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I have the next exercise, but first, the definition that I have of limit points (for a better understanding).

Definiton: Let $\{y_k\}$ be a sequence such that $\{y_k\}\subset \mathbb{R}^n$. $w$ is a limit point of $\{y_k\}$ if there exist a subsequence $\{y_{k_{j}}\}\subseteq\{y_k\}$ such that $\lim\limits_{j\rightarrow\infty}y_{k_j}=w$.

and altenatively, the next:

Let $S$ be a subset of $\mathbb{R}^n$. A point $x$ in $\mathbb{R}^n$ is a limit point of $S$ if every open ball such that contains $x$, contains at least one point of $S$ different from $x$ itself

Ok, here is the problem:

Proof that doesn't exist a sequence $\{x_k\}\subset\mathbb{R}^2$ such that their only limit points are all the points in $\mathbb{Q}\times\mathbb{Q}$

My proof is by contradiction. We suposse that there exist such sequence and we call $\{x_k\}$. Then, we want to proof if all the points in $\mathbb{Q}\times\mathbb{Q}$ are limit points of $\{x_k\}$, then, all the points in $\mathbb{R}^2$ are limit points of $\{y_k\}$.

Proof

Let $\varepsilon>0$ and $y_0\in\mathbb{R}^{2}$. By the density of $\mathbb{Q}\times\mathbb{Q}$ in $\mathbb{R}^2$, we know that $\mathbb{Q}\times\mathbb{Q}\cap B_{\varepsilon}(y_0)\neq\emptyset$. Then, at most a point $z_0\in\mathbb{Q}\times\mathbb{Q}\cap B_{\varepsilon}(y_0)$

By the fact that $z_0\in\mathbb{Q}\times\mathbb{Q}$, then, by the hypothesis we know that there exist a subsequence $\{x_{k_j}\}\subseteq\{x_k\}$ such that $\lim\limits_{j\rightarrow\infty}x_{k_j}=z_0$. Now, consider the next image, fot a better understanding of the next inequalities.

The inequalities

Take $\delta=\varepsilon-d(z_0,y_0)>0$. We claim that $B_{\delta}(z_0)\subseteq B_{\varepsilon}(y_0)$

Let $a\in B_{\delta}(z_0)$, then, $$d(a,y_0)\leq d(a,z_0)+d(z_0,y_0)<\delta+d(z_0,y_0)=\varepsilon-d(z_0,y_0)+d(z_0,y_0)=\varepsilon$$Hence, $d(a,y_0)<\varepsilon$ and thus, $a\in B_{\varepsilon}(y_0)$.

Finally, $B_{\delta}(z_0)\subseteq B_{\varepsilon}(y_0)$ and by the convergence of the sequence, $\{x_{k_j}\}\cap B_{\delta}(z_0)\neq\emptyset$ and then $\{x_k\}\cap B_{\varepsilon}(y_0)\neq\emptyset$

Thus, all the points of $\mathbb{R}^2$ are limit points of $\{x_k\}$ and we can conclude that doesn't exist a such sequence.

My question is, really there exist a sequence such that all their limit points are all the points in $\mathbb{R}^2?$ Can we have the correspondence rule?

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Yes there really can exist a sequence whose limit points is equal to $\mathbb{R}^2$. Snake through the rational grid. This generates a bijection between natural numbers and rational coordinates. The sequence will have all points in $\mathbb{R}^2$ as a limit points.

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    $\begingroup$ Then, in $\mathbb{R}^n$, there exist a sequence such that all their limit points are $\mathbb{R}^n$. Wow, it is so exciting. $\endgroup$ – Carlos Jiménez Dec 11 '16 at 0:26

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