Proof that if $X_n \Rightarrow X$ and $\rho(X_n,Y_n) \Rightarrow 0$, then $Y_n \Rightarrow X$

Question

I am looking at the proof of the following Theorem.

Suppose that $(X_n,Y_n)$ are random elements of the product metric space $S \times S$. If $X_n \Rightarrow X$ and $\rho(X_n,Y_n) \Rightarrow 0$, then $Y_n \Rightarrow X$.

The proof utilises the following inequality.

If $F$ is closed and $F_\epsilon$ is defined as $[x:\rho(x,F)\le \epsilon]$, then $$P[Y_n \in F] \le P[\rho(X_n, Y_n)\ge \epsilon] + P[X_n \in F_\epsilon].$$ How do we achieve this inequality?

Answer 1

It follows from the very definition of $F_{\epsilon}$ that

$$y \in F, x \notin F_{\epsilon} \implies \varrho(x,y) \geq \epsilon$$

for any $x,y \in S$. Thus,

$$\begin{align*} \{Y_n \in F\} &= \{Y_n \in F, X_n \in F_{\epsilon}\} \cup \{Y_n \in F, X_n \notin F_{\epsilon}\} \\ &\subseteq \{X_n \in F_{\epsilon}\} \cup \{\varrho(X_n,Y_n) \geq \epsilon\} \end{align*}$$