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?
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 $\forall y. (y\not\in x)$.
Counterexample: Consider $N = 29$ and $M = 20$ with $N \cap M = 20$