Do the point-open and compact-open topologies coincide on the space of continuous functions from $[0, 1]$ to $\mathbb{R}$, i.e. on $C([0, 1], \mathbb{R})$? If not, what would be a clear and simple counterexample?

I'm aware of the classic example that pointwise convergence does not imply uniform convergence (see https://en.wikipedia.org/wiki/Pointwise_convergence#Properties) but is there a simpler, topological argument?

  • $\begingroup$ The set $C([0,1],(0,1))$ is an open subspace of $C([0,1],\Bbb R)$ with the compact-open topology, but it's not open when the set is equipped with the point-open topology. $\endgroup$ – Stefan Hamcke Jun 21 '15 at 15:47
  • $\begingroup$ By the way, the point-open topology is just the topology inherited by the product topology on $\prod_{[0,1]}\Bbb R$. $\endgroup$ – Stefan Hamcke Jun 21 '15 at 15:48
  • $\begingroup$ Thank you, Stefan. Sorry if this is obvious, how do you prove that the set $C([0, 1], (0, 1))$ is not open when $C([0, 1], \mathbb{R})$ is equipped with the point-open topology, what's the actual argument? $\endgroup$ – Paul Jun 24 '15 at 8:20

The set $U=C(I,(0,2))$ is open in $C(I,\Bbb R)$ with the compact-open topology, and it contains the constant map $c\equiv1$. If you show that no basic open neighborhood $B$ of $c$ with respect to the point-open topology is contained in $U$, then this proves that the compact-open topology is finer than the point-open topology. Such a $B$ looks like $$\{g:I\to\Bbb R\mid g(x_1)\in(a_1,b_1)\}\cap\dots\cap\{g:I\to\Bbb R\mid g(x_n)\in(a_n,b_n)\}$$ with $a_1,\dots,a_n<1<b_1,\dots,b_n$. Can you show that there is a map $f\in B$ sending some point $x\in I$ to a value outside of $(0,2)$ ?

  • $\begingroup$ Thank you very much. I believe I can find such a map: one such map would be given by a Lagrange polynomial passing through $(x_1, 1), \ldots (x_n, 1)$ and $((x_1 + x_2)/2, 3)$, right? $\endgroup$ – Paul Jun 26 '15 at 8:13
  • $\begingroup$ @Paul: Or just the map which is constant on $[0,x_1]\cup[x_2,1]$, and on $[x_1,x_2]$ it is $$x\mapsto 3-2\frac{|x_1+x_2-2x|}{x_2-x_1}$$ assuming that $x_1<x_2<\dots<x_n$. $\endgroup$ – Stefan Hamcke Jun 26 '15 at 13:46

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