Prove that $X_n(Y_n-c) \xrightarrow{P} 0$ I'm stuck with this exercise.
Let $X,X_n,Y_n (n=1,2,\ldots)$ random variables such that $X_n \Rightarrow X$ and $Y_n \xrightarrow{P} c$, where $c$ is constant. Prove that $X_n(Y_n-c) \xrightarrow{P} 0$.
I know that for all $\delta > 0$, there exist $M > 0$ and $N \in \mathbb{N}$ such that
$$P(|X_n| \geq M) \leq \delta \hspace{0.5cm} \forall n \geq N.$$
Then, I want to prove that
$$\lim_{n \to \infty}P[|X_n(Y_n-c)| \geq \epsilon ]=0,$$
Then for $N \in \mathbb{N}$ that for all $n \geq N$ and for $M >0$
$$P[|X_n(Y_n-c)| \geq \epsilon ] \leq P[|X_n| \geq M ] + P[|(Y_n-c)| \geq \epsilon/M] \leq \frac{\delta}{2}+\frac{\delta}{2}=\delta,$$
since the the first I mentioned and $Y_n \xrightarrow{P} c$.
Is this correct? In fact I'm not completely sure why I can take the first inequality but I feel this is the correct way. Any help?
 A: An explanation to your specific question: inequalities of the form $P(A \cap B) \geq 1-P(A) -P(B)$ and $P(A \cup B) \leq P(A) + P(B)$ emerge from basic properties such as
$$
P(A) + P(B) = P(A \cup B) - P(A \cap B)
$$
and
$$
P(A) \leq P(B), \quad A \subset B.
$$
It then depends on context what you want to handle, for example, if you have $C = \{X + Y > t\},$ then you can consider $A = \{X > t/2\}$ and $B = \{Y > t/2\},$ then $A^\complement \cap B^\complement \subset C^\complement$ so $C \subset A \cup B$ and $P(C) \leq P(A) + P(B).$ Similarly with $C = \{XY > ab\},$ $A = \{X > a\}$ and $B = \{Y > b\}$ (assuming everything is positive), then the same argument as with the sum shows that $P(C) \leq P(A) + P(B).$
Also, you approach is correct but I think it is easier to work the other way around. That is, to prove $P(|X_n(Y_n - c)| \leq \varepsilon) \to 1.$ Indeed,
$$
\liminf P(|X_n (Y_n - c)| \leq \varepsilon) \geq \liminf P(|Y_n - c| \leq \frac{\varepsilon}{M}, |X_n| \leq M) \geq 1 - \delta,
$$
for every $\delta > 0.$ (It is well-known that if $P(A_n) \to 1$ and $P(B_n) \geq \delta$ then $\liminf P(A_n \cap B_n) \geq 1 - \delta.$ Anyway, to prove it just notice that $P(A \cap B) \geq 1 - P(A) - P(B).$) Obviously, this and what you wrote are essentialy the same and it is just a matter of taste which one is preferred.
