Assume that $\{X_t:t\geq0\}$ is a continuous-time Markov chain with values on a discrete (countable, but possibly not finite) set $S$.
Notation: for $i,j\in S$ and $t\geq0$,
$p_{ij}(t):=\mathrm{P}\{X_t=j|X_0=i\}$.
I want to prove that, for fixed $i$ and $j$ in $S$, the existence of a $s>0$ so that
$p_{ij}(s)>0$
implies
$\forall t>s, p_{ij}(t)>0$.
I've tried lots of things (within my limited knowledge on the subject) and researched as far as I could, but I can't figure it out and I'm extremely frustrated because it seems really easy.
I'll greatly appreciate detailed answers, since I'm afraid of not understanding the intermediate steps.
EDIT: As I predicted, it's actually quite simple, but I still want to understand all the details (see comments below).
The easiest way to see this is that in a continuous-time Markov chain, you always have a positive probability of staying put in a state for any amount of time.
(The time it takes to leave a state has the $\operatorname{Exp}(\lambda)$ distribution, where $\lambda$ is the sum of all transition rates out of that state, and the exponential distribution is unbounded.)
If we make up the notation $p_i^*(t)$ to denote the probability of staying put in state $i$ for $t$ units of time, then a lower bound on $p_{ij}(t)$ is $p_{ij}(s) \cdot p_j^*(t-s)$: we definitely end up going from $i$ to $j$ in $t$ seconds if we go from $i$ to $j$ in $s$ seconds and then stay put for $t-s$ seconds more. Both factors are positive, so the result is positive.