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Let $\{x_n\}$ be "any" sequence containing all rationals. I have to prove that every real number is the limit of some subsequence. I know that rationals are dense in real. But, are not the order of the rationals in the sequence creating problem here ? How to pick rationals from this sequence.

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Pick any real number you want, call it $r$. Then let pick the first rational in your sequence that has difference with $r$ of less than $1$, say your number was $x_{n_1}$. Then pick the next rational that appears after $n_1$ in your sequence that has difference with $r$ of less than $1/2$. This can be done since you are just ignoring the first $n_1$ numbers of your sequence, and there are infinite many rationals that are within $1/2$ units from $r$. Call this number $x_{n_2}$. Can you continue the process?

Daniel Montealegre
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Suppose that $A$ is any real number, and suppose that $x_{n_k}$ is the subsequence (that we will construct).

Suppose we have found the terms $x_{n_1}, x_{n_2}, \dots, x_{n_m}$ and we show that it is possible to choose a term closer to $A$ than previous terms of the subsequence:

Let $\epsilon$ be the smallest distance between $A$ and any of the (finitely many) terms of the subsequence chosen so far - and also less than $1/2^m$, and note that there are infinitely many rationals in the interval $(A-\epsilon/2, A+\epsilon/2)$. This infinite collection of rationals exist somewhere amongst the sequence $\{x_n\}$, so there must be one of the $x_n$ with $n > n_m$ in the interval, and this gives the next term of the subsequence.

Since the choice of the first few terms is irrelevant for the limit, and it is clear that the subsequence converges to $A$, we are done.

Old John
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I think that i've managed another way to show this result.

We will use two facts: $1)$ Rationals numbers are dense in the reals and $2)$ if $(x_n)$ is a sequence in $\mathbb{R}$ such that every open ball (centered at $a$) contains terms $x_n$ of the sequence for arbitrarely large $n$, then $a$ is the limit of some subsequence of $(x_n)$.

Let $\phi: \mathbb{N} \rightarrow \mathbb{Q} $ be any function that enumerates all the rational numbers. Then $\phi(\mathbb{N}) = \mathbb{Q}$ and since the rationals are dense in the reals, it follows that $\overline{\phi(\mathbb{N})} = \mathbb{R}$. Choose any $a\in \mathbb{R}$. Therefore, for every $\epsilon>0 $ it is true that $B(a;\epsilon)\cap \phi(\mathbb{N}) \neq \emptyset$. Take any of those open balls, say $B(a;\epsilon_0).$ It remains to prove that $B(a;\epsilon_0)$ contains infenitely many terms of the sequence $\phi_n$. Hence, supposing by contradiction that $B(a;\epsilon_0)$ contains only finitely many terms of $(\phi_n)$, say: $\phi_{n_1},\dots ,\phi_{n_m}$, take $r = \mbox{min}\{d(a,\phi_{n_1}),\dots, d(a,\phi_{n_m})\}$ and set $B' = B(a;r)$. Therefore, by construction $B'$ is a ball in $\mathbb{R}$ which contains no rational number, a contradiction with the fact that the rationals are dense in the reals.

By the arbitrary choice of the open ball, it follows $B(a;\epsilon)$ contains infinetely many terms of of $\phi_n$ for every $\epsilon>0$. Therefore, using $2)$ we have proven that $a$ is the limit point of some subsequence of $\phi$.

user2345678
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