Limit of subsequence We want to find a sequence (Sn) such that for all scalars "a" [0,1], there exists a subsequence of (Sn) such that the limit of this subsequence = "a".
I am completely stuck. Is there a method that has already been proven to have subsequences with limits = all infinitely many value between 0 and 1?
 A: The rational numbers inside of $[0,1]$ are dense in $[0,1]$ and also they are countable. Therefore we can make a sequence $\{q_n\}$ that enumerates all of the rational numbers in $[0,1]$.
The density of the rationals tells us that for every $a\in [0,1]$ there is a subsequence of $\{q_n\}$ that converges to $a$.

When we say that a set is countable, that means there is a 1-1 correspondence with the integers. Thus there exists some function $f:\mathbb{N} \to \mathbb{Q}$ for which given any $q \in \mathbb{Q}$ there is an integer $n$ satisfying $f(n)=q$. The identification does not need to be explicit.
It can be made explicit. For instance, we can order the rationals by the sum of their numerators and denominators. So we count $0/1$ first, then $1/1$. Next comes $1/2$ and $2/1$ etc. And we delete copies on the way.
It's much easier using the $f$ we described above. We don't need an explicit enumeration for this problem.
Let's take $a \in [0,1]$. Now suppose that $M$ is an integer large enough so that $I_M=(a-1/M, a+1/M) \subset [0,1]$. If $a=1$, then we simply consider $I_M=(a-1/M, 1]$. By the density of the rationals, there is some rational number $p_1$ in $I_M \cap \mathbb{Q}$. This rational number has a corresponding integer $n_1$ such that $f(n_1)=p_1$.
Now consider $I_{M+1}$. We wish to find a rational number $p_2$ in $I_{M+1} \cap \mathbb{Q}$ for which it's corresponding integer $n_2$ (i.e. $f(n_2)=p_2$) is larger than $n_1$. Remember, for subsequences we need the index to be increasing. Note that there are an infinite number of rational numbers in $I_{M+1}$, but only a finite number of integers less than $n_1$. Therefore, there exists a rational number $p_2$ in $I_{M+1} \cap \mathbb{Q}$ for which its corresponding integer, $n_2$, is larger than $n_1$.
Iterate on this argument, and this produces a subsequence of $\{q_n\}$ converging to $a$.
