Let $G$ be a group acting transitiviely on a set $\Omega$. A nonempty subset $\Delta$ of $\Omega$ is called a block for $G$ if for each $x \in G$ either $\Delta^x = \Delta$ or $\Delta^x \cap \Delta = \emptyset$.
(Separation Property) Suppose that $G$ is a group acting transitively on a set $\Omega$ with at least two points, and that $\Delta$ is a nonempty subset on $\Omega$. Show that $\Delta$ is not a block $\Leftrightarrow$ for each pair of distinct points $\alpha,\beta \in \Omega$ there exists $x \in G$ such that exactly one of $\alpha$ and $\beta$ lies in $\Delta^x$. In the case that $G$ is finite, show that the condition can be strengthened to: $\alpha \in \Delta^x$ but $\beta \notin \Delta^x$ for some $x \in G$.
My Attempt: i) Suppose $\Delta$ is not a block, then $|\Delta| > 1$. As $\Delta$ is not a block, there exists some $g \in G$ such that $$ \Delta^g \cap \Delta \ne \emptyset \quad\mbox{and}\quad \Delta^g \ne \Delta. $$ This show we have some $\alpha \in \Delta^g \cap \Delta$ and some $\beta$ contained in exactly one of $\Delta$ and $\Delta^g$ as they are different. If $\beta \in \Delta$, then $\alpha \in \Delta^g$ and $\beta\notin \Delta^g$ and the condition is fulfilled with $x = g$, otherwise $\beta \notin \Delta$ and the condition is fulfilled with $x = 1$.
ii) Now suppose $\Delta$ is a non-trivial block. Then $|\Delta| > 1$ and for each $\alpha, \beta \in \Delta$ and $x \in G$ we have either $$ \alpha, \beta \in \Delta^x \quad\mbox{or}\quad \alpha, \beta \notin \Delta^x $$ as we have either $\Delta^x = \Delta$ or $\Delta^x \cap \Delta = \emptyset$, where the first case implies $\alpha, \beta \in \Delta^x$ and the second case implies $\alpha, \beta \notin \Delta^x$. So by fixing some $\alpha, \beta \in \Delta$ we see that the condition is never fulfilled.
My Questions:
1) If we consider the trivial case in part ii) of the above proof. Then if $\Delta = \Omega$, as $\{ \Delta^x : x \in G \}$ partitions $\Omega$ and no $\Delta^x$ could be empty, we have $\Delta^x = \Omega$ for each $x \in G$, hence the condition also fails, but for the other trivial case, i.e. if $\Delta$ is the trivial block $\Delta = \{ \gamma \}$ for some $\gamma \in \Omega$, then by transitivity $\{ \Delta^x : x \in G \} = \{ \{ \beta \} : \beta \in \Omega \}$ and then for each distinct $\alpha, \beta$ we certainly find a withness to separate them, for example choose $x \in G$ such that $\alpha = \gamma^x$, then $\Delta^x = \{\alpha\}$ and $\beta \notin \Delta^x$.
But in the exercise it is nowhere mentioned that the trivial case should be excluded, so is this an error in the exercise or have I understood something wrong?
2) The additional part if $G$ is finite.
In my oppinion we can always find $x \in G$ such that $\alpha \in \Delta^x$ and $\beta \notin \Delta^x$, also if $G$ is infinite. For if by the condition we have $\alpha \notin \Delta^x$ and $\beta \in \Delta^x$, then also as $\{ \Delta^y : y \in G \}$ partitions $\Omega$ we have some $y \in G$ such that $\alpha \in \Delta^y$. But then as $\Delta^x \ne \Delta^y$ they must be disjoint as different sets of the partition, hence $\beta \notin \Delta^y$.
Have I overlooked something, I do not see why finiteness should be essential here?
The exercise is from the well-known book Permutation Groups by J.D. Dixon and B. Mortimer.