Does every set have a power set? While reading the probability space in Wikipedia, I'd found the usual formulation is a triplet, which is ${\displaystyle (\Omega ,{\mathcal {F}},P)}$.
Upon my understanding, the middle ${\mathcal {F}}$ is a power set of $\Omega$ which will be allocated with real-valued probabiilty by $P$.
If every set in this nature has power set, there might be no necessity of introduction of ${\mathcal {F}}$ I guess however, I've never thought of a set which doesn't have its power set.
Is there any set that doesn't have power set? or if not, which means every set has its power set, is there any plausible reason that ${\mathcal {F}}$ is introduced in probability formulation? 
 A: This $\mathcal{F}$ is actually not the power set, but a sigma algebra. The power set is a sigma algebra and is often used, but sometimes probability theory requires smaller subsets of the power set in order to properly define the problem. 
A: I think the notation here is due to the fact that a probability space is, is particular, a finite measure space. So, in the more general sense, the set $\mathcal{F}$ does not need to be the power set, it is sufficient that the set $\mathcal{F}$ satisfy the axioms of a $\sigma$-algebra of subsets of $\Omega$. And yes, every set has a power set, it is guaranteed by the power set axiom in set theory. If you're interested, study some measure theory, which provides the mathematical tools for developing rigorous probability.
A: In standard mathematics, every set has a power set. This is encoded in the Axiom of Power Set. However, your confusion lies with the definition of a probability space, not with set theory.
The set $\mathcal F$ in a probability space $(\Omega, \mathcal F, P)$ is not necessarily the power set of $\Omega$. The set $\mathcal F$ is a subset of the power set $\mathcal P(\Omega)$. This $\mathcal F$ is required to be a so-called sigma algebra, which tells you that it shares some properties in common with the full power set, but it need not be the full power set at all.
In particular, for any $\Omega$, you can take $\mathcal F = \{\emptyset, \Omega\}$, and this will be a sigma algebra on $\Omega$. Unless $|\Omega| \leq 1$, it will not be the power set.
