# supremum over closure

###### 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, we have $-1. Since $x_{0}\in\overline{A}$, there exists $x_{1}\in A$ with $-\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}, we have $\frac{-\varepsilon}{2}. Since $x_{2}\in\overline{A}$, there exists $x_{3}\in A$ with $-\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 SupremumOverClosure 2013-03-22 17:08:22 2013-03-22 17:08:22 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Theorem msc 06A05 msc 26A15