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In this discussion: https://math.stackexchange.com/a/131987/114618 One of the posts argues that the sets $E = \{(x, y) \in \mathbb R^2 : y \ge 1/x\text{ and }x > 0\}$ and $F = \{(x, y) \in\mathbb R^2 : y \ge -1/x\text{ and }x < 0\}$ are both closed, but that their algebraic sum $$E + F = \{(x, y) \in \mathbb R^2 : y > 0\}$$ is not closed.

My question is: why would $F$ be closed in the first place? If I take the sequence $(x_n , y_n) = (1/n , n+1)$, with $n=1, 2, 3, ...$, this sequence would converge to a point on the y-axis (ie. where $x=0$), but that point would not be in $F$. Given that $F$ does not contain one of its limit points, how can we argue that it is closed?

Community
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user114618
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4 Answers4

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The problem is that there is no point on the axis that the sequence $(x_n , y_n) = (1/n , n+1)$ converges to. Your sequence approaches the axis, yes, but that is not enough for convergence - there is no limit point of that sequence.

If you lok at any sequence in the region $F$ that does converge to a point, then you will indeed find that the point is in $F$.

Old John
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  • So, OldJohn does not think that $(0,+\infty)$ is a point in $\mathbb R^2$. I agree with him. – GEdgar Dec 09 '13 at 14:16
  • @GEdgar Thanks! So long as I am not dealing with the Riemann sphere, I like to keep $\infty$ out of things :) – Old John Dec 09 '13 at 14:21
  • Thank you. Is this a proven result or is this still the subject of academic debate? Are the points $(0 , +{\infty})$, $(0 , -{\infty})$, $(+{\infty} , 0)$, $(-{\infty} , 0)$, $(+{\infty} , +{\infty})$ and $(-{\infty} , -{\infty})$ part of the set $\mathbb R^2$? – user114618 Dec 09 '13 at 14:29
  • Most people would not consider any of those points to be in $\mathbb{R}^2$. I certainly can't think of any textbook which would do so, so I wouldn't think it is the subject of academic debate. – Old John Dec 09 '13 at 14:35
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The $x$-coordinate of your sequence converges to $0$. But the $y$-coordinate does not converge.

J126
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$A=\{1+1,2+{1\over 2},3+{1\over 3},\dots\}$

$B=\{-1,-2,-3,\dots\}$

$A+B=\{1,{1\over 2},\dots\}$

Myshkin
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In addition to the answers pointing out that the sequence $(n+1)$ does not converge, let me mention that $$F=\{(x,y)\colon xy\le -1, x\le0\},$$ which shows that $F$ is closed, since $(x,y)\mapsto xy$ and $(x,y)\mapsto x$ are continuous.

Carsten S
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