Suppose we take the union of a finite collection $\cup\mathbf{C}$. How do we find a universe that is big enough (is there a "formula" which, given the elements of a collection $\mathbf{C}$, outputs a universe which is big enough?) Is there always such a universe? What if the collection is infinite?
Note: By big enough, I mean that the universe does not exclude results, e.g. suppose $ U =\mathbb{N}$ then $\cup \{\{1,2\}, \{3 ,\pi\}\} = \{x \in \mathbb{N} :x\in \{1,2\} \lor x\in \{3, \pi\}\} =\{1,2,3\}$. $\pi$ is excluded because the universe $\mathbb{N}$ is not big enough.
Background:
My textbook states: "Let $X$ be a set and suppose $\mathbf{C}$ is a collection of subsets of $X$ (i.e. $\mathbf{C} \subseteq \mathcal{P}(X)$). Then $\cup \mathbf{C}=\{x \in X : \exists C\in \mathbf{C}(x\in C)\}$." I note that some textbooks do not include the restriction $\mathbf{C} \subseteq \mathcal{P}(X)$. Is this restriction given so that $X$ is big enough? Is it implicit in other definitions?
This seems to imply the universe is just the set $X$ for which $\mathbf{C} \subseteq \mathcal{P}(X)$. For example, let $X=\{0,1\}$. Then $\forall C\in\{\{{0}\},\{1\}\}(C\in\mathcal{P}(X))$ so $\mathbf{C} \subseteq \mathcal{P}(X)$, and thus the universe is $X$.
Moving up a level, we have $\forall C\in \{\{\{{0}\}\},\{\{1\}\}\}(C\in\mathcal{P}\mathcal{P}(X))$ so $\mathbf{C} \subseteq \mathcal{P}\mathcal{P}(X)$, and thus the universe is $\mathcal{P}(X)$.
Finally, $\{\{0\},\{\{1\}\}\} \subseteq \mathcal {P}\{0,\{1\}\}$, and thus $\{0,\{1\}\}$ is a universe.
In standard (ZF/ZFC) set theory it is explicitly an axiom that whenever $C$ is a set, the union $\cup C$ also exists as a set.
This axiom means that we're allowed to consider $\cup C$ as a legitimate set even though we cannot describe it as $\{x\in X : \cdots\}$ for any $X$ we already know.
This is an essential axiom of ZF; it cannot be derived from simpler (generally accepted) principles.
If $\cup \mathbf C$ is not known to be a set, the union may not exist, and we may need ad-hoc cleverness to find an $X$ that works.
(Indeed, we can prove that if $\cup \mathbf C$ is a set, then $\mathbf C$ is itself a set -- namely $\mathbf C = \{x\in\mathcal P(\cup\mathbf C)\mid x\in \mathbf C\}$ which is a set by the axioms of Power Set and Selection. Thus if $\mathbf C$ is known to be a proper class, we also know that $\cup\mathbf C$ is not a set).