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I would like to find a sequence of real numbers $(a_n)_ninmathbbN$ with this property: for any $LinmathbbR$ there is a subsequence $a_k_n$ such that $$lim_ntoinfty a_k_n = L$$ Does such a sequence exist?

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 $a_n_k to L$.

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.

Rephrasing:

Consider $a_n$, $n=1,2,3,3,....,$ the sequence of rational numbers. Recall that $mathbbQ$ is countable, hence can be written as a sequence $a_n$, $nin mathbbN$.

Let $L in mathbbR$.

Since $mathbbQ$ is dense in $mathbbR$, we can construct a subsequence $a_n_k$ that converges to $L$.

Consider a general topological space $(X, mathscrT)$ and a sequence of points $x in X^mathbbN$. 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:

$$(forall V, n)(V in mathscrV_mathscrT(x) wedge n in mathbbN implies (exists m)(m geqslant n wedge x_m in V ))$$

where $mathscrV_mathscrT(x)$ symbolizes the filter of neighbourhoods of point $x$ induced by the topology $mathscrT$. 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 $sigma: mathbbN rightarrow M$ such that for each $t in M$ the fibre $sigma^-1(t)$ be infinite.

Consider such a surjection $sigma: mathbbN rightarrow T$ and define the sequence $t=(sigma(n))_n in mathbbN$ (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 $(X, mathscrT)$ 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 $(X, mathscrT)$ 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 $(mathbbR, mathscrO)$, where $mathscrO$ denotes the (usual) order topology.