supremum over closure

Theorem 1.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be a continuous function and $A\subseteq\mathbb{R}$ . Then $\displaystyle\sup_{x\in A}f(x)=\sup_{x\in\overline{A}}f(x)$ , where $\overline{A}$ denotes the closure of $A$ .

Proof.

The theorem is clearly true for $A=\emptyset$ . Thus, it will be assumed that $A\neq\emptyset$ .

Since $A\subseteq\overline{A}$ , we have $\displaystyle\sup_{x\in A}f(x)\leq\sup_{x\in\overline{A}}f(x)$ .

Suppose first that $\displaystyle\sup_{x\in\overline{A}}f(x)=\infty$ . Let $r\in\mathbb{R}$ . Then there exists $x_{0}\in\overline{A}$ with $f(x_{0})\geq r+1$ . Since $f$ is continuous, there exists $\delta>0$ such that, for any $x\in\mathbb{R}$ with $-\delta<x-x_{0}<\delta$ , we have $-1<f(x)-f(x_{0})<1$ . Since $x_{0}\in\overline{A}$ , there exists $x_{1}\in A$ with $-\delta<x_{1}-x_{0}<\delta$ . (Recall that $x\in\overline{A}$ if and only if every neighborhood of $x$ intersects $A$ .) Thus, $f(x_{1})-f(x_{0})>-1$ . Therefore, $f(x_{1})>f(x_{0})-1\geq r+1-1=r$ . Hence, $\displaystyle\sup_{x\in A}f(x)=\infty$ .

Now suppose that $\displaystyle\sup_{x\in\overline{A}}f(x)=R$ for some $R\in\mathbb{R}$ . Let $\varepsilon>0$ . Then there exists $x_{2}\in\overline{A}$ with $f(x_{2})\geq R-\frac{\varepsilon}{2}$ . Since $f$ is continuous, there exists $\delta^{\prime}>0$ such that, for any $x\in\mathbb{R}$ with $-\delta^{\prime}<x-x_{0}<\delta^{\prime}$ , we have $\frac{-\varepsilon}{2}<f(x)-f(x_{2})<\frac{\varepsilon}{2}$ . Since $x_{2}\in\overline{A}$ , there exists $x_{3}\in A$ with $-\delta^{\prime}<x_{3}-x_{2}<\delta^{\prime}$ . Thus, $f(x_{3})-f(x_{2})>\frac{-\varepsilon}{2}$ . Therefore, $f(x_{3})>f(x_{2})-\frac{\varepsilon}{2}\geq R-\frac{\varepsilon}{2}-\frac{% \varepsilon}{2}=R-\varepsilon$ . Hence, $\displaystyle\sup_{x\in A}f(x)\geq R$ .

In either case, it follows that $\displaystyle\sup_{x\in A}f(x)=\sup_{x\in\overline{A}}f(x)$ . ∎

Note that this theorem also holds for continuous functions $f\colon X\to\mathbb{R}$ , where $X$ is an arbitrary topological space. To prove this fact, one would need to slightly adjust the proof supplied here.

Title	supremum over closure
Canonical name	SupremumOverClosure
Date of creation	2013-03-22 17:08:22
Last modified on	2013-03-22 17:08:22
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	9
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 06A05
Classification	msc 26A15