I am interested in a simple condition for a continuous monoid on a topological space to be topologically transitive. My setting is as follows:
Let $X$ be a $2$nd countable topological space and let $M$ be a separable topological monoid. I assume also that $X$ is a Baire space. I have a right continuous action $\alpha:M\times X\to X$. The orbit of a point $x\in X$ is simply $orb(x):=\{ \alpha(m,x): m\in M \}$. I say that $X$ is topologically transitive if $\overline{orb(x)}=X$. I also denote $\alpha(m,A):=\{ \alpha(m,a): a\in A \}$ and $\alpha^{-1}(m,A):= \{ y\in X: \alpha(m,y)\in A \}$.
I know that for group actions, when $M$ is a group and has inverses, that the action is topologically transitive if and only if for every two nonempty open sets $U_1,U_2\subseteq X$, there exists a $g\in M$ such that $U_1 \cap \alpha(g,U_2)\neq \emptyset$.
My question is whether the same condition holds for monoids? I can see that topological transitivity would be equivalent to having for any $U_1,U_2\subseteq X$ elements $m_1,m_2\in M$ such that $\alpha^{-1}(m_1,U_1)\cap \alpha^{-1}(m_2,U_2)\neq \emptyset$.
The proof of this statement goes as follows. Since $X$ is topologically transitive, there exists some $x_0\in X$ such that $\overline{orb(x)}=X$. Hence, for any open $U_1,U_2\subseteq X$ we have that $\overline{orb(x)}\cap U_i \neq \emptyset$ for $i\in\{1,2 \}$. So there exists $m_i \in M$ such that $\alpha(m_i,x_0)\in U_i$. Since $\alpha$ is continuous, we know that $\alpha^{-1}(m_i,U_i)$ is a neighbourhood of $x_0$ for $i\in \{1,2\}$.
When $M$ is a group, $\alpha(m,\cdot)$ is a homeomorphism for all $m\in M$ and we have $\alpha^{-1}(m,U_i)=\alpha(m^{-1}, U_i)$. Hence, we can say that $U_1\cap\alpha(m_2^{-1}m_1,U_2)\neq \emptyset$.
I am unsure what I am missing, but I saw somewhere that the original condition should also hold for monoids. I would appreciate anyone pointing out what I am missing.
** Later edit:**
Following @Carlyle's comments I notice that I omitted some assumptions on $X$ and $M$. Namely that $X$ is a $2$nd countable Baire space and $M$ is separable. The following is an adapted argument from a paper by Mike Hochman about genericity in topological dynamics.
Let $\{ U_i \}_{i=1}^\infty$ be a countable base of $X$ and let $\{m_j\}_{j=1}^\infty \subseteq M$ be a dense countable subset of $M$. The assumption on intersection, implies that $\cup_{i=1}^\infty \alpha^{-1}(m_i,U)$ is dense in $X$, for any open nonempty $U\subseteq X$.
We know that $x\in X$ has dense orbit if $x\in \cup_{j=1}^\infty \cup_{i=1}^\infty \alpha\big( m_j, \alpha^{-1}(m_i,U)\big)$ for all $U$ open. Hence the set of points with dense orbits are given by $T:=\cap_{\ell=1}^\infty \cup_{j=1}^\infty\cup_{i=1}^\infty \alpha\big( m_j, \alpha^{-1}(m_i,U_\ell)\big)$. Since it is a $G_\delta$ dense set in a Baire space, $T$ is dense and not empty.