I need to prove that an irreducible Markov chain cannot have an absorbing state.
It looks quite intuitive and I suppose it directly follows from the definitions. Here's what I'm thinking.
On the contrary, suppose there exist an irreducible Markov chain with an absorbing state $i$. From the definition of irreducibility, it follows that all states communicate with each other. Therefore, there exists a state $j$ s.t. $i \rightarrow j$ and $j \rightarrow i$ are true. But since $p_{ii} = 1$ ($p_{ij}$ being the transition probability from state $i$ to $j$. ), therefore, $i \rightarrow j$ doesn't hold true.
Is there any other way to prove it?
So if $i $ is an absorbing state then there is "no way to exit it", but there is always a way to exit any state in an irreducible markov chain with more than only one state, a contradiction. I think it is as simple as it can get, other prooves would probably rely on this contradiction anyways.