The set at which a function's oscillation is at least $\epsilon$ is closed

Question

Let $f$ be a bounded function on a compact interval $J$, and let $I(c,r)$ denote the open interval centered at $c$ of radius $r>0$. Let $osc(f,c,r)=\sup|f(x)-f(y)|$, where the supremum is taken over all $x,y\in J\cap I(c,r)$, and define the oscillation of $f$ at $c$ by $osc(f,c)=\underset{r\rightarrow 0}{\lim}osc(f,c,r)$. Clearly, $f$ is continuous at $c\in J$ if and only if $osc(f,c)=0$.

Prove that for every $\epsilon>0$, the set of points $c$ in $J$ such that $osc(f,c)\geq \epsilon$ is compact.

My Proof Attempt:

Proof. Let the assumptions be as above. Let $\epsilon > 0$. Define \begin{equation*} D_\epsilon=\{c\in J: osc(f,c)\geq \epsilon\} \end{equation*} It suffices to show that $D_\epsilon$ is closed as any closed subset of a compact set, in this case namely $J$, is also a compact set. Let $(x_n)_n\subset D_\epsilon$ such that $x_n\rightarrow x\in J$. We will prove that $x\in D_\epsilon$ and, thus, $D_\epsilon$ would be closed.

Claim: Since $x_n\in D_\epsilon$, that implies that for sufficiently small $\varrho$, \begin{equation*} osc(f,x_n,\varrho)\geq \epsilon \end{equation*} We will prove this claim by the following:

Let $r>r'>0$ and $x\in J$, then $J\cap I(x,r')\subset J\cap I(x,r)$. This means that $osc(f,x,r)\geq$ $osc(f,x,r')$ since $osc(f,x,r)$ is defined to be the supremum over the set $\{d(f(x),f(y)): x,y\in J\cap I(x,r)\}$. But any $x,y\in J\cap I(x,r')$ must also be in $J\cap I(x,r)$. Thus, $osc(f,x,r)$ cannot increase as $r$ is made smaller. If $osc(f,x)\geq \epsilon$, It cannot be that for "really small" $r>0$ that $osc(f,x,r)<\epsilon$. Because then the limit cannot be $\geq \epsilon$. This proves the Claim.

Now, given any $r>0$, either $(x-\frac{r}{2},x)\subset J\cap I(x,r)$ or $(x,x+\frac{r}{2})\subset J\cap I(x,r)$, or both. WLOG, suppose $(x-\frac{r}{2},x)\subset J\cap I(x,r)$ and denote it as $I$. Then $\exists N\in \mathbb{N}$ such that for all $n\geq N,$ it follows that $x_n\in I$. As $I$ is open, we can choose $\varrho'$ sufficiently small such that $B_{\varrho'}(x_n)\subset I$. Set $\rho=max\{\varrho, \varrho'\}$. Then \begin{equation*} osc(f,x,r)\geq osc(f,x_n,\rho)\geq \epsilon \end{equation*} Since we choose $N$ and $\rho$ for any given $r>0$, it follows that $osc(f,x)\geq \epsilon$ and $x\in D_\epsilon$. This completes our proof.

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Any corrections of the proof or comments on style are very much appreciated. Thank you guys for your time.

Answer 1

Your proof is ok, but maybe it's easier to show that the complement,

$D^c_\epsilon=\{x\in J: osc(f,x)< \epsilon\},$ is open.

If $x\in D^c_\epsilon$, there is a $\delta>0$ such that $\sup\{|f(z)-f(w)|:z,w\in (x-\delta,x+\delta)\}<\epsilon.$

Now suppose $y\in (x-\delta/3,x+\delta/3)$. Then, $B(y):=(y-\delta/2,y+\delta/2)\subseteq (x-\delta,x+\delta)$, so that $|f(z)-f(y)|<\epsilon$ as soon as $z\in B(y).$ From this it follows that $osc(f,y)<\epsilon$ and therefore, that $y\in D_{\epsilon}^c$.

Thus, we have found a ball about $x$, namely $ (x-\delta/3,x+\delta/3)$ which is contained in $D_{\epsilon}^c,$ and so $D_{\epsilon}^c$ is open.