Oscillation less than epsilon gives property near x

Question

I'm working through a Real Analysis textbook and came across this problem that I am having trouble making rigorous

Let $f:[a,b]\rightarrow \mathbb{R}$ be bounded, $\epsilon>0$ and $\omega_f(x) < \epsilon$ for all $x \in [a, b]$. Show for each $x_0 \in (a, b)$ there exists $h > 0$ such that $x, y \in B_h(x_0) \implies |f(x) − f(y)| < \epsilon$.

The book defines oscillation ($\omega_f(x_0)$) to be $\limsup_{x\rightarrow x_0}f(x)-\liminf_{x\rightarrow x_0}f(x)$.

I see the general idea that the oscillation being less than $\epsilon$ tells us that near $x_0$, the function cannot change too much, but I'm not seeing how to rigorously make that connection.

This is pretty standard. You need to use Heine Borel theorem. — Paramanand Singh, Feb 06 '21 at 02:25
In general when you need to extend some local property (dealing with a point) of a function to a global property (dealing with an interval) then you use Heine Borel theorem. — Paramanand Singh, Feb 06 '21 at 02:28

score 5 · Accepted Answer · answered Feb 06 '21 at 02:24

We have for all $z \in [a,b]$,

$$\lim_{\delta \to 0+}\sup\{f(x) :x \in B_\delta(z)\cap[a,b]\} - \lim_{\delta \to 0+}\inf\{f(x) :x \in B_\delta(z)\cap[a,b]\}\\:= \omega_f(z) < \epsilon, $$

and there exists $\delta_z $ such if that $x,y \in B_{2\delta_z}(z)\cap[a,b]$, then $|f(x) - f(y)| < \epsilon$.

Since $[a,b]$ is compact and $\{B_{\delta_z}(z):z \in [a,b]\}$ is an open cover, there is a finite subset of intervals $B_{\delta_k}(z_k)$ where $k = 1, \ldots, n$ and

$$\tag{*}[a,b] \subset \bigcup_{k=1}^n B_{\delta_k}(z_k)$$

For any $x \in [a,b]$ there exists by (*) a $j$ such that $x \in B_{\delta_j}(z_j)$. Let $\delta = \min(\delta_1,\ldots, \delta_n)$ and suppose for $y \in [a,b]$ we have $|y-x| < \delta$. Then we have $|x - z_j| < \delta_j < 2\delta_j$ and

$$|y-z_j| \leqslant |y-x| + |x- z_j| < \delta + \delta_j < 2\delta_j$$

Thus, $x,y \in B_{2\delta_z}(z_j)\cap[a,b]$ and it follows that $|f(x) - f(y)| < \epsilon$.

To summarize, for any $\epsilon > 0$ there exists $\delta > 0$ such that for all $x,y \in [a,b]$ such that $|x-y|< \delta$ we have $|f(x) - f(y)| < \epsilon$.

Finally, for any $x_0 \in (a,b)$ there exists $h_0>0$ such that $B_{h_0}(x_0) \in (a,b)$ since $(a,b)$ is open. Let $h = \min (h_0,\delta/2)$. If $x,y \in B_h(x_0)$, then we have both $x,y \in [a,b]$ and $|x-y|< \delta$, and, hence, $|f(x) - f(y)| < \epsilon$.

Oscillation less than epsilon gives property near x

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