# How could we show that the set is equal to the empty set?

I want to show that the intersection of any inductive set is empty since every inductive set contains the empty set.

I thought that we could do it like that:

We know that $B$ is an inductive set. So:

$$\varnothing \in B \wedge \forall x(x \in B \rightarrow x' \in B)$$

$$y \in \bigcap B \leftrightarrow \forall b \in B: y \in b$$

Since $\varnothing \in B$ we get that $y \in \bigcap B \leftrightarrow y \in \varnothing$.

But since there is no $y$ such that $y \in \varnothing$ we conclude that we cannot find a $y$ such that $y \in \bigcap B$.

But it isn't right, since we cannot just take one set to get the equivalence, right?

How else could we do this?

• What is $x'$ in this $x' \in B$? Or is this supposed to be $\forall x' \in x: x' \in B$ or equivalently $x \subseteq B$? – Thorsten Dec 31 '14 at 1:50
• With $x'$ I mean the next element of $x$. $x'=x \cup \{x\}$. @Thorsten – evinda Dec 31 '14 at 1:51

## 2 Answers

This looks right. So let me reorder your argument:

Now your argument: we know $\emptyset \in B$. So for an arbitrary $y$ we have two directions to prove:

• if $y\in \bigcap B$ it follows $y \in \emptyset$ (By your property for $\bigcap$).
• if we have $y\in \emptyset$, then we have a contradictory assumption (by the definition of the emptyset), and so we have $y \in \bigcap B$ in particular.

So in total: $\forall y (y \in \bigcap B \leftrightarrow y \in \emptyset)$. And by the axiom of extensionality, we have $\bigcap B = \emptyset$.

Alternatively you can prove it as follows: We know that the emptyset $\emptyset$ is the unique set with the property $\forall y(y\not\in \emptyset)$. Assume there is an element $y \in \bigcap B$, so we have $y \in \emptyset \in B$ by the property of $\bigcap B$, a contradiction. So $\forall y(y\not\in \bigcap B)$, and $\bigcap B = \emptyset$ by uniqueness.

EDIT: And to answer the title question "How could we show that the set is equal to the empty set?": You show that a set $x$ is equal to the emptyset by proving that $\forall y. (y\not\in x)$.

• Thorsten Could you explain me further how we could prove using the property of $\bigcap$ that if $y \in \bigcap B$ then we have that $y \in \varnothing$? – evinda Dec 31 '14 at 2:26
• You already stated the property $\forall x\ (x \in \bigcap B \leftrightarrow \forall b \in B: x \in b)$. So instantiate the outer quantifier with $y$, take the direction from left to right and instantiate the right quantifier with $\emptyset$, and you get: $y \in \bigcap B \rightarrow y \in \emptyset$. – Thorsten Dec 31 '14 at 2:53

Counterexample: Consider $N = 29$ and $M = 20$ with $N \cap M = 20$

• In mathematics, the intersection A ∩ B of two sets A and B is the set that contains all elements of A that also belong to B (or equivalently, all elements of B that also belong to A), but no other elements – Don Larynx Dec 31 '14 at 1:48
• @Don Larynx $\bigcap B=\bigcap_{A\in B} A$. – Tim Raczkowski Dec 31 '14 at 1:52
• @Don Not quite. In set theory, everything is a set, so the elements of a set are sets as well, and if $A$ is a non-empty set, $\bigcap A$ is defined. There is no need to have two sets, or to consider the intersection of an inductive set with something else. – Andrés E. Caicedo Dec 31 '14 at 2:07
• @evinda $\bigcap B$ can be the empty set, if none of the elements of $B$ have a common element. For example if $\emptyset\in B$ then $\bigcap B=\emptyset$. – Tim Raczkowski Dec 31 '14 at 2:10
• Yes, if your definition of inductive set requires that it has the empty set as an element $\bigcap B=\emptyset$ is true. – Tim Raczkowski Dec 31 '14 at 2:14