MHB Why does it suffice to show that X is an inductive set?

  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Set
Click For Summary
To demonstrate that the elements of natural numbers are indeed natural numbers, it suffices to show that the set X, defined as {n ∈ ω: ∀y ∈ n (y ∈ ω)}, is inductive. This is based on the principle that if a subset X of ω contains 0 and is closed under the successor operation, then X must equal ω. Since X is a subset of ω, proving its inductiveness confirms that all elements n in ω satisfy the condition ∀y ∈ n (y ∈ ω). Thus, the conclusion follows that all elements of natural numbers are natural numbers. The argument effectively utilizes the properties of inductive sets to establish this relationship.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hi! (Smile)

We want to show that the elements of the natural numbers are natural numbers, i.e. $(n \in \omega \wedge x \in n) \rightarrow x \in \omega$

Could you explain me why, in order to show this, it suffices to show that $X=\{ n \in \omega: (\forall y \in n)(y \in \omega)\}$ is an inductive set? (Worried)
 
Physics news on Phys.org
I am looking again at the proof..

It suffices to show that $X=\{ n \in \omega: (\forall y \in n) (y \in \omega)\}$ is inductive because we know that $X \subset \omega$ and because of the following sentence:

For each subset $X$ of $\omega$ it holds the following:

$$0 \in X \wedge (\forall n (n \in X \rightarrow n' \in X)) \rightarrow X=\omega$$
we conclude that $\{ n \in \omega: (\forall y \in n) (y \in \omega)\}=\omega$ and thus $(\forall y \in n) (y \in \omega)$ holds for all $n \in \omega$, i.e. the elements of natural numbers are natural numbers, right? (Smile)
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
13
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K