Why is zero the only infinitesimal real number?

I am currently reading Elementary Calculus: An Infinitesimal Approach by H. Jerome Keisler and was wondering if someone could help me with an aspect treated in the book.

On page 24 he says a number $\varepsilon$ is said to be infinitely small or infinitesimal if $$-a< \varepsilon < a$$ for every positive real number $a$. He then says the only real number that is infinitesimal is zero.

I really don't get that. What I understand is that in order for a number to be considered infinitely small it has to be bigger then $-a$ and smaller then $a$. Well if I take $a$ to be $-2$ that means that $-1$ would be infinitesimal since it is bigger than $-2$ but smaller then $2$. So then how can zero be the only real number that satisfies that condition?

Answer

Your example of taking $a$ to be $2$ and concluding that $1$ is infinitesimal since it is between $-2$ and $2$ is not a good example.

The reason for this is that the definition of an infinitesimal $\varepsilon$ is that $-a \leq \varepsilon \leq a$ for every positive real number $a$. You just picked some positive real number. This has to be true for every positive real number. That means $\varepsilon$ needs to be in $[-2, 2]$ and in $[-1, 1]$ and in $[-\frac{1}{2}, \frac{1}{2}]$ and in $[-\frac{1}{1000000}, \frac{1}{1000000}]$, and so on. That same $\varepsilon$ has to be in all of these at the same time to be an infinitesimal.

The only real number that satisfies that it is between $-a$ and $a$ for every real $a > 0$ is $\varepsilon = 0$.

So any number $\varepsilon$ other than $0$ that satisfies $-a \leq \varepsilon \leq a$ for every $a > 0$ real cannot itself be a real number, but there are plenty of infinitesimals that aren't real numbers. As we discussed, $0$ is the only number that's both real and infinitesimal.