supremum over closure

Theorem 1.

Let $f\mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{R}$ be a continuous function and $A\mathrm{\subseteq }\mathrm{R}$. Then $\underset{x\mathrm{\in }A}{\sup }\mathit{}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}\underset{x\mathrm{\in }\overline{A}}{\sup }\mathit{}f\mathit{}\mathrm{(}x\mathrm{)}$, where $\overline{A}$ denotes the closure of $A$.

Proof.

The theorem is clearly true for $A=\mathrm{\varnothing }$. Thus, it will be assumed that $A\ne \mathrm{\varnothing }$.

Since $A\subseteq \overline{A}$, we have $\underset{x\in A}{sup}f(x)\le \underset{x\in \overline{A}}{sup}f(x)$.

Suppose first that $\underset{x\in \overline{A}}{sup}f(x)=\mathrm{\infty }$. Let $r\in ℝ$. Then there exists $x_0\in \overline{A}$ with $f(x_0)\ge r+1$. Since $f$ is continuous, there exists $\delta >0$ such that, for any $x\in ℝ$ with , we have . Since $x_0\in \overline{A}$, there exists $x_1\in A$ with . (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, $f(x_1)>f(x_0)-1\ge r+1-1=r$. Hence, $\underset{x\in A}{sup}f(x)=\mathrm{\infty }$.

Now suppose that $\underset{x\in \overline{A}}{sup}f(x)=R$ for some $R\in ℝ$. Let $\epsilon >0$. Then there exists $x_2\in \overline{A}$ with $f(x_2)\ge R-\frac{\epsilon }{2}$. Since $f$ is continuous, there exists ${\delta }^{\prime }>0$ such that, for any $x\in ℝ$ with , we have . Since $x_2\in \overline{A}$, there exists $x_3\in A$ with . Thus, $f(x_3)-f(x_2)>\frac{-\epsilon }{2}$. Therefore, $f(x_3)>f(x_2)-\frac{\epsilon }{2}\ge R-\frac{\epsilon }{2}-\frac{\epsilon }{2}=R-\epsilon$. Hence, $\underset{x\in A}{sup}f(x)\ge R$.

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

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