Every set is a union of singleton sets Is it possible to prove this statement? What axioms are necessary to conclude that every set on, for instance, $\mathbb{R}$ can be expressed as a union of singleton sets (sets containing exactly one element)?
Let $X=\{x : P\}$ where $P$ is a defining property of the elements of $X$, for instance $P$ can be $x>2$.
Then I can simply state that $$X = \bigcup_{P} X_i$$ where $\{X_i : i \in \mathbb{R_+}\}$ is a collection of all singleton sets of $\mathbb{R}$.
By the way, there are sets of bigger cardinality than of cardinality of all real numbers. In this case, using $\mathbb{R_+}$ as indexes for all singleton sets from such sets wouldn't be sufficient, right?
 A: Maybe I'm losing some obvious, but I think that in your informal proof you are using at least:
Replacement: The set $Z = \{\{x\}:x\in X\}$ is the image of $X$ by the function  $x\longmapsto \{x\}$.
Union: $\bigcup Z$ is a set.
Extensionality: $X = \bigcup Z$ because both have the same elements.
A: First of all, given two sets $A$ and $B$ we say that
$$\tag{1}
A=B
\quad\Longleftrightarrow\quad
A\subseteq B
~\text{ and }~
B\subseteq A
$$
That is, the sets coincide if every element of $A$ is in $B$ and every element of $B$ is in $A$.
As @MSE says, the intuition is that
$$\tag{2}
  M = \bigcup_{x\in M}\{x\}
\qquad\text{(to be proven)}
$$
To prove this, we shall show that both inclusions hold.
It is convenient to define
$$\tag{3}
\mathscr M := \bigl\{\{x\}:~x\in M\bigr\}
$$
and rewrite $(2)$ as
$$\tag{4}
  M = \bigcup_{S\in \mathscr M}S
\qquad\text{(equivalent to $(2)$, to be proven)}
$$
Now, let $x\in M$, then $\{x\}\in\mathscr M$ by definition $(3)$, and therefore
$
\{x\}\subseteq\bigcup_{S\in\mathscr M}S
$,
which implies
$
x\in\bigcup_{S\in\mathscr M}S
$.
Since $x\in M$ is arbitrary, this proves that
$$
  M \subseteq \bigcup_{S\in \mathscr M}S
$$
Conversely, let $x\in\bigcup_{S\in \mathscr M}S$, then $x\in S$ for some $S\in\mathscr M$.
By definition $(3)$ of $\mathscr M$, there exists $y\in M$ such that $S=\{y\}$.
Since $x\in\{y\}$, it follows that $x=y\in M$.
Again, from the arbitrarity of $x\in\bigcup_{S\in \mathscr M}S$ it follows that the other inclusion holds as well
$$
  \bigcup_{S\in \mathscr M}S \subseteq M
$$
This proves the sought identity $(4)$ and hence $(2)$.
A: Let $M$ an unspecified set. Then, $$M=\bigcup_{x\in M}\{x\}.$$
A: I don't think the union axiom is needed.
You just need, for every set X, something to guarantee the existence of all singletons of the elements of X.
You can then see that X contains every element of all those singletons and nothing else, and conclude that, by definition of union, it's the union of all the singletons.
The union axiom guarantees the existence of particular union sets (the ones containing all elements of the elements of a set); it's not needed to allow the existence of sets that happen to be the union of other sets.
Example: if we accept the infinity axiom and the empty set axiom, we can say the set {{},{{}}} exists; for either the pair axiom or restricted comprehension, {{}} and {{{}}} also exist.
Let a set be an union of a family of sets if it contains all of the elements of the sets of the family and nothing else.
We can then see that {{},{{}}} is union of {{}} and {{{}}}.
So, if I didn't miss anything, my answer to:
"What's needed to prove that any set is union of the family of singletons of its elements?"
is:
"Either the pair axiom, or the axiom of restricted comprehension."
edit: or, as suggested in a comment of Eric Wofsey on another answer, "the existence of any nonempty set together with Replacement"
A: It looks like you need:


*

*Pairing — existence of unordered pairs, hence existence of singletons,

*Union — obviously the union of a set has to exist,

*Replacement, or Power set + Comprehension (= Separation),

*Extensionality.


Clearly both 1. and 2. are required.
Given a set $X$, somehow you need to prove existence of $Y = \{\{x\}\mid x\in X\}$. That's where 3. comes in. Using Replacement, this is a set, because it's the image of the set $X$ under the function $x\mapsto \{x\}$. Another way to prove its existence: $\mathcal{P}(X)$ exists by Power set, hence by Comprehension so does the subset of things in it that are singletons: $Y = \{s\in \mathcal{P}(X) \mid (\exists u\in s)\, (\forall v\in s)\, u=v\}$ exists.
Now it's clear that $(\forall x)\,(x\in X\leftrightarrow x\in \bigcup Y)$, so by extensionality $X = \bigcup Y$.
A: Following the axioms in Kunen : Set Theory, and avoiding Power : (1) Pairing. $\ \forall x\;\exists y\; (y=\{x\})..\;$(2) Extensionality. $ \forall x\;\exists ! y\;(y=\{x\})$. (3) Replacement. $ (\forall A \; \forall x\in A \; \exists ! y\;y=\{x\}) \implies (\forall A \;\exists B\; \forall x\in A\; (\{x\}\in B)\;$. (4) Comprehension : Taking $A$ and $B$ as in (3), $\exists C=\{y\in B: \exists ! x\in A\; (y=\{x\})\; \}.$ (5). Union. Taking $A$ and $C$ as in (4), $\cup C=\{x : x\in A\}\;$. (6). Extensionality again. From (5) and (4), $\cup C=\{x :x\in A\}=A.$ I think that step (5) implicitly assumes Extensionality. Then the question is whether we can leave out any of these axioms, and add some other(s) and still succeed. We can omit Replacement with Power in step (3), and  re-write step (4) as $\exists C=\{y\in P(A):\exists ! x\in A \; x\in y\}\;$. Trying to omit any others makes me cross-eyed.
