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I was working on a problem and wanted to use limit points and their sequences to show that a set is open (by showing it's complement is closed). I got through the whole thing only to realize I didn't show that "the complement contained all of its limit points" but rather that "every point in the complement was a limit point" which is now obvious to me is definitely not the same thing. I added an image of the problem and my (wrong) proof. I'm now stuck scratching my head over how one could use limit points to show a set is closed/open... If I can't figure it out, I'll probably resort to using open balls or something. Thanks!

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amazonprime
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  • you have started with any point of $A^c$. whereas to prove it is closed using limit point concept you should show that $D(A^c)$ is contained in $A^c$ – Meow Jan 27 '16 at 06:25
  • rather here even proving that set of adherent points of $A^c$ is $A^c$ itself will also be enough. – Meow Jan 27 '16 at 06:26

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You just need to show that if $x(n) \to x$, and $x(n) \in A^c$ for all $n$, then $x \in A^c$.

Use continuity of addition to show this is true here.

copper.hat
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A set $X \subseteq \mathbb{R}^n$ is closed if it contains all of its limit points, so to show that $X$ is closed take a limit point $x \in \mathbb{R}^n$ of $X$ and use the fact that there exists a sequence $(x_n)_n$ in $X$ converging to $x$ to show that $x \in X$.

In your case if $(x_n,y_n) \to (x,y)$ and $(x_n,y_n) \in A$ for all $n \in \mathbb{N}$, then $x_n + y_n \geq 0$, and then by the usual limit rules (i.e. continuity of $+$, monotony) it is not difficult to show $x + y \geq 0$, that is, $(x,y) \in A$.

Matthias Klupsch
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To prove $A^c$ is closed, P.T. $cl(A^c)=A^c$ instead of looking at the derived set. Hint : take a point $(a,b)$ of cl($A^c$) and use the fact that there exists sequence $(x_n,y_n)$ of points of $A^c$ converging to it.(u dnt actually need to find one). then by usual limit rules prove that $a+b\leq 0$ So it belongs $A^c$.

Meow
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  • One thing I don't understand is why the fact that the sequence converging to the point (a,b) that we pick implies that (a,b) is in the set. In the problem, the origin is not part of the set A, yet I can easily find a sequence of points in A that converge to (0,0), but I only see that this makes (0,0) a limit point of A, not that (0,0) is an element of A. – amazonprime Jan 27 '16 at 16:20
  • the point will be in set if it satisfies the given condition. the points of seqnc are already in the set so they stisfy the condn – Meow Jan 28 '16 at 17:09