It often happens that we want to adjoin a 'new' element $\infty$ to a set $X$, e.g. when constructing the Alexandroff compactification, or adding "infinity" to the real numbers or any ordered set. This is not a problem, because there is no set that contains all sets, or because by the axiom of foundation we can take any element of some nonempty subset of $X$ (or $X$ itself should $X$ only contain the empty set).
It becomes tricky when we want to do this for many sets at once; e.g. when defining a functor: we have to choose an element for each set, and choice is scary. I figured that we can take $\infty=\{X\}\cup\bigcup X$: if $X$ is a singleton $\{a\}$ then $a\neq X$ by foundation. If $X$ is not, $\infty$ strictly contains its elements.
Then I started wondering if:
For a fixed set $I$, does there exist a canonical choice of $I$ distinct elements (i.e. an injective map with domain $I$) not in $X$, for all sets $X$?
If $I=\{1,\ldots,n\}$ is finite and well-ordered we can iterate the construction of $\infty$ above.
My first idea was to take something like $\infty \cup\{i\}$ for each $i\in I$, but we may not be able to distinguish those when $i=X$ or when $i$ is an element of $X$ or of a subset of $X$. The problem lies in those sets $X$ that are related to $I$ through its elements or their elements or .. etc.
Even Foundation is not necessary.
One can easily prove that for every set $X$, there is some $Y\subseteq X$ such that there is no element of $X$ of the form $(Y,x)$, for any $x$. (This involves a standard Russell-like proof.) Moreover, this $Y$ is canonically defined for $X$, once we fix a definition for $(Y,x)$, so for example, the Kuratowski pairing would give us a canonical $Y$.
Now given $I$, simply pick this $Y$ and consider $\{Y\}\times I$. This set will be disjoint from $X$ and clearly have size $I$.