5
$\begingroup$

Given this set $A$: $$A =\left\{ \, (x,y) \mid x = 1/n, \ |y| \le n, \ n \in \mathbb{N} \, \right\} \subset \mathbb{R}^2; $$

I'd love to find the interior, closure, set of limit points, set of contour points (I doubt this is a correct English term) and similar things of this nature.

Set A

The interior is most likely empty. However it seems that one might need to explicitly add $y$ axis when specifying closure (that is $[A] = A \cup(\{0\} \times\mathbb{R})$ - kinda odd notation) and the same goes for set of limit points. It that correct? Is there some additional trickery?

$\endgroup$
  • 3
    $\begingroup$ The "trickery" is that you need to prove these things formally using their definitions. You can then be sure that there is no additional trickery. $\endgroup$ – Lord Soth Apr 9 '13 at 21:12
  • $\begingroup$ @LordSoth Fair enough. However I'm not too sure about the proper way to do this hence I would be grateful if somebody could demonstrate a formal proof in question for limit points. $\endgroup$ – Pranasas Apr 9 '13 at 21:21
  • 1
    $\begingroup$ What do you mean by contour point? $\endgroup$ – Stefan Hamcke Apr 9 '13 at 21:22
  • $\begingroup$ @StefanH. Whose every neighborhood contains both a point of $A$ and a point of $\mathbb{R}^2 \setminus A$. I don't know a proper English word for it. $\endgroup$ – Pranasas Apr 9 '13 at 21:31
  • 2
    $\begingroup$ I don't know if there is an official name. In German it is called Randpunkt which would be boundary point in English. $\endgroup$ – Stefan Hamcke Apr 9 '13 at 21:39
2
$\begingroup$

You got it all right intuitively.

Let $P=(x,y)$ be a limit point of $A$, i.e. all neighborhoods of $P$ intersects $A\setminus\{P\}$. In other words, there is a (non quasi-constant) sequence in $A$ that converges to $P$ (pick the $n$th element from the disk around $P$ of radius $1/n$, e.g.).

So, assume $A\ni (a_n,b_n)\to (x,y)$. That is, $a_n\to x$ and $b_n\to y$. Now the first coordinate of elements of $A$ form the set $\{1/k\,\mid\,k\in\Bbb N\}$, so, either $a_n\to 1/k$ for some $k$, or $a_n\to 0$. In the former case $a_n$ must be quasi-constant, i.e. $a_n=1/k$ for $n>n_0$ for some $n_0$, and then for these $|b_n|\le k$ must hold, so $(x,y)\in A$ will follow. This shows that the limit points $\subseteq A\cup(\{0\}\times\Bbb R)$.

For the other inclusion: if $(x,y)\in A$, then, one can find a sequence on the line segment of $(x,y)$ that converges there.

Finally, for $(0,r)$, let $N:=\left\lceil\, |r|\,\right\rceil\,\in\Bbb N$, and consider for example the sequence $(a_n,b_n)\in A$ with $$a_n:= \frac1{N+n},\quad\quad b_n:=r\,.$$ This is not quasiconstant and tends to $(0,r)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.