I don't understand why $n \log \log n $ is not $\Theta (n)$.
Suppose we give $n$ a value of $10,000$. Then $n \log \log n$ is $6020.6$. So isn't $n \log \log n$ upper- and lower-bounded by $n$, as $n \log \log n \geq Cn$?
I don't understand why $n \log \log n $ is not $\Theta (n)$.
Suppose we give $n$ a value of $10,000$. Then $n \log \log n$ is $6020.6$. So isn't $n \log \log n$ upper- and lower-bounded by $n$, as $n \log \log n \geq Cn$?
As $n$ grows bigger, the ratio between $n\log\log n$ and $n$, namely $\log\log n$, tends to infinity. Hence it is not possible to upper bound $n\log\log n \leq Cn$ for any constant $C$. If you stare at this inequality, you discover that the constant $C$ must satisfy $\log\log n \leq C$ for all $n$, yet the function $\log\log n$ is not bounded.
You could not narrow $f(x) = x\log \log x$ with $ g = n $, because inequalities that must be met for this, never occur. Remember that the logarithm function is always increasing. So, If We had it, $f$ should perform the next inequalite for some constants $c$ and $C$. $$ cn\leq f \leq Cn$$
Let's see that, for example a right side of these inequalities and see the contradiction.
$$ \begin{align} x\log \log x \leq cx\\ e^{x\log \log x }\leq e^{cx}\\ e^x\log x \leq e^{cx}\\ \log x \leq e^c \end{align} $$ So, you are saying that $\log x \leq k$, for any $x\in R$, that's not true.
Simply, because $\log\log n$ is not a constant! actually, $$\lim_{n\to \infty} \log\log n = \infty$$
while for some function to be $\Theta(n)$ it is required that it is bounded by $cn$ for some constant $c$.
tl;dr you didn't look at large enough $n$'s. try $n$ = Graham's number.