# $\lim \sup_{t\rightarrow \infty} \frac{W_t}{\sqrt{t}}$ question

Why does this hold ? :

$$\lim \sup_{t\rightarrow \infty} \frac{W_t}{\sqrt{t}} \geq \lim \sup_{t\rightarrow \infty} \frac{W_{\lfloor{t}\rfloor}}{\sqrt{\lfloor{t}\rfloor}}$$

With $W_t$ a Brownian Motion

I can see that $\frac{1}{\sqrt{t}}\geq \frac{1}{\sqrt{\lfloor{t}\rfloor}}$ but what do we know in order to compare $W_{\lfloor{t}\rfloor}$ and $W_t$ ? Is that when $t\rightarrow \infty$ we consider $W_{\lfloor{t}\rfloor}$ and $W_t$ equal? if it's the case why?

This isn't really to do with Brownian motion but rather to do with the definition of $\lim \sup$. One definition of $\lim \sup_{t \to \infty} f(t)$ is $\lim \sup_{t \to \infty} f(t) = \sup E$ where $E$ $$E = \{x : \mbox{ there is an increasing sequence } t_n \to \infty \mbox{ such that } f(t_n) \to x \}$$ That is $E$ is the set of all subsequential limits of the sequence $f(t)$. One can define the $\lim \sup$ of a sequence in terms of subsequential limits similarly.
This has nothing to do with the properties of $W_t$.
It follows from the fact that for $a_t$ a real valued sequence $\limsup_{t\rightarrow \infty}a_t\geq \limsup_{n \rightarrow \infty} a_n$.
Since, if $a$ is an accumulation point of $a_n$, then it also an accumulation point for $a_t$.