Is there a countable dense set $A$ in $ X$ with $f_\lambda(A)\subseteq A$, for a family of continuous maps $f_\lambda$ on compact metric space $X$?

Question

Let $X$ and $\Lambda$ be compact metric spaces. Consider continuous map $f:\Lambda\times X\to X$. Is there a countable dense set $A\subseteq X$ such that for every $\lambda\in \Lambda$, we have $f_\lambda(A)\subseteq A$.

Answer 1

The claim is false. Consider the complex multiplication $S^1\times S^1\to S^1$ and note that for any proper subset $A\subseteq S^1$ there is $z$ such that $zA\not\subseteq A$. We can find such $z$ by taking $vw^{-1}$ for $v\not\in A$ and $w\in A$.

So we need stronger assumptions. One might be: $X$ has a countable dense subset and $\Lambda$ is countable. None of them has to be compact.

In that case, let $D$ be a countable dense subset of $X$. For any $\lambda\in\Lambda$ and $n\in\mathbb{N}$ define

$$D(n,\lambda)=f_\lambda^n(D)$$

where $f_\lambda^n$ is $f_\lambda$ composed with itself $n$ times. Note that $f_\lambda^0$ is the identity. Let

$$A=\bigcup_{n\geq 0, \lambda\in\Lambda}D(n,\lambda)$$

I leave as an exercise that it satisfies our conditions.