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Find all limit points of given sets:

$A = \left\{ (x,y)\in\mathbb{R}^2 : x\in \mathbb{Z}\right\}$

$B = \left\{ (x,y)\in\mathbb{R}^2 : x^2+y^2 >1 \right\}$

I don't know how to do that. Are there any standard ways to do this?

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    A couple of standard ways to get started: 1) Draw pictures. 2) Review the definition and any previous examples you have seen of finding limit points. 3) Make a guess.2012-05-24

3 Answers 3

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1) Note that A is closed, therefore for every point limit $z$ of A we have that $z\in A$ ,also note that every point of A is limit point because $(a, b)\in A$ la sequence $(a_n) =\{(a, b+\frac{1}{n})\}$ is of points of A and $(a_n)\rightarrow (a, b)$.

2) The set of the points limit of B is $B\,\,'=\{x^2 +y^2=1\}$.

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    The set of limit points for $B$ is not $\{x^{2}+y^{2}=1\}$: that's its boundary. The set of limit points is $\{x^{2}+y^{2}\geq 1\}$.2012-05-25
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1) Is set A closed or not? If it is we're then done, otherwise there's some point not in it that is a limit point of A

2) As before but perhaps even easier.

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Here are a few hints. I would start by drawing a picture and think "what would be the closure of these sets, i.e. the smallest closed subset of $\mathbb{R}^{2}$ containing them?". If the sets are closed already, then this task is completed.

For A. Note that $\mathbb{Z}$ is a closed subset of $\mathbb{R}$ (can you show it?), the projection to the first component $pr_{1}:\mathbb{R}^{2}\to\mathbb{R}$ is continuous, and that the preimage of a closed set under a continuous function is closed. Can you conclude something to the set $A$ from all these?

For B. Again, what would be the smallest closed set containing $B$? Take a guess (you will probably guess correctly if you think for a while), and show that any point outside this "guess" has a strictly positive distance to $B$.