# Connectedness of sequence spaces (countable products) in different metrics

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 $$(X,d)$$?

• Is it clear that the second topology does not depend on the choice of $D_0$?
– YCor
Commented May 22, 2019 at 12:57
• 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.
– user85365
Commented May 22, 2019 at 14:18
• 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 $i<n$. Commented May 23, 2019 at 13:36
• 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 $X^{\mathbb{Z}}$.
– user85365
Commented May 24, 2019 at 14:11
• But your space $X$ is compact and so is its countable power. Commented May 26, 2019 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.
• Yes, this solves the problem for $(X^{\mathbb{Z}},D_1)$, and consequently also for $(X^{\mathbb{Z}},D_2)$! Thank you!