5

I am trying to follow a proof in Hijab's: Introduction to Calculus and Analysis.

He gives the definition of inductive set:

enter image description here

And then there is this proof where he shows that there is no $n$ between $1$ and $2$.

enter image description here

I am confused in multiple levels here.

  • What is happening in this proof? I guess it's not clear to me what information we actually know and what is the pathway to prove it.
  • By picking $S=\{1\}\cup \{n\in N: n \geq 2\}$ isn't he assuming what we want to prove?
Red Banana
  • 23,956
  • 20
  • 91
  • 192
  • 2
    What the author is describing is a proof that $(1, 2)\cap \mathbb{N} = \varnothing$, not a proof of Theorem 1.3.1 .The idea is that the set $S\subset \mathbb{N}$ is also inductive and thus must be $S$ by Theorem 1.3.1. Since $S$ doesn't contain any element of $(1, 2)$ by construction, neither does $\mathbb{N}$. – anomaly Mar 06 '23 at 00:46
  • @anomaly I know that this isn't a proof of theorem 1.3.1. See the question title. – Red Banana Mar 06 '23 at 00:53
  • 1
    What is the author's definition of $\Bbb N$? – Robert Shore Mar 06 '23 at 02:00

3 Answers3

2

What is actually known:

  1. What does $1, 2, +$ mean
  2. $1+1=2$
  3. What is $\mathbf{N}$, What is $\mathbf{R}$, and what $<, =$ mean
  4. How we define an inductive set

The author picks $S = \{1\} \cup \{n \in \mathbf{N}, n \geq 2\}$ but they have not assumed that
there is no $x \in\mathbf{N} $ such that $x \neq 1 ; x < 2$.

That's what they want to prove.

whoisit
  • 3,069
1

The author is doing the following:

  1. Create a set $S$ that includes $1$ and all natural numbers greater than or equal to $2$.

  2. Demonstrating that $S$ is inductive per the definition.

  3. Concluding that $S = \mathbb{N}$.

  4. Noting that by construction, $S$ contains no elements between $1$ and $2$, and hence concluding that there are no natural numbers between $1$ and $2$.

ConMan
  • 24,300
1

By picking $S=\{1\}\cup\{n∈\mathbb{N}:n≥2\}$ isn't he assuming what we want to prove?

The answer is no, they are not assuming what the want to prove, the reason why is very simple, the sentence

Let $S=\{1\}\cup\{n∈\mathbb{N}:n≥2\}$

Is not an assumption, it is just a definition. You may be confused because you are interpreting this as the author assuming that $S=\mathbb{N}$, based on the statement of the previous theorem using $S$ as the name for the set which turns out to be $\mathbb{N}$, but that $S$ is not some fixed, predefined mathematical object, it is just a placeholder name for possible sets that you might try and apply the theorem to.

Carlyle
  • 2,807
  • 2
  • 22