A Basic Truth of Set Theory?

Question

Forgive me in advance if this question seems ridiculous.

Let $X$ be a set. Then "$X\neq\bigcup_{x\in X}\{x\}$ is false" is a true statement.

This statement says that a set is always equal to the union of its elements(points).

Is this axiomatic or can this actually be proven by by something even more obvious than the statement?

Answer 1

When you say "provable" you need to specify some axioms from which this to be proved.

If we consider the axioms of $\sf ZFC$, which are considered more or less the canonical set theory nowadays, then the answer is that yes. We can prove that.

First of all we need to show that if $X$ is a set then $\{\{x\}\mid x\in X\}$ is a set. This is true because $S=\{x\}\mid x\in X\}$ is a definable subset of $\mathcal P(X)$, and the latter exists from the axiom of power set. So using the separation axiom schema we conclude that $\{\{x\}\mid x\in X\}$ is a set.

Now the union $\bigcup_{x\in X}\{x\}$ is the union of $S$. This union exists as a consequence of the axiom of union, which states that if $A$ is a set, then $\{b\mid\exists a\in A:b\in a\}$ is also a set. Call this union of singletons $Y$.

Finally, $X=Y$ due to the axiom of extensionality. Given $x\in X$, $\{x\}$ was an element of $S$, therefore $x\in Y$, and on the other hand if $x\in Y$, then there is some $s\in S$ such that $x\in s$, but if $s\in S$ then $s=\{x'\}$ for some $x'\in X$. So if $x\in s$ it follows that $x=x'$ and therefore $x\in X$. $\square$

Answer 2

The claim that the statement "$X\neq \bigcup_{x\in X}{x}$ is false" is not true. What I believe you mean is $X\neq \bigcup_{x\in X}{\{x\}}$.

For example, take $X=\{\{0\},\{1\}\}$. Then $\bigcup{X}=\{0,1\}\neq X$.

However, the claim that "$X\neq \bigcup_{x\in X}{x}$ is false" is true. To see this, notice that $\{y\}\subset \bigcup_{x\in X}{\{x\}}$ for every $y\in X$, so that $y\in \bigcup_{x\in X}{\{x\}}$ for every $y\in X$. This shows us that $X\subset \bigcup_{x\in X}{x}$. To show the converse, notice that if $y\in \bigcup_{x\in X}{\{x\}}$, then by definition there is $\{z\}$ such that $y\in \{z\}$ with $z\in X$. But then $y=z$, and thus $y\in X$. This shows us that $\bigcup_{x\in X}{x}\subset X$.

Putting these two together gives us $\bigcup_{x\in X}{\{x\}}=X$ is true (which is equivalent to the statement you gave).

The reason the original statement was false was due to the fact that writing $\bigcup_{x\in X}{x}$ means the collection of all elements in $x$ for $x\in X$.

Answer 3

Did you mean "is it the case that for any set $X$, $X=\bigcup_{x \in X} \{ x \}$"? If so, then at least in ZF or an extension thereof, this is more like a definition than an axiom. Just expand definitions:

$$y \in \bigcup_{x \in X} \{ x \} \Leftrightarrow (\exists x \in X) y \in \{ x \} \Leftrightarrow (\exists x \in X) y = x \Leftrightarrow y \in X.$$

If instead you meant "is it the case that for any set $X$, $X=\bigcup_{x \in X} x$", then this is false. For example, when $X = \{ \emptyset \}$, the resulting union is $\emptyset$, which is a different set ($X$ has an element, $\emptyset$ does not).

In a set theory other than ZF or an extension thereof, everything should go through as long as all of the objects are even defined, but they needn't be defined. For example, in NF, when $X$ is a set, $\{ \{ x \} : x \in X \}$ need not be a set, and so we need not be able to take its union.

Answer 4

You can prove the desideratum by the axiom "$\neg \neg \phi \Rightarrow \phi$" and using a usefull trick: To prove that $A=B$, prove that $A\subseteq B$ and $B\subseteq A$.

The first, suppose $y\in X$, then $\{y\}\subseteq \bigcup_{y\in X} \{y\}$ as $\{y\}$ is one of the sets being united, so $y\in \{y\} \subseteq \bigcup_{y\in X} \{y\}$.

The second, suppose $y\in \bigcup_{y\in X} \{y\}$then there is some $y'$ st $y\in \{y'\}$ and $y'\in X$, so $y=y'$ and then $y\in X$ as desired.