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One of the propositions given in Advanced Calculus of a Single Variable is: If |a - b| < ε for each ε > 0 then a = b.

Would not this conflict with the epsilon-delta definition of limit? The epsilon-delta definition of limit states that: for every ε>0, there exists a δ>0, such that if 0<|x−a|<δ, then |f(x)−L|<ε. Here, even though it is for any ε>0 that |f(x)−L|<ε (where 0<|x−a|<δ for a certain δ), it is not possible for the inequality to become an equality (i.e. |f(x)−L|=ε). Would not ε be something similar to an infinitesimal as it represents a state of f(x) approaching but never reaching L? If so, will this conflict with the former proposition?

I hope what I am asking makes sense. Thank you!

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    Note that $f(x)$ is not a single number, whereas the $a$ in $|a-b|<\varepsilon$ is. – Mark S. Sep 29 '23 at 17:38
  • Also, if there exists an $\epsilon \gt 0$ such that no matter how close you restrict $x$ to $a$ you can't get $|f(x)-L| \lt \epsilon$, all that means is that the limit of $f$ at $a$ is not equal to $L$, (or may not even exist). – JonathanZ Sep 29 '23 at 17:59
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    In the $\epsilon,\delta$ definition of a limit, the definition starts with a given function $f(x)$, a given real number $a$, and a given real number $L$. The definition then proceeds by giving a meaning to this sentence: The limit as $x$ approaches $a$ of $f(x)$ equals $L$. So, if you are proposing that the proposition from that book conflicts with the $\epsilon,\delta$ definition of a limit, the natural question for you to ponder is this: What function $f(x)$, number $a$, and number $L$ are you using to demonstrate the conflict? – Lee Mosher Oct 03 '23 at 12:46

2 Answers2

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If $\varepsilon>0$ is infinitesimal as you suggest, then indeed it will typically be impossible to choose a standard $\delta>0$ to guarantee the implication "if 0<|x−a|<δ, then |f(x)−L|<ε" that you mentioned. However, one will still be able to choose a nonstandard $\delta>0$ to guarantee the implication (for continuous $f$, of course). The existence of infinitesimals is not incompatible with the axioms of the real numbers, so long as one works in a richer st-$\in$-language rather than the traditional $\in$-language of set theory; see e.g., this answer.

Mikhail Katz
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If you remove all of the parts involving $\delta$ from the epsilon-delta definition, then you indeed get something like the first proposition:

For every $\varepsilon > 0$, $\lvert f(x) - L \rvert < \varepsilon$.

This would imply (by the first proposition) that $f(x) = L$.

(It would not imply that $\lvert f(x) - L \rvert = \varepsilon$. Think again: how is that equation relevant to anything else in the question?)

If $f(x) = L$ for all $x$ then $x$ is a constant function. So a definition like this would only be useful for finding the limit of a constant function, which is hardly even worth defining.

That's why we need $\delta$. Moreover, it's why we need to introduce "some" $\delta$ and why we need to write "some" $\delta$ after we write "for all $\varepsilon$", because when it is written that way it allows us to specify a different $\delta$ for each $\varepsilon$ whenever we want to.

In the epsilon-delta definition, therefore, there is never a time when we choose two particular numbers $a$ and $b$, such as $a = f(x)$ for some $x$ and $b = L$, and say that $\lvert a - b \rvert < \varepsilon$, that is, $\lvert f(x) - L \rvert < \varepsilon$, for every $\varepsilon$. We always have $\delta$ getting in the way of that notion.

Remember that in the definition of a limit, $\lvert f(x) - L \rvert < \varepsilon$ only has to be true when $0 < \lvert x - a \rvert < \delta$. This implies that no matter what $x$ you might want to choose at which to evaluate $f(x)$, there is always a $\delta$ that forbids you to use that $x$. Obviously, $x = a$ is ruled out, and if you choose any other $x$, then $\delta = \frac12\lvert x - a \rvert$ will rule out that $x$ too.

Indeed, in a typical proof of a limit by the epsilon-delta definition, we are always looking at an ever-shrinking set of values of $x$ for smaller and smaller values of $\varepsilon$, there is no value of $x$ that is always in that set, and there is no particular $f(x)$ that needs to satisfy $\lvert f(x) - L \rvert < \varepsilon$ for every $\varepsilon$.

David K
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