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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?

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    What does translation mean in an arbitrary compact metric space?2010-08-08
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    I meant that translating the sequence (i.e., on $\mathbb{N}$) does not affect the limit.2010-08-08
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    Ah, I see. I would call that operation "shift", or something like that, but I believe that your terminology is also reasonably standard. Sorry for not picking up on it.2010-08-08

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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.

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    Thanks! I think you're right about no translation-invariant ultrafilters: that would imply that a U-limit of real numbers could be made translation-invariant, and then that would be a problem by the following argument (which I found by googling). Consider the sequence s=1,0,1,0... It's U-limit is either zero or one by definition. The shift is also 1-s so has to have a U-limit of one or zero, the opposite as of s, meaning that U-limits are not translation-invariant.2010-08-08
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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 $[0,1]^C$.