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Let $\Omega\subset\mathbb{R}^n$ be open, $\mathscr{C}(\Omega,\mathbb{R})$ the Fréchet space of real-valued continuous functions on $\Omega$ endowed with the compact-open topology, and $\mathscr{C}_u(\Omega,\mathbb{R})\subset\mathscr{C}(\Omega,\mathbb{R})$ the linear subspace of uniformly continuous real-valued functions on $\Omega$.

Is $\mathscr{C}_u(\Omega,\mathbb{R})$ some member of the Borel hierarchy of subsets of $\mathscr{C}(\Omega,\mathbb{R})$? For instance, is $\mathscr{C}_u(\Omega,\mathbb{R})$ a $G_\delta$ set, or a $F_\sigma$ set, in the compact-open topology of $\mathscr{C}(\Omega,\mathbb{R})$?

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Let $(K_n)_{n\in\mathbb N}$ be a compact exhaustion of $\Omega$ (that is, every compact set is contained in some $K_n$) and define $$ A_{n,m,k}=\lbrace f\in \mathscr C(\Omega,\mathbb R): \sup\lbrace |f(x)-f(y)|: x,y \in K_n, d(x,y)<1/m\rbrace < 1/k\rbrace.$$ This set is open with respect to the semi-norm $\|f\|_n=\sup\lbrace |f(x)|:x\in K_n\rbrace$ which easily follows from the triangle inequality. Hence $$ \mathscr C_u(\Omega,\mathbb R)= \bigcap_{k\in\mathbb N} \bigcup_{m\in\mathbb N} \bigcap_{n\in\mathbb N} A_{n,m,k}$$ is (at least) $G_{\delta \sigma \delta}$.

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Replacing $<1/k$ by $\le 1/k$ in the definition of $A_{n,m,k}$ one gets closed sets, hence the space of uniformly continuous functions is also $F_{\sigma,\delta}$. – Jochen Wengenroth May 4 '12 at 12:22
OK, I got the idea. It is clear that $\mathscr{C}_u(\Omega,\mathbb{R})$ can be written as above. To show that $A_{n,m,k}$ is open, one notices that $f\in A_{n,m,k}$ implies that $\sup_{x,y\in K_n:d(x,y)<\frac{1}{m}}|f(x)-f(y)|\leq\frac{1}{k}-\epsilon$ for some $0<\epsilon<\frac{1}{k}$. If $g\in\mathscr{C}(\Omega,\mathbb{R})$ satisfies $|f(x)-g(x)|\leq\frac{\epsilon}{3}$ for all $x\in K_n$, then one has by an $\frac{\epsilon}{3}$ argument that $\sup_{x,y\in K_n:d(x,y)<\frac{1}{m}}|g(x)-g(y)|\leq\frac{1}{k}-\frac{\epsilon}{3}<\frac{1}{k}‌​$, as desired. – Pedro Lauridsen Ribeiro May 4 '12 at 13:06

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