Let $(X_n)_{n\geq1}$ be a sequence of random variables weakly converging to $X$ ($X_n\Rightarrow X$) as $n\rightarrow \infty$. I am wondering if this implies that $X_n-X\Rightarrow 0$ as $n\rightarrow \infty$? I know the converse is true via continuous mapping, but I would like to know if the two statements are actually equivalent? I believe it should be true as we could use the Skorokhod representation theorem to get the almost sure convergence of the difference to $0$, and thus the weak convergence of the difference to $0$, but I am not sure if this is correct? Any ideas or comments would be greatly appreciated.
edit: my reasoning using SRT was as follows: $X_n\Rightarrow X$ so there exist copies $Y_n$ and $Y$ with respective laws equal to those of $X_n$ and $X$ on another probability space (say $(\Omega,\mathcal{A}, \mathbb{Q}))$ such that $Y_n\rightarrow Y$ $\mathbb{Q}$-a.s. As $Y\rightarrow Y$ $\mathbb{Q}$-a.s, then $Y_n-Y\rightarrow 0$ $\mathbb{Q}$-a.s and therefore $Y_n-Y\Rightarrow 0$, and $X_n-Y\Rightarrow 0$. There must be a mistake here somewhere, but I can't seem to see where?
The answer is no. Weak convergence does not require the $X_n$ and $X$ to live on the same probability space. Even if they all do, here's a counterexample. Toss a fair coin once and let $X$ be the number of heads seen, let $Y$ be the number of tails seen (so $X+Y=1$), and define $X_n=Y$ for all $n$. Then everybody has the same distribution, so $X_n$ converges weakly to $X$. However, the difference $X_n-X=1-2X$ does not converge weakly to zero; it takes values $-1$ and $1$ with equal probability.