I am having difficulty proving the following result. Thanks for any help in advance.
Let $\{F_{n}, n\geq 1\}$ be a sequence of distribution functions and suppose that $\{F_{n}, n\geq 1\}$ is tight. Prove for every sequence of constants $\{c_{k}, k\geq 1\}$ in $(-\infty, 0)$ with $\lim_{k\to\infty} c_{k} = -\infty$ that
$\lim_{k\to\infty} \sup_{n \geq 1} F_{n}(c_{k}) = 0$
I think I know what I should do by the end of the proof. I believe that I can conclude that there exists some K such that for all $k \geq K$, $\epsilon$ > $F_{n}(b) \geq \sup_{n\geq 1}F_n(c_{k}) \geq 0$ where $\epsilon > 0$ and b is some appropriately chosen constant that I can choose thanks to the tightness supposition; however, I am unsure how to use this supposition beyond the definition.
EDIT: {$F_{n}$} is tight if $\lim_{a\to\infty} \sup_{n \geq 1} \int_{[|x| \geq a]} dF_{n}(x) = 0$ i.e. if $\lim_{a\to\infty}\sup_{n \geq 1} P(|X_n| \geq a) = 0$.
From your definition of tightness, all you need to do is see that if $c_k<0$, $F_n(c_k)\leq\mathbb{P}[|X_n|>-c_k]$. So $\lim_{a\rightarrow\infty}\sup_{n\geq 0}\mathbb{P}[|X_n|>a]=0\Rightarrow\lim_{k\rightarrow\infty}\sup_{n\geq 0}F_n(c_k)=0$.