1
$\begingroup$

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 $\mathscr{C}(\Omega,\mathbb{R})$?

$\endgroup$

1 Answer 1

1
$\begingroup$

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) $G_{\delta \sigma \delta}$.

$\endgroup$
2
  • $\begingroup$ 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 $F_{\sigma,\delta}$. $\endgroup$ May 4, 2012 at 12:22
  • $\begingroup$ 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. $\endgroup$ May 4, 2012 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.