MHB There is a unique inductive set

[evinda]

Hi! (Nerd)

Sentence:

There is a unique inductive set that is contained in each inductive set.

Proof:

Let $A$ be an inductive set (we know that there is such a set from the axiom of infinity)

and we define:

$$B=\{ X \subset A: X \text{ is an inductive set}\}$$
($B$ is a set, because if $X \subset A$ where $X$ is an inductive set then $X \in \mathcal{P}(A)$. So, $B$ is a set.)

The set $B \neq \varnothing$ since $A \in B$. So, the set $\bigcup B$ is defined. We will show that $\bigcap B$ has the desired properties.

$\bigcup B$ is an inductive set
(from the Remark: Let $B$ be a nonempty set of inductive sets. Then $\bigcap B$ is an inductive set.)

We will show that $\bigcap B$ is contained in each inductive set.
Let $C$ be an inductive set. Then, from the Remark, the set $A \cap C$ is an inductive set and obviously $\bigcap B \subset A \cap C \subset C$.

It remains to show the uniqueness.
Let $c,d$ sets, so that each of them is contained in each inductive set.
Then $c \subset d$
and $d \subset c$, so $c=d$.Could you explain me how we concluded that $A \cap C$ is an inductive set and how that $\bigcap B \subset A \cap C \subset C$ ? (Thinking)

 

[evinda]

The set $A \cap C$ is an inductive set, because $\{ A, C \}$ is a non-empty set of inductive sets and so $\bigcap \{ A, C \}=A \cap C$ is an inductive set, right?

But how do we conclude that $\bigcap B \subset A \cap C$? (Thinking)

 

[Siron]

The set $A \cap C$ is an inductive set, because $\{ A, C \}$ is a non-empty set of inductive sets and so $\bigcap \{ A, C \}=A \cap C$ is an inductive set, right?

Correct.

But how do we conclude that $\bigcap B \subset A \cap C$? (Thinking)

Note that $A \cap C \subset A$.

 

[evinda]

Note that $A \cap C \subset A$.

From this, we have that $A \cap C \in B$.
How can we conclude that $\bigcap B \subset A \cap C$ ? (Thinking)

 

[Siron]

From this, we have that $A \cap C \in B$.
How can we conclude that $\bigcap B \subset A \cap C$ ? (Thinking)

By definition $\bigcap B$ is the intersection of all elements in $B$. To formulate it less formally, it's the set that all elements in $B$ have in common. Thus all elements in $B$ contain $\bigcap B$. In particular, since $A \cap C \in B$ it has to contain $\bigcap B$.

 

[evinda]

By definition $\bigcap B$ is the intersection of all elements in $B$. To formulate it less formally, it's the set that all elements in $B$ have in common. Thus all elements in $B$ contain $\bigcap B$. In particular, since $A \cap C \in B$ it has to contain $\bigcap B$.

So, is it like that? (Thinking)

$$\forall x ( x \in \bigcap B \leftrightarrow (\forall b \in B) x \in b)$$

$$\forall x(x \in \bigcap B \leftrightarrow x \in A \cap C)$$

 

[Siron]

Yes.

Suppose $z \in \bigcap B$ then we have $\forall X \in B: z \in X$ by definition. As $A \cap C \in B$ we conclude $z \in A \cap C$, that is, we've proved $\bigcap B \subset A \cap C$

 

[evinda]

Yes.

Suppose $z \in \bigcap B$ then we have $\forall X \in B: z \in X$ by definition. As $A \cap C \in B$ we conclude $z \in A \cap C$, that is, we've proved $\bigcap B \subset A \cap C$

A ok... But I think that we cannot write it like that:

$$\forall x(x \in \bigcap B \leftrightarrow x \in A \cap C)$$

because that would mean that $\bigcap B=A \cap C$.
Or am I wrong? (Thinking)

 

[Siron]

A ok... But I think that we cannot write it like that:

$$\forall x(x \in \bigcap B \leftrightarrow x \in A \cap C)$$

because that would mean that $\bigcap B=A \cap C$.
Or am I wrong? (Thinking)

Indeed. I didn't notice that.

 

[evinda]

Having shown that $\bigcap B \subset A \cap C$ have we shown that $\bigcap B$ is contained in each inductive sets, since $A,C$ have been picked randomly and so $A \cap C$ is also an arbitrary set? Or am I wrong? (Thinking)

 

[Siron]

Having shown that $\bigcap B \subset A \cap C$ have we shown that $\bigcap B$ is contained in each inductive sets, since $A,C$ have been picked randomly and so $A \cap C$ is also an arbitrary set? Or am I wrong? (Thinking)

You're correct. The inductive set $C$ was picked arbitrary and it was shown that $\bigcap B \subset C$. Hence, $\bigcap B$ is contained in every inductive set.

 

[evinda]

You're correct. The inductive set $C$ was picked arbitrary and it was shown that $\bigcap B \subset C$. Hence, $\bigcap B$ is contained in every inductive set.

I see, thanks a lot! (Happy)