continuity, supremum/infimum

Question

I came across this but I'm still struggling to understand the concept fully. Specifically, how do we know $\text{lub}\{f(x)\mid a-\epsilon<x<a+\epsilon\}-\text{glb}\{f(x)\mid a-\epsilon<x<a+\epsilon\}$ decreases as $\epsilon$ decreases?

Answer 1

Because as $\varepsilon$ decreases you're taking bounds over smaller sets. Anything that is an upper or lower bound for the larger set must necessarily also be an upper or lower bound for any subset. So at a minimum, we know that the difference is non-increasing as $\varepsilon$ gets smaller.

If $f$ is continuous at $a$ (as the answer to the linked question assumes), then the difference of the bounds must in fact go to $0.$ If the difference were to go to some $\mu \gt 0$, then any open set around $a$, no matter how small, would have points $y, z$ such that $\vert f(y)-f(z) \vert \gt \frac{\mu}{2}.$

That means for $\varepsilon \lt \frac{\mu}{4}$ it's impossible to find $\delta$ such that $\vert x-a \vert \lt \delta \Rightarrow \vert f(x)-f(a) \vert \lt \varepsilon$, which contradicts the assumption that $f$ is continuous at $a$.

Answer 2

If $A \subset B$ and $B$ is bounded above or below then $A$ is also bounded above or below and $\sup A \le \sup B$ and $\inf A \ge \inf B$.

Why? If $A$ is bounded above then there is a $u$ that is an upper bound of $B$. So $u \ge b; \forall b \in B$. So for all $a \in A \subset B$ we know $a \in B$ so $u \ge a$ so $u \ge b; \forall a \in A$. SO $u$ is an upperbound of $A$ so $A$ is bounded above.

Now let $s = \sup B$ the $s$ is an upper bound of $B$ but $A\subset B$ so $s$ is an upper bound of $A$. So the least upper bound of $A$ must be less than or equal to $A$. So $\sup A \le \sup B$.

The exact same argument is true the $\inf A \ge \inf B$.

So $\sup A \le \sup B$ a and $\inf A \ge \inf B$ so

$A\subset B$ and $A,B$ bounded above and below then $\sup A - \inf A \le \sup B-\inf A$.

Now if $\epsilon_1 < \epsilon_2$ and if $A =\{f(x)\mid a-\epsilon_1<x<a+\epsilon_1\} $ and $B= \{f(x)\mid a-\epsilon_2<x<a+\epsilon_2\} $ then (as $a-\epsilon_2 < a-\epsilon_1< a +\epsilon_1 < \epsilon_2$) we have $A \subset B$ and

our result follows.