Proof step in Steps in Commutative Algebra I am trying to solve this exercise in Sharp's Steps in Commutative Algebra
(I did not copy the question itself, just the first part of the exercise to prevent solutions of it):

For me we should have $a-b\notin I^t$ above the red line.
I need help.
Thanks in advance.
 A: I  think your mistake lies in negating the sentence of the form $``\exists x : P(x)$ is true". The expanded form of the sentence in the book is


There exists a greatest $t_0 \in \Bbb{N}$ such that $(a-b) \in I^{t_0}$.


What is the negation of this sentence? Well first we need to understand what it means! It means that for any $t > t_0$, we have $(a-b) \notin I^t$. Thus the negation of the sentence in the book is:


There is no greatest $t_0 \in \Bbb{N}$ such that $(a-b) \in I^{t_0}$. 


In other words,  here is the situation that is going to happen. Start with $a-b$ which is an element of $R = I^0$. Note by definition $I^0$ is equal to $R$.
The must be some $t_1 > 0$ such that $(a-b) \in I^{t_0}$ otherwise this $0$ would be the greatest element in $\Bbb{N}$ such that $(a-b) \in I^0$ and $a-b \notin I^t$ for all $t > 0$.
Having chosen $t_1$, we can again choose another $t_2$ such that $a-b \in I^{t_2}$ again by repeating the argument in the paragraph before. Continuing this process ad infinitum shows we have a sequence of natural numbers
$$t_0,t_1,t_2,\ldots $$
such that $a-b \in I^{t_k}$ for all $k \geq 0$. In summary:


The sentence "There is no greatest $t_0 \in \Bbb{N}_0$ such that $a-b \in I^{t_0}$" implies the sentence "There is a sequence $t_0,t_1,\ldots $ such that $a-b \in I^{t_k}$ for all $k \geq 0$.


A: Assume there is no greatest $t$ such that $a-b\in I^t$, then $a-b\in I^{t_k}$ for some increasing sequence $\{t_k:k\in\mathbb{N}_0\}$. In this case
$$
a-b\in\bigcap\limits_{k=1}^\infty I^{t_k}\subset\bigcap\limits_{n=1}^\infty I^n
$$
so $\bigcap_{n=1}^\infty I^n\neq 0$. Contradiction, hence there is the greatest $t\in\mathbb{N}_0$ such that $a-b\in I^t$.
