My question concerns a quite elementary problem in set-theoretic topology: Assume that $(X,d)$ is a compact metric space. Consider the infinite product $X^{\mathbb{Z}}$ of all (two-sided) sequences in $X$. Besides the product topology, there are at least two other topologies on $X^{\mathbb{Z}}$ induced by metrics which are very natural:

(1) The topology induced by the sup-metric: $D_1(x,y) = \sup_{n\in\mathbb{Z}}d(x_n,y_n)$.

(2) The topology induced by the metric $D_2(x,y) = \sup_{n\in\mathbb{Z}}D_0( \theta^n x,\theta^n y)$, where $D_0$ is any metric on $X^{\mathbb{Z}}$ that induces the product topology and $\theta:X^{\mathbb{Z}} \rightarrow X^{\mathbb{Z}}$ is the left shift operator that sends a sequence $(x_n)_{n\in\mathbb{Z}}$ to $(x_{n+1})_{n\in\mathbb{Z}}$ ($\theta^n$ is the $n$-th iterate of $\theta$). The metric $D_0$ may be given by $D_0(x,y) = \sum_{n\in\mathbb{Z}}\frac{1}{2^{|n|}}d(x_n,y_n)$.

My question is: Are there any known characterizations of connectedness of $(X^{\mathbb{Z}},D_i)$, $i=1,2$, in terms of the properties of $(X,d)$?

  • $\begingroup$ Is it clear that the second topology does not depend on the choice of $D_0$? $\endgroup$ – YCor May 22 '19 at 12:57
  • $\begingroup$ No, that's actually not clear. In any case, the topology induced by the sup-metric is always finer than the topology induced by any of the metrics $D_2$: The identity map from $(X^{\mathbb{Z}},D_1)$ to $(X^{\mathbb{Z}},D_2)$ is continuous. There is an argument showing that it is strictly finer, but at the moment I don't remember it. $\endgroup$ – C. Kawan May 22 '19 at 14:18
  • $\begingroup$ I hope that the space $(X^{\mathbb Z},D_1)$ is connected if and only if for any $\varepsilon>0$ there exists $n\in\mathbb N$ such that any points $x,y$ in the metric compact space $(X,d)$ can be lined by a chain of points $x=x_0,\dots,x_n=y$ such that $d(x_{i},x_{i+1})<\varepsilon$ for every $i<n$. $\endgroup$ – Taras Banakh May 23 '19 at 13:36
  • $\begingroup$ Thank you for this very helpful hint. The property that you define (call it uniform connectedness) seems to be a necessary condition for the connectedness of the sequence space with metric $D_1$, but maybe not sufficient. There is an exercise in the book "General Topology" by Engelking (Ex. 6.1.D) that indicates that $X^{\mathbb{Z}}$ should be compact to draw the desired conclusion that uniform connectedness of $X$ implies connectedness of $X^{\mathbb{Z}}$. $\endgroup$ – C. Kawan May 24 '19 at 14:11
  • $\begingroup$ But your space $X$ is compact and so is its countable power. $\endgroup$ – Taras Banakh May 26 '19 at 11:31

As far as I can see, $(X^\mathbb{Z}, D_1)$ is connected if and only if $X$ is connected. It is clear that $X$ is a continuous image of $(X^\mathbb{Z}, D_1)$, so we only need to verify that the latter is connected if $X$ is.

Assume $X$ connected and consider the subspace $F \subset X^\mathbb{Z}$ consisting of sequences that take only finitely many distinct values. Since $X$ is totally bounded, $F$ is everywhere dense, so we only need to prove that $F$ is connected.

If we take arbitrary $a, b \in F$, there is a finite partition $P$ of $\mathbb{Z}$ such that both $a$ and $b$ are constant on each element of $P$. We can then define a mapping $i: X^P \to F$ by $i(x)(n) = x([n]_P)$, which has $a$ and $b$ in its image. If both spaces are endowed with the $\sup$ metric, this is easily seen to be an isometric embedding and because $X^P$ is connected, $a$ and $b$ lie in the same component of $F$. Thus $F$ is connected.

I have not thought about $D_2$, but if, as you suggested in a comment, there are continuous surjections $(X^\mathbb{Z}, D_1) \to (X^\mathbb{Z}, D_2) \to X$, that takes care of that.

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  • $\begingroup$ Yes, this solves the problem for $(X^{\mathbb{Z}},D_1)$, and consequently also for $(X^{\mathbb{Z}},D_2)$! Thank you! $\endgroup$ – C. Kawan Jun 4 '19 at 7:16

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