Hitting times in a Markov-chain are pairwise disjoint Let $(X_n)_{n\geq 0}$ be a Markov-chain with state space $E$.
Let $\tau_z=\inf\{n\geq 1: X_n=z\}$ be the smallest time where the Markov chain reaches the state $z\in E$ and $\{\tau_z=n\}=\{(X_0,...,X_{n-1})\in B, X_n=z\}$  with  $B=E\times(E\setminus\{z\})^{n-1}$.
Then the events $\{\tau_z=n\}$ and $\{\tau_z=k\}$ with $n\neq k$ are disjoint. My attempt:
If $k<n$ then $X_k=z$ because $\{\tau_z=k\}$, but $X_i\neq x$ with $i<n$ because $\{\tau_z=n\}$, so
$P(\{(X_0,...,X_{n-1})\in B, X_n=z\}\cap \{(X_0,...,X_{n-1})\in B, X_k=z\})=0$
Is this correct ?
 A: You have only argued that the intersection $\{\tau_z=n\}\cap\{\tau_z=k\}$ is a null set, i.e. that $P(\{\tau_z=n\}\cap\{\tau_z=k\})=0$, not that it is empty, i.e. $\{\tau_z=n\}\cap\{\tau_z=k\}=\emptyset$.
I'll show a very general result. Let $R\in\mathcal R$ be a random object, that is $R:\Omega\rightarrow\mathcal R$ is measurable, $\Omega$ is a probability space and $\mathcal R$ is a measurable space. Then for any measurable sets $\mathcal E,\mathcal F\in\mathcal R$ with $\mathcal E\cap\mathcal F=\emptyset$ we have $\{R\in\mathcal E\}\cap\{R\in\mathcal F\}=\emptyset$.
Proof: By definition we have $\{R\in\mathcal E\}\cap\{R\in\mathcal F\}=R^{-1}(\mathcal E)\cap R^{-1}(\mathcal F)=R^{-1}(\mathcal E\cap\mathcal F)=\emptyset$, since we always have $f^{-1}(A)\cap f^{-1}(B)=f^{-1}(A\cap B)$.
Now, take $R=\tau_z$, $\mathcal E=\{n\}$ and $\mathcal F=\{k\}$. This means: Your question is not related to hitting times or Markov chains, it is an example of a very general result, namely that the preimage of the intersection is the intersection of the preimages, in this case empty.
