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It is an axiom of ZFC that $\cup x$ exists for any set $x$, where $\cup$ denotes union. But how does one prove that for any non-empty set $x$, the intersection $\cap x$ exists? We need $x$ to be non-empty, because $\cap \emptyset$ would be the set of everything, which does not exist in ZFC.

user107952
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3 Answers3

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It is a subset of $\cup x$. Recall that $$\cup x=\{ z\mid \exists y(z\in y\wedge y\in x)\}.$$ The intersection is $$\cap x=\{z\in \cup x\mid \forall y( y\in x\rightarrow z\in y)\}.$$ This is a set by the Axioms of Union and Separation.

In fact, this definition allows you define the intersection of an empty family; the issue is that it is equal to $\varnothing$ (since $\cup\varnothing = \varnothing$), so then you run into issues with associativity.

Arturo Magidin
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  • What do you mean by issues with asscosiatvity ? – Vivaan Daga Nov 20 '21 at 18:41
  • @Lobic: Associativity implies that if $I$ is an index, and $I_1$ and $I_2$ form a partition of $I$, then $\cap_{i\in I}x_i = (\cap_{i\in I_1} x_i)\cap (\cap_{i\in I_2}x_i)$. But then you need both parts to be nonempty for this to hold, so it is not "full associativity" the way it is, for example, with sums (where the empty sum is $0$) or products (where the empty product is $1$). – Arturo Magidin Nov 20 '21 at 19:33
  • but how to prove that this set is indeed the intersection ? (I appreciate your response very much) – Vivaan Daga Nov 21 '21 at 05:54
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    @Lobic:It's a definition. You don't prove definitions. – Arturo Magidin Nov 21 '21 at 06:10
  • I mean how to prove this: ∀x [x!= ∅ → ∃z z = {w : (∀y ∈ x)(w ∈ y)] – Vivaan Daga Nov 21 '21 at 06:32
  • @Lobic: It's in the post. "By the Axioms of Union and Separation". And you do not need to exclude the empty set. – Arturo Magidin Nov 21 '21 at 06:39
  • What im asking is why the set intersection x proven to exist in the post satisfies the properties of z. – Vivaan Daga Nov 21 '21 at 06:51
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    @Lobic Because it does. Because, by construction an element is in the set if and only if it lies in every element of $x$. You are literally asking why the set satisfies its defining property. And the answer is that it does because the property defines it. – Arturo Magidin Nov 21 '21 at 07:09
  • so the empty set is a difference according to your definition the intersection is just empty ; however if such a z existed then it must be V which is not a set in zfc – Vivaan Daga Nov 21 '21 at 07:20
  • and there is a difference becuause the set you construct's elements must also be in union x – Vivaan Daga Nov 21 '21 at 12:26
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    @lobic There is no element in the intersection which is not is not in the union. So, no. At this point, this is both too long for comments, and too far from the original post. This is not a place to engage in discussions, so you are misusing the comments. – Arturo Magidin Nov 21 '21 at 18:15
  • Last comment please (I think this is an important question) which book uses this def of Intersection for the empty set? most books I know just leave it undefined; as you say a disadvantage of this definition is that some properties of intersection are lost – Vivaan Daga Nov 22 '21 at 03:44
  • @Lobic: As you can see here, others agree they've seen it, but cannot recall a particular author's textbook. The most common ones do not seem to include it. – Arturo Magidin Nov 22 '21 at 04:06
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Assuming $U = \bigcup x$ is defined and using the separation axiom we get $$ I := \bigcap x, \qquad \forall z:z \in I \leftrightarrow (z \in U)\land (\forall y(y \in x \to z\in y) $$ where the formula here is $\phi(x,y,z) =\forall y(y \in x \to z\in y)$

Physor
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Exercise 4.6 of Chapter 1, Section 4 (p.15) of the third edition (1999) of Introduction to Set Theory by K. Hrbacek and T. Jech asks precisely that.

The Axiom Schema of Comprehension on p.8 states: Let $\mathbf{P}(x)$ be a property of $x$ (and, possibly, of other parameters). For any set $U$, there is a set $V$ such that $x\in V$ if and only if $x\in U$ and $\mathbf{P}(x)$.

When $S\neq\emptyset$, the existence of $\bigcap S$ follows from the Schema of Comprehension as follows:

If $S\neq\emptyset$, there exists $A\in S$. Consider the property $\mathbf{P}(x):x\in X\text{ for all }X\in S$. By the Schema of Comprehension applied to $A$ and $\mathbf{P}$, there is a set $V$ such that $x\in V$ if and only if $x\in X$ for all $X\in S$. But this is precisely the definition of $\bigcap S$. That is, $V=\bigcap S$ and we are done.

John
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