# Topology for Divergent Sequence

Let $a_n$ be a sequence of real numbers and define the "Cesàro limit" of $a_n$ to be

$C \lim\limits_{n\rightarrow \infty}{a_n} = \lim\limits_{N\rightarrow\infty}{\frac{1}{N}} \sum\limits_{n=1}^{N}{a_n}$.

I'd like to give $\mathbb{R}$ a topology that gives rise to this mode of convergence. However, if we take the closure of a set as the set of all limit points of that set (i.e. $Cl(S)$ is set of real numbers which are Cesàro limits of sequences of numbers in $S$), then it's clear to me that $Cl(S)$ is the (closure of) the convex hull of $S$. This ensures that this closure operator cannot come from a topology, since $Cl(S\cup T) \neq Cl(S)\cup Cl(T)$ whenever $Cl(S)\cup Cl(T)$ is not connected.

My question is then this. Is there a generalization of "topology" that would allow such a thing? Is there anything useful that can be gained from this definition?

The only topology on $\mathbb R$ for which the Cesaro mean is a limit point of every Cesaro-convergence sequence is the indiscrete topology.
Let $U$ be a non-empty open set in our topology. Choose $x\in U$. For any $y\in\mathbb R$, consider the sequence $$\{x+y,x-y,x+y,x-y,\cdots\}$$ This sequence is Cesaro-convergent to $x$. Thus $U$ must contain a tail of the sequence. Discarding any finite number of terms still leaves us with $x-y,x+y\in U$. Since $y$ is arbitrary, it follows that $U=\mathbb R$. Thus the only non-empty open set is $\mathbb R$, which means the topology is given by the indiscrete topology: $\{\varnothing,\mathbb R\}$.