If $\lim_{x\to a}f(x)=L$, is the set $\{\delta\in\mathbb{R}:0<|x-a|<\delta\implies |f(x)-L|<\varepsilon\}$ always an interval?

Question

Here's an interesting problem I stumbled upon while I was pondering on the $(\varepsilon,\delta)$ definition of a limit:

Suppose that $\lim_{x\to a}f(x)=L$ ($f$ is a real function). Fix some $\varepsilon>0$ and let $S_\varepsilon$ be the set $$\{\delta\in\mathbb{R}:|f(x)-L|<\varepsilon\text{ is true for every $x\in\text{dom}[f]$ satisfying }0<|x-a|<\delta\}$$ Does $S_\varepsilon$ always contain an interval? If so, can we do better and conclude that $S_\varepsilon$ is an interval that is either $(0,b)$, $(0,b]$, or $(0,\infty)$ for some $b>0$?

I’m pretty sure the answer to the first question is yes. Here's my argument:

If $\lim_{x\to a}f(x)=L$, then $S_\varepsilon$ is not empty because the $(\varepsilon,\delta)$ definition of a limit implies that there is a $\delta>0$ such that

$$|f(x)-L|<\varepsilon\text{ is true for every }x\in\text{dom}[f]\text{ satisfying } 0<|x-a|<\delta$$

If we then fix any $\delta_1\in S_\varepsilon$, we can infer that every $s\in(0,\delta_1)$ will be an element of $S_\varepsilon$, since

\begin{align} x\in\text{dom}[f]\text{ and }0<|x-a|<s\text{ and }s<\delta_1 &\implies x\in\text{dom}[f]\text{ and }0<|x-a|<\delta_1\\ &\implies|f(x)-L|<\varepsilon \end{align} It follows $(0,\delta_1)\subseteq S_\varepsilon$, so $S_\varepsilon$ contains an interval.

Does this argument work? If so, is the stronger claim mentioned in the problem statement also demonstrably true? After spending some time visualizing the $(\varepsilon,\delta)$ definition for constant functions and functions which are bounded/unbounded at the endpoints of their domain, I suspect that the answer is yes. Unfortunately, I have no idea how to prove this. Any help is appreciated.

Answer 1

Looks like you have the gist of it, but you might be bogged down by some unnecessarily complicated notation. How about this:

Let $\delta > 0$ be such that for any $x$ that satisfies the equation \begin{equation} \tag{1} \left| x - a \right| < \delta, \end{equation} we have \begin{equation} \tag{2} \left|f(x) -L \right| < \varepsilon \end{equation}

Let $0<\delta' < \delta$, if $\left| x- a \right| < \delta'$, then, since $\delta' < \delta$, we have $\left| x -a \right| < \delta' < \delta$ i.e. $x$ satisfies equation (1). This implies that equation (2) also holds.

In particular, this implies that if $\delta > 0$ is in $S_\varepsilon$, then $\left(0, \delta \right) \subset S_\varepsilon$. EDIT: Call this Fact 1.

EDIT: $S_\varepsilon$ is an interval:

Ether $S_\varepsilon$ has a supremum or not. If the supremum $s$ exists, then by Fact 1 we have that $S_\varepsilon$ is either $(0, s)$ or $(0, s]$.

If $S_\varepsilon$ does not have a supremum, for any $M > 0$ there is an element $\delta > M$ in $S_\varepsilon$. Hence, by Fact 1, the set $S_\varepsilon$ contains the interval $(0, M)$ for arbitrarly large $M$'s. Thus $S_\varepsilon = (0,\infty)$.