I've got a question about mathematical induction.
$\bigg(P(0) \land \big[\forall n\in \mathbb{N},\;\; P(n)\implies P(n+1)\big]\bigg)\implies \forall n\in \mathbb{N}, \; P(n)$
for the inductive step usually we say
"Let any $n\in \mathbb{N}$ and we assume that $P(n)$ holds and we will show that P(n+1) holds"
But today I have seen this on an another forum
"We assume there exists $n$ such that $P(n)$ holds..."
I told them that is formally wrong, because we will prove that $P(n)\implies P(n+1)$ only for a specific $n$
What is your point of view, Am I right?
Example we want to prove that $\forall n\ge 1, \quad 2^n>n$
Induction step ($\forall n\ge 1,\;\; P(n)\implies P(n+1)$)
(*) Let any $n\ge 1$ and we assume that $2^n>n$ holds
$2^{n+1}>n+1\iff 2^n+2^n>n+1$ as $2^n>n$ and $2^n>1$, we conclude that$2^{n+1}>n+1$ holds
So we have proved that $\forall n\ge 1,\;\; P(n)\implies P(n+1)$ which means $\forall n\ge 1,\;\; P(1)\implies P(2)\implies \cdots \implies P(n)\implies P(n+1)\cdots$
It remains the base step because we know noting about $P(1)$, $P(2)$ and ...
$P(1) :2^1>1$ so $P(1)$ is true
So we can conclude that $\forall n\ge 1, \quad 2^n>n$
If somone says for the induction step (*)
*"We assume there exists $n\ge 1$ such that $2^n>n$ holds..."*
this "there exists" annoys me
After the base case, that is $\exists n_0$ such that $P(n_0)$ holds, for the induction step we assume as induction hypothesis that for some $n$ $P(n)$ holds and we need to show that $P(n) \implies P(n+1)$.
Note that when we prove the induction step we need to check that $P(n) \implies P(n+1)$ holds for $n\ge n_0$ otherwise we need to check again the base case.
For example when we prove by induction that $2^n\ge n^2$ we have that for the base case $n=0$ works but the induction step holds only for $n\ge 3$ then we need to prove the base case for $n\ge 3$ otherwise the proof by induction is wrong.