Sequence such that every subsequence can have a different real limit I would like to find a sequence of real numbers $(a_n)_{n\in\mathbb{N}}$ with this property: for any $L\in\mathbb{R}$ there is a subsequence $a_{k_n}$ such that $$\lim_{n\to\infty} a_{k_n} = L$$ Does such a sequence exist?
 A: Just arrange the set of rational numbers in a sequence $\{a_n\}$. Given any real number $L$ and any positive integer $n$ there are infinitely many rationals in $(L-\frac 1 n, L+\frac 1 n)$. Pick $a_{n_1}$ in this interval with $n=1$. Then consider the case $n=2$. You can surely find $n_2 >n_1$ such that $a_{n_2} \in (L-\frac 1 2, L+\frac 1 2)$. Use induction to construct a subsequence $a_{n_k}$ such that $|a_{n_k}-L| <\frac 1 k$ for all $k$.  Then $a_{n_k} \to L$.
A: Take the sequence that sweep the interval $[-1,1]$ by $1/2$ steps, then the interval $[-2,2]$ by $1/2^2$ steps, then the interval $[-n,n]$ by $1/2^n$ steps... and so on.
You’ll be able to prove that every real is a limit point of that sequence.
A: Rephrasing:
Consider $a_n$, $n=1,2,3,3,....,$ the sequence of rational numbers. Recall that $\mathbb{Q}$ is countable, hence can be written as a sequence $a_n$, $n\in \mathbb{N}$.
Let $L \in \mathbb{R}$.
Since $\mathbb{Q}$ is dense in $\mathbb{R}$, we can construct a subsequence $a_{n_k}$ that converges to $L$.
A: Consider a general topological space $(X, \mathscr{T})$ and a sequence of points $x \in X^{\mathbb{N}}$. One says that point $t \in X$ is adherent to the sequence $x$ (some authors use the terminology 'cluster-point', but I don't fancy it so much) if:
$$(\forall V, n)(V \in \mathscr{V}_{\mathscr{T}}(x) \wedge n \in \mathbb{N} \implies (\exists m)(m \geqslant n\ \wedge x_m \in V ))$$
where $\mathscr{V}_{\mathscr{T}}(x)$ symbolizes the filter of neighbourhoods of point $x$ induced by the topology $\mathscr{T}$. In a more descriptive fashion, $t$ is adherent to sequence $x$ if any neighbourhood of $t$ contains terms of arbitrarily high rank from the sequence $x$. If the filter of neighbourhoods of $t$ admits a countable base, then $t$ can be expressed as the limit of a subsequence of $x$. Therefore, in a space satisfying the First Axiom of Countability (i.e. all points have a countable base of neighbourhoods), the points adherent to a given sequence $x$ can be equivalently characterised as limits of subsequences of $x$.
Now, if the space $X$ is non-empty and separable, let us fix a certain dense subset $T \subseteq X$. As $X$ is non-empty, so must $T$ be. It is not difficult to show that a non-empty countable set $M$ admits a surjection $\sigma: \mathbb{N} \rightarrow M$ such that for each $t \in M$ the fibre $\sigma^{-1}(\{t\})$ be infinite. 
Consider such a surjection $\sigma: \mathbb{N} \rightarrow T$ and define the sequence $t=(\sigma(n))_{n \in \mathbb{N}}$ (which is actually the graphic of map $\sigma$). The condition on the cardinality of the fibers ensures that any element $x \in T$ is the limit of a (constant) subsequence of $t$. If we furthermore assume that the space $(X, \mathscr{T})$ is $T_1$, then any $x \in X \setminus T$ will be an accumulation point of $T$ and thus necessarily adherent to sequence $t$.  
To conclude, given a topological space $(X, \mathscr{T})$ that is non-empty, separable, first countable and $T_1$, one can always find a sequence $t$ of points within the space such that each element $x \in X$ be expressible as the limit of a subsequence of $t$. This applies in particular to the topological space $(\mathbb{R}, \mathscr{O})$, where $\mathscr{O}$ denotes the (usual) order topology.
