Recently in something that I'm studying, I needed to know if the following map is transitive: $\sigma: M^{\mathbb{N}}\to M^{\mathbb{N}}$ the unilateral shift, where $M$ is a uncountable compact metric space. We are considering in $M^{\mathbb{N}}$ the topology of the cylinders.

1.The case in which $M$ is a finite set is a known fact. The answer is yes and the solution consists in construct a $\underline{x}\in M^{\mathbb{N}}$ just pasting all possible finite sequences of elements of $M$.

2.The second natural case is when $M$ is infinite countable, the solution is similar to the previous case

3.Given the method of item 1, in order to establish the transitivity, the solution of my problem seems to be false, because it seems to me that it will be necessary a non-countable number of iterates to go through all opens of $M^{\mathbb{N}}$

4.Require that $M$ has countable base as hypothesis can help in order to establish transitivity?

• The numbers $1/n$ together with $0$ is compact in the usual metric and countable. – Benjamin Steinberg Sep 28 '14 at 22:26
• you are absolutely right ! I got confused , I forgot that the hypothesis is necessary $M$ has no isolated points... – user11178 Sep 28 '14 at 22:46

Topological transitivity of the shift map $M^{\mathbb N} \to M^{\mathbb N}$ does indeed follow if you are willing to assume that $M$ has a countable basis, by a proof which is pretty similar to the proof for the case that $M$ is finite. Just paste together a sequence $\underline x$ such that, for each finite sequence of basis elements $B_0$, $B_1$, …,$B_K$ there exists $i$ such that $x_i \in B_0$, $x_{i+1} \in B_1$, …, $x_{i+K} \in B_K$.
As Lee Mosher noted if $M$ is a compact metric space, then the shift map on the product space $X=M^\mathbb{N}$ has a dense orbit and hence it is transitive. Let me add a few remarks: There are two notions known as "transitivity". Let $X$ be a topological space and $T\colon X\to X$ be a continuous map. We say that
• $T$ is transitive if for any nonempty open sets $U,V\subset X$ there is $n\ge 0$ such that $T^{-n}(U)\cap V\neq\emptyset$;
• $T$ has dense orbit if for some $x\in X$ the set $\{T^n(x):n=0,1,2,\ldots\}$ is dense in $X$.
The shift map on the product space $X=M^\mathbb{N}$ is always topologically transitive in the former sense. The proof follows basically from the definition of the product topology. If $M$ is separable and of second category then the exitsence of a dense orbit follows from the topological transitivity. In particular, if $M$ is a compact metric space, then both notions of transitivity agree for the shift map on the product space.