MHB Understanding the Inductive Set Intersection: $\bigcap B = \{0\}$ Explained

[evinda]

Hi! (Mmm)

• If $B$ is a nonempty set, the elements of which are inductive sets then $\bigcap B$ is an inductive set.
  
• If $B$ is an inductive set, it isn't necessarily true that $\bigcap B$ is inductive.
  
  For example, if $B=\omega$ then $\bigcap \omega=\{ 0 \}$

Could you explain me why, if $B=\omega$ then $\bigcap \omega=\{ 0 \}$?
It is like that:
$$0=\varnothing \\ 1=\{0\} \\ 2=\{0,1\} \\ \dots \dots$$It doesn't hold that $0 \in 0$, right? (Thinking)
But then why is it $\bigcap \omega=\{ 0 \}$

 

[Evgeny.Makarov]

Yes, if you take von Neumann definition of natural numbers, then $\bigcap\omega=\varnothing$ because $\varnothing\in\omega$. But $\varnothing$ is not an inductive set either because $\varnothing\notin\varnothing$.

 

[evinda]

Yes, if you take von Neumann definition of natural numbers, then $\bigcap\omega=\varnothing$ because $\varnothing\in\omega$. But $\varnothing$ is not an inductive set either because $\varnothing\notin\varnothing$.

Could we also say that $\bigcap \omega= \varnothing$ since considering all the elements of $\omega$ we cannot find a common element, because of the fact that $0=\varnothing$ does not contain any element?
Or am I wrong? (Thinking)

 

[Evgeny.Makarov]

Yes, you are right.

 

[evinda]

Yes, you are right.

Nice! (Whew) Thanks a lot! (Happy)

 

[evinda]

Could you also give me an other example of an inductive set $B$ such that $\bigcap B$ is not an inductive set? (Thinking)

 

[Evgeny.Makarov]

The intersection of any inductive set is empty since every inductive set contains the empty set.

 

[evinda]

The intersection of any inductive set is empty since every inductive set contains the empty set.

Doesn't this imply that if $B$ is an inductive set, it is never true that $\bigcap B$ is inductive since $\varnothing \notin \varnothing$ ? (Thinking)

 

[Evgeny.Makarov]

Yes, it does.

 

[evinda]

Yes, it does.

So, could we justify it like that? (Thinking)

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 $y \in \bigcap B$.

 

[Evgeny.Makarov]

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

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

How did you arrive at the last equivalence?

 

[evinda]

How did you arrive at the last equivalence?

Since $y \in \bigcap B \leftrightarrow \forall b \in B: y \in b$ I thought that we could also write this equivalence for each $b$ seperately and taking $b=\varnothing$ we would arrive at it. So can't we do it like that? (Thinking)

 

[Evgeny.Makarov]

Since $y \in \bigcap B \leftrightarrow \forall b \in B: y \in b$ I thought that we could also write this equivalence for each $b$ seperately and taking $b=\varnothing$ we would arrive at it.

It would be nice for you to learn the laws of reasoning, or inference rules, (especially since you are studying logic) and refer to them when you explain your reasoning. Saying "I could also write this equivalence for each $b$ separately" is not really an explanation. It is not clear when you can or cannot write something separately.

So, you use
\[
y \in \bigcap B \leftrightarrow \forall b \in B: y \in b
\]
to conclude
\[
y \in \bigcap B \leftrightarrow y \in \varnothing.
\]
In a similar way, do you use the equivalence
\[
p\text{ is prime}\leftrightarrow (\forall n\;n\mid p\to n=1\lor n=p)
\]
to conclude
\[
p\text{ is prime}\leftrightarrow (5\mid p\to 5=1\lor 5=p),
\]
which, if $p>5$, can be rewritten as
\[
p\text{ is prime}\leftrightarrow \neg(5\mid p)\ ?
\]

 

[evinda]

It would be nice for you to learn the laws of reasoning, or inference rules, (especially since you are studying logic) and refer to them when you explain your reasoning. Saying "I could also write this equivalence for each $b$ separately" is not really an explanation. It is not clear when you can or cannot write something separately.

So, you use
\[
y \in \bigcap B \leftrightarrow \forall b \in B: y \in b
\]
to conclude
\[
y \in \bigcap B \leftrightarrow y \in \varnothing.
\]
In a similar way, do you use the equivalence
\[
p\text{ is prime}\leftrightarrow (\forall n\;n\mid p\to n=1\lor n=p)
\]
to conclude
\[
p\text{ is prime}\leftrightarrow (5\mid p\to 5=1\lor 5=p),
\]
which, if $p>5$, can be rewritten as
\[
p\text{ is prime}\leftrightarrow \neg(5\mid p)\ ?
\]

Oh sorry! That what I said makes no sense... (Tmi)
Could you give me a hint how else we could show that $\bigcap B=\varnothing$ ? (Thinking)