Banach limits in metric spaces Let $X$ be a compact metric space.  Given a sequence $x_n \in X$ and an ultrafilter $\mathcal{U}$ on $\mathbb{N}$, we can define the "Banach limit" of ${x_n}$ with respect to $\mathcal{U}$. This limit is the unique element of $X$ such that every neighborhood of it contains a $\mathcal{U}$-large collection of the sequence ${x_n}$.  However, this process is not necessarily translation-invariant. When $X$ is a closed interval in the reals, then one can define a Banach limit by using an ultrafilter and taking the ultrafilter-limit of the Cesaro means.
Is there a way to define a translation-invariant "limit" with reasonable properties for sequences taking values in an arbitrary compact metric space?
 A: Let $X = \lbrace 0,1\rbrace$ be the compact metric space with exactly two points.  Then the space of sequences is just $\Sigma_2^+ = \lbrace 0,1 \rbrace^\mathbb{N}$, and an ultrafilter $\mathcal{U}$ defines a Banach limit in a fairly straightforward way:  for $i=0,1$, write $A_i = \lbrace n \in \mathbb{N} \mid x_n = i \rbrace$, and $\lim_\mathcal{U} x_n = i$ if and only if $A_i \in \mathcal{U}$.  This limit is translation invariant if and only if the ultrafilter itself is, so your question reduces to the existence of a translation invariant ultrafilter on $\mathbb{N}$.
I believe that no such ultrafilter exists, and hence the answer to your question is "no"; however, I'm not particularly knowledgeable about ultrafilters, so I welcome correction or confirmation on this last point.
A: Every Tychonoff space (and consequently every compact metric space) X can be embedded into the topological power $[0,1]^C$ (the functions from C to [0,1] with the usual product topology, i.e., the pointwise convergence).
If you work in $[0,1]^C$, then you can define the "limit" of a sequence $(f_n)$ by putting $L(f_n)(c)=\mathcal{U}-lim f_n(c)$, i.e. you can work pointwise. This limit is obviously shift-invariant and every convergent sequence is convergent in this sense, too.
However, I do not like calling such a thing Banach limit, since it is different from what you have seen in the reals. The usual Banach limit respects the usual linear structure of reals. If you use this construction, then this structure is lost by arbitrarily choosing the embedding into the cube $[0,1]^C$.
