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THIS QUESTION IS NOT A DUPLICATE OF THIS ONE BUT IT IS SIMPLY A $\bf\underline{CLARIFICATION}$ QUESTION OF THIS ANSWER SO THAT DO NOT CLOSE IT PLEASE!!!

Given a set $X$ the Axiom of union ensures the existence of a set $\bigcup X$ such that $y\in\bigcup X$ if and only if there exists $Y\in X$ such that $y\in Y$ so that it is usual to put $$ \bigcap X:=\Big\{y\in\bigcup X:y\in Y\,\forall Y\in X\Big\} $$ Now if $\bigcup\emptyset$ was not empty then there would be exists $y\in\bigcup \emptyset$ and thus in particiular there would be exists $Y\in\emptyset$ such that $y\in Y$ but this is impossible so that $\bigcup\emptyset $ is empty: however if $\bigcap\emptyset$ existed and did not contain an element $y$ then there would be exists $Y\in\emptyset$ such that $y\notin Y$ but this is impossible so that if $\bigcap\emptyset$ existed then it would contain any set so that it would be possible to define an universal set $$ U:=\Big\{x\in \bigcap\emptyset:x\,\text{set}\Big\} $$ which unfortunately does not exists and so we conclude that $\bigcap\emptyset $ does not exists.

Now in this Brian M. Scott's answer it is stated that the identity $$ X\equiv\Big\{x\in X:x\in\bigcap\emptyset\Big\} $$ is true by a vacuosly trust because (as above just explained) if $x\in X$ was not in $$X^*:=\Big\{x\in X:x\in\bigcap\emptyset\Big\}$$ then there would be exists $Y\in\emptyset$ such that $x\notin Y$ and this is impossible: howeve now I suspect that the above identity does not holds because if $\bigcap\emptyset$ does not exists then the set $X^*$ must be empty or rather it is impossible to define it with respect $\bf ZF$ axioms so that I thought to ask a clarification question where someone explain to me which is my confusion. So could someone help me, please?

Antonio Maria Di Mauro
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  • With the definition you have given of $\bigcap X$, we have $\bigcap \emptyset = \emptyset$. So this identity holds in ZF only for $X = \emptyset$. It is not always true with your definition that $x \in \bigcap X$ iff $x \in Y$ for each $Y \in X$, and indeed in ZF there is no set with this second property in the case $X = \emptyset$. – Izaak van Dongen Jul 04 '22 at 10:02
  • @IzaakvanDongen Sorry, but $\bigcap\emptyset$ it do not surely exists: indeed, $\emptyset\subseteq X$ for any set $X$ and so $\bigcap X\subseteq\bigcap \emptyset$ so that you can see that if $\bigcap\emptyset$ was empty then $\bigcap X$ would be empty for any set $X$ and this is clearly false: e.g. if you take $$X:=\Big{\big{{\emptyset}\big}\Big}$$ then $$\bigcap X={\emptyset}\neq\emptyset$$ So what can you say about? – Antonio Maria Di Mauro Jul 04 '22 at 10:09
  • With your definition of intersection ($\bigcap X = {y \in \bigcup X : (\forall Y \in X)(y \in Y)}$), it's not always true that $A \subseteq B$ implies $\bigcap B \subseteq \bigcap A$. Like I said, this definition of intersection does not have all the properties that you might expect in the case of the empty set. And there is indeed no set in ZF that does have all the properties you expect. You can either define $\bigcap \emptyset$ but it doesn't behave like the intersection, or you say it can't be defined - two sides of the same coin, and a bit of a technicality really. – Izaak van Dongen Jul 04 '22 at 12:29
  • Umm...Sorry but I do not understand why with respect my definition the implication $$A\subseteq B\Rightarrow\bigcap B\subseteq\bigcap A$$ does not always holds: indeed, many texts (e.g. into the text Introduction to set theory by Karell Hrbacek and Thomas Jech) define $\bigcap X$ in this way requiring that $X$ is NOT empty. So could you explain better, please? Could you give a counterexample? – Antonio Maria Di Mauro Jul 04 '22 at 13:18
  • The implication does indeed hold when $A$ and $B$ are not empty... but if you have only defined $\bigcap X$ when $X$ is non-empty, what do you mean when you write $\bigcap \emptyset$? What do you mean exactly when you say "if $\bigcap \emptyset$ exists"? Maybe the answer you're looking for is "if you don't define $\bigcap \emptyset$ in ZF first, then the putative "identity" isn't even a sentence of the language of set theory, so it makes no sense to ask whether or not it holds in ZF". There is indeed no set in ZF whose membership condition is "$(\forall Y \in \emptyset)(x \in Y)$". – Izaak van Dongen Jul 04 '22 at 13:38
  • @IzaakvanDongen Okay, I now understand: however I cannot answer precisely because I just asked this question for ask clarification about the meaning of $\bigcap\emptyset$ as you can see reading it. I knew that $\bigcap\emptyset$ does not exists as I explain into the first part of the question and so when I read Brian M. Scott's answer I was surprised by the identity $$X=X^*$$ so that I thought to ask a clarification question as I first said, that's all. So did I explain what I want mean? – Antonio Maria Di Mauro Jul 04 '22 at 13:42

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It is not too hard to write down a formal proof of $$ (\exists y)(y\in x) \Rightarrow (\exists z)(\forall u)\bigl(u\in z \Leftrightarrow (\forall y)(y\in x\Rightarrow u\in y)\bigr) $$ starting from the first few axioms of ZF; in words: if $x$ is a nonempty set then there is a set $z$ that consists of exactly the sets that are members of all members of $x$. By extensionality any two $z$s that meet this requirement are identical. This allows one to define a new function symbol $\bigcap$ that satisfies $$(\forall x)\Bigl((\exists y)(y\in x)\Rightarrow(\forall u)\bigl(u\in\bigcap x \Leftrightarrow (\forall y)(y\in x\Rightarrow u\in y)\bigr)\Bigl)$$ (There is no need to assume existence of the union you can apply separation (or comprehension) to an element of $x$.) Thus ZF proves that for every nonempty $x$ there is indeed a unique set that we want to denote $\bigcap x$.

ZF also proves $$(\forall y)(y\notin x)\Rightarrow (\forall u)(\forall y)(y\in x\Rightarrow u\in y)\qquad(*) $$ and $$ (\forall z)(\exists y)(y\notin z) $$ These two combine to show that ZF proves that there is no $z$ that can be denoted $\bigcap\emptyset$. That is, if you write down the formula $\phi(x,z)$ that is equivalent to (or better defines) $z=\bigcap x$: $$ (\forall u)\bigl(u\in z \Leftrightarrow (\forall y)(y\in x\Rightarrow u\in y)\bigr) $$ Then ZF proves $$ (\exists y)(y\in x)\Rightarrow (\exists! z)\phi(x,z) $$ and $$ (\forall y)(y\notin x)\Rightarrow \neg(\exists z)\phi(x,z) $$ So the answer to the question is "no", because $\bigcap\emptyset$ is something that ZF proves the nonexistence of, so it makes no sense to use it in a definition.

On the other hand you can tweak the formula in $(*)$ a bit to get what you want (if it still is what you want): given a set $X$ if you apply comprehension thus $$ \{u\in X: (\forall y)(y\notin x)\Rightarrow (\forall y)(y\in x\Rightarrow u\in y)\} $$ the result will be $X$.

This is what many people use as an excuse to (informally) stretch the meaning of $\bigcap$ a bit and abbreviate the previous expression to $$ \{u\in X:u\in\bigcap\emptyset\} $$ and conclude that that set is equal to $X$.

hartkp
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