I was wondering about the following problem:


Assume we have a state space $S:=\mathbb{Z}$ and a Markov chain, such that we can go from any state $x$ to some state $y$ with positive probabilities, i.e. $p\_t(x,y)>0$ for any $t >0 $ and $x,y \in S.$ 


Let $T\_0^x$ be the hitting time to go from state $x$ to $0$.


If we know that $P(\lim\_{x \rightarrow -\infty} T\_0^x<\infty)=1.$ So we can almost surely go to state $0$ from -infinity. Does this imply that $\liminf\_{x \rightarrow -\infty} p\_t(x,0)>0,$ for all $t>0$ i.e. does this imply that we go to state $0$ in every finite time step? 


The thing is that we a priori only know that we can go to $0$ in some finite time, but not necessarily in any finite time (with some positive probability.)



Finite hitting time implies hits at any finite time?
It's not clear what condition you are imposing, but I think the following example indicates that the answer is no, there are positive times so that you don't have to get to $0$ with uniformly bounded probability: 


Arrange the states in a tree with root $0$ with disjoint paths of length $n$ connected to the root for each $n\in \mathbb{N}$. On a path of length $n$, let the rate of movement toward the root be $n$. If $0 \lt t\lt 1$ and we start at the leaf of a path of length $n$, the distribution at time $t$ will be concentrated about $t$ of the way along the path. So, the limit inferior of the probability of reaching $0$ by time $t$ is $0$.


