Finding limit points for these sets Here's my resoning for finding limit points for some sets. Could you guys read it and see if it's all good? <3
$$\{(x,y)\mid \ x^2+y^2<1\}$$
For this set, its kinda simple to see that every point can be a limit point, because every open ball with center in a point of this set, will have another point diferente from the center.
$$\{(x,1)\mid x=\frac{1}{N}, n\in \mathbb N\}$$
Besides having some holes (not the complete real line), given a point of this set, I can always construct an open ball, and then I try the points $\frac{1}{10}, \frac{1}{100}\cdots \frac{1}{1000000000}$ until I find one that is inside the ball. So every point of this set is a limit point.
$$\{(x,y)\mid x,y\in \mathbb Q\}$$
Same reasoning as before, but now in two dimensions. Would be nice to have a proof. Can somebody help me?
$$\{(x,y)\mid x,y\in \mathbb N\}$$
Here I can't always draw an open ball that contains another point except the one in the center, because I'm working with the integers. If I construct a ball with radius $\frac{1}{2}$ then no onter point is in it.
$$\{(x,y)\mid x=1, 1<y<2\}$$
Every point of this set is a limit point, by the same reasons...
 A: First, I am assuming that given a set $S \subset R^2$, we are looking for limit points of $S$ that are in $R^2$, not just in $S.$
I don't follow your arguments. Take for example the second question. It is not clear what you mean when you say, given a point of this set, I can always construct an open ball, and then I try the points $1/10,1/100⋯1/1000000000$ until I find one that is inside the ball.
In fact, none of the points in the set in the second example (let's call that set $S$) are limit points of that set. This is because for any point $x$ in that set, I can find an open ball in $R^2$ whose intersection with $S$ is $\{x\}$. (Prove this!) But the point $(0,1)$ is a limit point of that set.
A hint for the third question: Any open interval in $R^1$ contains rational points. This gives a hint for the fourth question: there are open intervals in $R^1$ that do not contain any integer points. You would have to extend these claims to similar statements about open balls in $R^2$.
A: I think we can find easily the limit points in complex plane or in $\Bbb R^2$ Plane,  as we find in case of Real line $\Bbb R$. 
I think the set set of limit points of set $\{(x,y)|x,y \in \Bbb N\}$  is empty set $\emptyset$ and hence it is closed set.
A: Let $S = \{ (x,y): x^2 + y^2 < 1 \} \subset \mathbb{R}^2$.

Firstly, let me define a slightly stronger term - Accumulation Point

A point $x$ is an accumulation point of a subset $A \subset \mathbb{R}^p$ if and only if every reduced neighborhood of $x$ has a nonempty intersection with $A$, that it, for every $r>0$ we have $B'(x,r)\cap A \neq \emptyset$.

We will denote the set of accumulation points of $S$ by $S'$.
Accumulation points are somewhat "stronger" than limit points, however we know that if a point is an accumulation point, the point is also a limit point. (The converse is not necessarily true).

So now, consider the following sketch of our region $S$.

Now, consider any arbitrary point $x \in S$ - I have drawn a few examples in the following sketch. Clearly for every $x \in S$ we have that for every $r>0$, $B'(x,r)\cap A \neq \emptyset$. Hence $S \subset S'$.

Now let us consider the boundary, that is, the set $S_b = \{ (x,y) | x^2 + y^2 = 1 \}$

Clearly for every $x \in S_b$ we have that $x \in S'$ since for every $r>0$, $B'(x,r)\cap A \neq \emptyset$. Thus $S_b \subset S' $
Now, let us lasty consider and point $x \in \{ (x,y): x^2 + y^2 > 1 \}$

For every $x \in \{ (x,y): x^2 + y^2 > 1 \}$ there exists an $r>0$ such that $B'(x,r) \cap A = \emptyset$. Thus $\{ (x,y): x^2 + y^2 > 1 \} \not\subset S'$.
From the above we may deduce that $S' = S\cup S_b = \{ (x,y): x^2 + y^2 \leq 1 \}$.
And thus we also have that $\{ (x,y): x^2 + y^2 \leq 1 \}$ are limit points of $S$.
