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Is it possible to find a sequence $\{a_n\} \in \mathbb{R}^n$ with the property that for any $x \in \mathbb{R}^n$, $\{a_n\}$ has a subsequence that converges to $x$?

I tried approaching it by first finding a sequence in $\mathbb{R}$ with the above property. Consider a sequence $\{b_n\} \in \mathbb{Q}$ that contains every rational number an infinite number of times. It should be possible to construct such a sequence because the rationals are countable. We also know by the construction of $\mathbb{R}$ that every real number $x$ has some sequence $\{x_i\} \in \mathbb{Q}$ that converges to it. And as each rational number will appear within $\{b_n\}$ an infinite times, we can certainly find $\{x_i\}$ as a subsequence of $\{b_n\}$.

I'm stuck when it comes to generalizing this process to $\mathbb{R}^n$. Perhaps we could consider all possible n-tuples of rational numbers and put these in a sequence so that each shows up an infinite number of times? Is that even countable? I'm unsure as to how I can state this rigorously. I'm only looking for a little insight, as I'd like to see the solution for myself.

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$\mathbb Q^n$ is countable and is dense in $\mathbb R^n$. Therefore every point is a limit point of the sequence given by the enumeration and hence a subsequence converging towards that point can be selected.

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Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and countable, the set of $n$-tuples of $\mathbb{Q}$, $\mathbb{Q}^n$, is also countable and dense in $\mathbb{R}^n$. Since it is countable, let $\{a_n\}_{n \in \mathbb{N}}$ denote a sequence enumerating these $n$-tuples of rational numbers. Then by denseness, this sequence has the desired property.