# Spivak Calculus: Chapter 2 question 9

I've been searching for an awnser but couldn't find anything satisfying. I have to use mathematical induction to prove the following:

Spivak's Calculus, chapter 2 question 9

Prove that if a set a $A$ of set of natural numbers contains $n_0$ and $k+1$ whenever it contains $k$, then $A$ contains all natural numbers $\ge n_0$

Book's solution

Let $B$ be the set of all natural numbers $l$ such that $n_0 -1 + l$ is in A.; Then 1 is in $B$, and $l+1$ is in $B$ whenever $l$ is in $B$, so $B$ contains all natural numbers, which means that $A$ containts all natural numbers $\ge n_0$

My questions:

Here is what I understand:

I get that $l =1$ is in $B$ since $n_0 -1 +1 = n_0 -1 + 1 = n_0 \in A$

However I dont understand why they assume that $l+1$ is in B whenever $l$ is in $B$..

I would appreciate a lot if someone could give a complete explanation of this assumption and how can you prove that $l+1$ is in B and how it leads to proving the statement (the question)

Thank you so much and sorry If I made some english mistakes!

• Seems like the proof would've been simpler than the book to say that since $n_0 \in A$, let $k=n_0$, so $n_0+1=n_1 \in A$. From there say: assume $n_i \in A$ then $n_i+1=n_{i+1} \in A$ as second step in induction. – Χpẘ Jul 28 '17 at 22:37

It follows from the assumption on the set $A$ that whenever $k$ is in $A$, then $k+1$ is in $A$.
In detail, assume that $l$ is in $B$. By definition of $B$, we have that $n_0-1+l$ is in $A$. Let $k=n_0-1+l$, so $k+1 = (n_0-1)+(l+1)$ is in $A$. By definition, this implies that $l+1$ is in $B$.
Note that whenever $k \in A$, $k+1 \in A$. As such, $n_0-1+l \in A$ implies $n_0+l \in A$. Therefore, $l+1 \in B$.
If $l\in B$ then $n_0-1+l\in A$. By definition of $A$ you can conclude that $n_0+l\in A$ and in the same form $n_0+l+1\in A$, this implies $l\in B$.