# Where did the author use the condition that $C$ contains $x$? (Michael Spivak's "Calculus on Manifolds")

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 $$C\subset\mathbb{R}^n-B$$.

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

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 $$x \in C$$.