right of the timestamp until an X appears.
$endgroup$
– Barry Cipra
Jan 11 at 16:56




$begingroup$
If your comment here was intended for dmtri, you can delete it (here) by hovering your cursor to the right of the timestamp until an X appears.
$endgroup$
– Barry Cipra
Jan 11 at 16:56












$begingroup$
It was meant for you as well. As you know, I made typos and you did well to make corrections. Thanks, once again!
$endgroup$
– Omojola Micheal
Jan 11 at 16:57




$begingroup$
It was meant for you as well. As you know, I made typos and you did well to make corrections. Thanks, once again!
$endgroup$
– Omojola Micheal
Jan 11 at 16:57












$begingroup$
OK. But I didn't even notice the typo!
$endgroup$
– Barry Cipra
Jan 11 at 17:00




$begingroup$
OK. But I didn't even notice the typo!
$endgroup$
– Barry Cipra
Jan 11 at 17:00











1












$begingroup$

I think you should have $3x+2y$ in the 3 line from the bottom....






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45


















1












$begingroup$

I think you should have $3x+2y$ in the 3 line from the bottom....






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45
















1












1








1





$begingroup$

I think you should have $3x+2y$ in the 3 line from the bottom....






share|cite|improve this answer









$endgroup$



I think you should have $3x+2y$ in the 3 line from the bottom....







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 16:31









dmtridmtri

1,4612521




1,4612521












  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45




















  • $begingroup$
    Thanks a lot! It was a typo! (+1)
    $endgroup$
    – Omojola Micheal
    Jan 11 at 16:45


















$begingroup$
Thanks a lot! It was a typo! (+1)
$endgroup$
– Omojola Micheal
Jan 11 at 16:45






$begingroup$
Thanks a lot! It was a typo! (+1)
$endgroup$
– Omojola Micheal
Jan 11 at 16:45




















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