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For $\delta>0$ let $$M(a,f,\delta)=\sup\{f(x):x\in A\text{ and }|x-a|<\delta\},\\ m(a,f,\delta)=\inf\{f(x):x\in A\text{ and }|x-a|<\delta\}.$$ The oscillation $o(f,a)$ of $f$ at $a$ is defined by $o(f,a)=\lim_{\delta\to 0} [M(a,f,\delta)-m(a,f,\delta)]$. This limit always exists, since $M(a,f,\delta)-m(a,f,\delta)$ decreases as $\delta$ decreases.

In the proof of Theorem 1-11 in "Calculus on Manifolds" by Michael Spivak, the author wrote as follows:

Let $C$ be an open rectangle containing $x$ such that ...

Where did the author use the condition that $C$ contains $x$?

Is the condition really necessary?

1-11 Theorem. Let $A\subset\mathbb{R}^n$ be closed. If $f:A\to\mathbb{R}$ is any bounded function, and $\varepsilon>0$, then $\{x\in A:o(f,x)\geq\varepsilon\}$ is closed.

Proof. Let $B=\{x\in A:o(f,x)\geq\varepsilon\}$. We wish to show that $\mathbb{R}^n-B$ is open. If $x\in\mathbb{R}^n-B$, then either $x\notin A$ or else $x\in A$ and $o(f,x)<\varepsilon$. In the first case, since $A$ is closed, there is an open rectangle $C$ containing $x$ such that $C\subset\mathbb{R}^n-A\subset\mathbb{R}^n-B$. In the second case there is a $\delta>0$ such that $M(x,f,\delta)-m(x,f,\delta)<\varepsilon$. Let $C$ be an open rectangle containing $x$ such that $|x-y|<\delta$ for all $y\in C$. Then if $y\in C$ there is a $\delta_1$ such that $|x-z|<\delta$ for all $z$ satisfying $|z-y|<\delta_1$. Thus $M(y,f,\delta_1)-m(y,f,\delta_1)<\varepsilon$, and consequently $o(y,f)<\varepsilon$. Therefore $C\subset\mathbb{R}^n-B$.

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    $\begingroup$ Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. $\endgroup$ Commented Oct 17, 2023 at 8:56
  • $\begingroup$ Thank you for your advice. I will edit my question now. $\endgroup$ Commented Oct 17, 2023 at 8:59

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Spivak wants to prove that $B$ is closed which means that $\mathbb R^n - B$ is open. By definition of "open" he has to show that for each $x \in \mathbb R^n - B$ there exists an open rectangle $C$ such that $$x \in C \subset \mathbb R^n - B \tag{1} .$$

In case $x \notin A$ this is obvious.

If $x \in A$ (i.e. if $x$ is in the domain of $f$), we know that $o(f,x) < \varepsilon$ because $x \notin B$. Hence there exists $\delta > 0$ such that $M(x,f,\delta)-m(x,f,\delta)<\varepsilon$. The question is how to find $C$ with property $(1)$. Certainly one has to take some $C$ with $x \in C$. Rectangles not containing $x$ are useless here. Spivak takes a "sufficiently small" such $C$ which is made precise by requiring that $|x-y|<\delta$ for all $y\in C$. He then proves that $C \subset \mathbb R^n - B$ which means that $o(f,y) < \varepsilon$ for all $y \in C$:

If $y\in C$ there is a $\delta_1 > 0$ such that $|x-z|<\delta$ for all $z$ satisfying $|z-y|<\delta_1$. Thus $M(y,f,\delta_1)-m(y,f,\delta_1)<\varepsilon$, and consequently $o(y,f)<\varepsilon$. Therefore $C\subset\mathbb{R}^n-B$.

In this argument it is essential to use the fact that $|z-y|<\delta_1$ implies $|x-z|<\delta$, and finding $\delta_1$ only works for $x \in C$.

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  • $\begingroup$ Thank you very much for your kind answer. $\endgroup$ Commented Oct 17, 2023 at 11:13
  • $\begingroup$ I cannot upvote now because I am new here. Sorry. $\endgroup$ Commented Oct 17, 2023 at 11:14

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