I was thinking of a way to represent a function as a union of cartesian products.

For example with A = {1,3} and B = { 2,5,6,7} the function f from A to B such that :

                         f = { (1,2>, <3,5>} 

could be represented, as the union of

  • the cartesian product : {1} cross {2}

and of

  • the cartesian product : {3} cross {5}.

For f = { <1,2> } U { <3,5> }.

But in order the sets {1} , {3} on the one hand and the sets {2} , {5} to be available, I think I would need the set of all singletons of A , and of B.

Do these sets of singletons have a name?

Is it even true in general that for every set S there exists a set of singletons having each one as element a different element of S.

I mean, is this true even if S is infinite?

  • 2
    $\begingroup$ Basically, you are searching for a function $f(x)= \{ x \}$, for every $x \in A$. $\endgroup$ – Mauro ALLEGRANZA May 27 '19 at 13:37
  • 1
    $\begingroup$ Thus, $f = \{ (x , \{ x \} ) \mid x \in A \} \subseteq A \times \mathcal P(A)$. $\endgroup$ – Mauro ALLEGRANZA May 27 '19 at 13:38
  • $\begingroup$ The name ... I do not know. $\endgroup$ – Mauro ALLEGRANZA May 27 '19 at 13:39
  • $\begingroup$ @MauroAllegranza. Thanks, this is the function I was looking for. $\endgroup$ – user654868 May 27 '19 at 14:15

Usually a function is thought of as a subset of the Cartesian product of the domain and the codomain with a particular property: $f:A\rightarrow B$ is a function when $f\subseteq A\times B$ and $\forall a\in A$, $\exists!b\in B$ such that $(a,b)\in f$. In your example, $f=\{(1,2),(3,5)\}$.

Now, if you are asking how exactly ordered pairs can be constructed from sets, the simplest way is having $(a,b):=\{a,\{a,b\}\}$. From this, because sets cannot contain themself, we can determine the order of $(a,b)$ by saying the first element in the ordered pair $(a,b)$ is the (unique) $x$ satisfying $x\in(a,b)$ and $\exists y\in(a,b)$ such that $x\in y$. Note that $a$ satisfies this, but for example $b$ fails at the first property, and $\{a,b\}$ fails at the second property.

Then finally, getting to your question, we can construct sets like the set of all singleton subsets, because we are able to construct the power set. For a set $X$, the set you wanted to know is just $\{x\in\mathcal P(X):\vert x\vert=1\}$. If $\vert x\vert=1$ is too high level an operation, we could write instead $\exists!y\in x$. In ZF we can construct the power set of any set, including infinite sets, so the answer to your question is yes, we can always construct such a set.


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