Why is the function for this statement, $f(x) = L$?

Question

Let $f$ be a function. Let $a, L \in \mathbb R$. Assume that $f$ is defined on some open interval around $a$, except maybe at $a$.

There exists $ > 0$ such that for every $ε > 0$, $$0 < |x - a| < \implies |f(x) - L| < ε$$

Could someone explain the meaning of this statement and how it results in the function being a horizontal line?

Answer 1

I claim that whenver $0<|x-a|<\delta$, $f(x)=f(a)$.

For if not, let $\varepsilon = \dfrac12|f(x)-f(a)|$. Then, according to the premise, $|f(x)-f(a)| < \varepsilon$, contradiction.

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Now, I claim that $f(x)=f(y)$ for all $x$ and $y$.

Let $N = \dfrac1\delta|x-y|$, and form a sequence $x,x+0.5\delta,x+\delta+,x+1.5\delta,x+2\delta,\cdots,x+N\delta$. Each consecutive term is the same when $f$ is applied, by the above section, i.e. $f(x)=f(x+0.5\delta)$, $f(x+0.5\delta)=f(x+\delta)$, etc.

Therefore, $f(x)=f(x+N\delta)$. Since $x+N\delta$ and $y$ differ by an amount less than $\delta$, by the above section, $f(x+N\delta)=f(y)$.

Therefore, $f(x)=f(y)$.

Answer 2

Consider the statement

$$\exists \delta > 0 \; \forall \varepsilon > 0, \; 0 < |x-a| < \delta \implies |f(x)-L| < \varepsilon $$

This statement is saying that, when we pick $x$ close enough to $a$, we can make $f(x)$ as close to $L$ as we like. The statement guarantees us that if we pick $x$ within a distance $\delta$ of $a$, we can pick $\varepsilon$ however small we want, and we should have $|f(x)-L| < \varepsilon$. The only way this is possible is if we indeed have $f(x) = L$.

Answer 3

Choose $x_0$ such that $0\lt|x_0-a|\lt\delta$ and $f(x_0)\ne L$ (if a such $x_0$ does not exist there is nothing to prove) Put $f(x_0)=L+h$ where $|h|\ne 0$ so you have $$|f(x_0)-L|=|L+h-L|=|h|\lt \epsilon$$ absurde for arbitrary small $\epsilon$. Thus $f(x)=L$ where it is defined.