In Pearl's Causality - Models, Reasoning and Inference (2009), he defines d-separation as follows:
Let $X\perp\!\!\!\perp Y |Z$ mean "$Z$ d-separates $X$ from $Y$".
But there seems to be a weird edge case that satisfies this criterion, but actually shouldn't: We can have a situation where $X\subset Z$. E.g. imagine the case where $X=Z=\{V\}$, and $Y=\{W\}$.
Then we want that $X\perp\!\!\!\perp Y |Z$ holds, because we want $X\perp\!\!\!\perp Y |Z$ to coincide with the probabilistic notion of conditional independence, and clearly $X$ and $Y$ are conditionally independent given $X$. However, consider a simple graph consisting of nodes $\{V,W\}$, with an edge $V\to W$. Here, the path between $V$ and $W$ is not blocked, $X$ and $Y$ are not d-separated.
What am I missing here?
When at least one of the two nodes is fixed by the set of conditions, then it is trivial that the two nodes shall be conditionally independent, whatever other conditions ($C$) may apply, by the definition of conditional probability.
$$\begin{align}\mathsf P(X{=}x, Y{=}y\mid X{=}x, C)~&=~\mathsf P(X{=}x\mid X{=}x, C)\,\mathsf P(Y{=}y\mid X{=}x, X{=}x, C)\\[1ex]&=~\mathsf P(X{=}x\mid X{=}x, C)\,\mathsf P(Y{=}y\mid X{=}x, C)\end{align}$$
But neither of those end nodes is a chain, fork, or collider on any path between them, so they do not meet any of the criteria to cause d-separation of any path.
If there are no other nodes on the net d-seperates those two nodes, then they will not be d-separated.
That is okay, because d-separation on a DAG does not coincide with conditional independence.
d-separation is sufficient but not necessary for conditional independence between two nodes.