Take the simple two state Markov process characterized by transitions $$ \begin{aligned} 0\rightarrow1 & \quad \text{ at rate } \quad \alpha\lambda \\ 1\rightarrow0 & \quad \text{ at rate } \quad \alpha. \end{aligned} $$
If we let $\alpha$ go to infinity, then we effectively have a stationary process. No matter what the initial distribution, the process at any future time has distribution $\pi.$ Is that correct? I know that $\pi=(\pi_0,\pi_1)=\left(\frac{1}{1+\lambda},\frac{\lambda}{1+\lambda}\right)$ irrespective of $\alpha.$
What if we introduce another state $2$ with transitions $$ \begin{aligned} 0\rightarrow2 & \quad \text{ at rate } \quad \gamma_{02} \\ 1\rightarrow2 & \quad \text{ at rate } \quad \gamma_{12} \\ 2\rightarrow1 & \quad \text{ at rate } \quad \beta \\ 2\rightarrow0 & \quad \text{ at rate } \quad \delta. \end{aligned} $$
Now call this $3$ state process $X_\alpha.$
Now what happens as we let $\alpha\rightarrow\infty?$ Presumably $X_\alpha\overset{\alpha\rightarrow\infty}{\longrightarrow} X$ where $X$ is the process which can be characterized with what follows.
We can think of $X$ as having only two states: $\{0,1\}$ and $2.$
The transitions of $X$ are $$\{0,1\}\rightarrow2 \quad \text{ at rate } \quad \gamma_{02} \pi_0+\gamma_{12} \pi_1$$ where $\pi_i$ is as described above, and $$2\rightarrow\{0,1\} \quad \text{ at rate } \quad \delta+\beta.$$
Question: Is it ok to think of $X$ in this way (as a two-state process)? When $X$ is not in state $2$, we know its distribution is $\pi$, yes?
What would be the best way to formalize the idea of $X_\alpha\overset{\alpha\rightarrow\infty}{\longrightarrow} X?$ I.e. what mode of convergence should be used?