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 $A\eq\\emptyset$ .

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.

单词	SupremumOverClosure
释义	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 $A\eq\\emptyset$ . 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.
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