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 $
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 $\ge n_0$
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!