Is it correct to say that ($\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \supseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$? 
Let $W_0, W_1, W_2, \dots$ be random variables on a probability space
  $(\Omega, \mathscr{F}, \mathbb{P})$ where
$$\sum_{k=0}^{\infty}P(|W_k|>k) <\infty$$
Prove that $$\limsup \frac{|W_k|}{k} \le 1 \ \text{a.s.} $$


I initially thought the conclusion meant $(**)$ when it really means $(*)$:
$$P(\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) = 1 \ \text{(*)}$$
$$P(\limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}) = 1 \ \text{(**)}$$

What I tried:
By the first Borel-Cantelli Lemma, we have $P(\limsup (|W_k| > k)) = 0$
$\to P(\limsup (|W_k|/k > 1)) = 0$
$\to P(\liminf (|W_k|/k > 1)) = 0$
$\to P([\liminf (|W_k|/k > 1)]^C) = 1$
$\to P(\limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}) = 1$
$\to P(\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) = 1$ QED assuming $(**)$ implies $(*)$.

Is it really the case that $(**)$ implies $(*)$? Here is why I think such:
$\forall \omega \in \Omega, \omega \in (\limsup \color{red}{(}|W_k|/k \le 1\color{red}{)})$
Then $\forall m \ge 1, \exists n \ge m$ s.t.
$\omega \in (\frac{|W_n(\omega)|}{n} \le 1)$
$\to \omega \in (\limsup \frac{|W_n(\omega)|}{n} \le \limsup 1 = 1)$
Now if $(\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \supseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$, then it follows by monotonicity that
$(**)$ implies $(*)$ QED.

However $(*)$ does not imply $(**)$ because it does not hold that $(\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \subseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$. Counterexample given in comments below. I guess the intuitive explanation is that
$\limsup \frac{|W_n(\omega)|}{n} \le \limsup 1$
--/--> $\frac{|W_n(\omega)|}{n} \le 1$
for similar reasons that
$x_n \le y_n \to \lim x_n \le \lim y_n$, if those limits exist (1)
but
$\lim x_n \le \lim y_n$ --/--> $x_n \le y_n$ (2), which makes sense:
(1) has an infinite number of statements which imply 3 statements (limsup, liminf, limsup=liminf)
(2) has 3 statements that claims to imply an infinite number of statements
Is that right?

Edit based on what Daniel Fischer said:
To prove $$\liminf \{\frac{W_k(\omega)}{k} \le 1\} \subseteq \{\limsup \frac{W_k(\omega)}{k} \le 1\}:$$
Suppose $$\omega \in \liminf \{\frac{W_k(\omega)}{k} \le 1\}$$
Then $\exists m \ge 1$ s.t. $\forall k \ge m$,
$$\omega \in \{\frac{|W_k(\omega)|}{k} \le 1\}$$
Since $\frac{|W_k(\omega)|}{k} \le 1 \to \limsup \frac{|W_k(\omega)|}{k} \le \limsup 1 = 1$, we have
$$\to \omega \in \{\limsup \frac{|W_k(\omega)|}{k} \le 1\}$$
 A: Let's try to get things straight. We have two conditions,
\begin{align}
P\left(\left\{\omega \in \Omega : \limsup_{k\to\infty} \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1\right\}\right) &= 1,\text{ and} \tag{$\ast$}\\
P\left(\limsup_{k\to\infty} \left\{\omega\in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \right\}\right) &= 1. \tag{$\ast\ast$}
\end{align}
The question is whether one condition implies the other. An inclusion between the sets $A := \left\{\omega \in \Omega : \limsup_{k\to\infty} \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1\right\}$ and $B := \limsup_{k\to\infty} \left\{\omega\in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \right\}$ would give us such an implication.
However, in general we have neither $A \subset B$ nor $B \subset A$. We can use the verbal description of the limes superior of sets to describe $B$ as
$$B = \left\{\omega \in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \text{ for infinitely many } k \right\}.$$
This implies $\liminf\limits_{k\to \infty} \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1$ for all $\omega \in B$, but it doesn't say anything about $\limsup\limits_{k\to \infty} \frac{\lvert W_k(\omega)\rvert}{k}$. We could for example have $W_k \equiv 0$ for odd $k$, and $W_k \equiv 5k$ for even $k$. Then we have $B = \Omega$ and $A = \varnothing$, since for this example $\limsup\limits_{k\to \infty} \frac{\lvert W_k(\omega)\rvert}{k} = 5$ for all $\omega$.
On the other hand, we could have $W_k \equiv k(1+2^{-k})$ for all $k$, then we have $A = \Omega$ and $B = \varnothing$, since $\frac{\lvert w_k(\omega)\rvert}{k} \to 1$ uniformly on $\Omega$, but $\frac{\lvert W_k(\omega)\rvert}{k} = 1 + 2^{-k} > 1$ for all $k$ and $\omega$.
Thus, we have no implication whatsoever between the conditions $(\ast)$ and $(\ast\ast)$.
But the Borel-Cantelli lemma yields something stronger than $(\ast\ast)$. Namely, from
$$\sum_k P(\{\omega\in \Omega : \lvert W_k(\omega) \rvert > k\}) < +\infty$$
we conclude that the set of $\omega$ such that $\lvert W_k(\omega)\rvert > k$ for infinitely many $k$ is a null set. In other words, for almost all $\omega$, we have $\lvert W_k(\omega)\rvert \leqslant k$ for all but finitely many $k$. In other words, from the Borel-Cantelli lemma we obtain
$$P\left(\liminf_{k \to \infty}\left\{\omega \in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1\right\}\right) = 1.\tag{$\ast{\ast}\ast$}$$
And now, we note that
$$C = \liminf_{k\to \infty} \left\{\omega \in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \right\} = \left\{\omega \in \Omega : \frac{\lvert W_k(\omega)\rvert}{k} \leqslant 1 \text{ for all but finitely many } k\right\}$$
is clearly a subset of $A$. Hence we have the implication $(\ast{\ast}\ast) \implies (\ast)$. Also it's clear that $C \subset B$, whence we have the implication $(\ast{\ast}\ast) \implies (\ast\ast)$.
A: The difference between the two formulas in the title is that the first one makes sense and the second one is meaningless.  The second one pretends to form the lim sup of some statements (not of numbers), and there's no such thing.
