Let $(X_n)$ be a random sequence and $X,Z$ be two random variables. I need to prove a proposition of the following form:
Proposition 1: If $X_n\to X$ in probability, then there exist a random sequence $(Y_n)$ such that $X_n Y_n\to Z$ in probability.
Suppose that I have proven the following proposition for any state $\omega\in\Omega$:
Proposition 2 : If $X_n(\omega)\to X(\omega)$, then there exist a nonrandom scalar sequence $(y_n)$ such that $X_n(\omega) y_n\to Z(\omega)$.
Question: Can I show that proposition 2 implies proposition 1?
My idea is to use the following fact: $X_n$ converges to $X$ in probability if and only if every subsequence of $X_n$ has a sub-subsequence that converges to $X$ almost surely.
The problem is that I don't know how to define $(Y_n)$ in terms of $y_n$ (state by state) to make it work. Any idea is greatly appreciated.
The question is badly stated but here the answer to your question based on the corrections suggested in your comments:
Just define $Y_n(\omega)=\frac {Z(\omega)} {X_n(\omega)}$ if $X_n(\omega) \neq 0$ and $0$ otherwise. Then $Y_n$'s are random variables. Note that if $X_n(\omega)=0$ for infintely many values of $n$ the $Z(\omega)=0$ so $X_n Y_n \to Z$ at such points. At other points $X_nY_n=Z$ after some stage. So we actually have $X_nY_n \to Z$ almost surely!