I came across this problem while preparing for a qualifying exam... not really sure where to start. Help is much appreciated :)

Question: Consider the sequence $f_n(x) = \sin(nx)$ in $C(\mathbb{R},\mathbb{R})$ where $\mathbb{R}$ has the usual topology. For which of the following topologies for $C(\mathbb{R},\mathbb{R})$ does the sequence converge?

(a) Uniform topology

(b) The topology of pointwise convergence (i.e. the point-open topology)

(c) The compact-open topology (under our assumptions this topology coincides with the topology of compact convergence on $C(\mathbb{R},\mathbb{R})$.

My thoughts so far: (b) seems easy, it will not converge because for a "nice" choice of $x$, the sequence will not converge. i.e. if $x = \frac{\pi}{2}$, then $\sin(nx)$ is the sequence $\{1, 0, -1, 0 , 1, \ldots \}$, which clearly does not converge.

(a) and (c) are giving me some difficulties...

  • $\begingroup$ By uniform topology do you mean the topology of uniform convergence? If so, your negative answer to (b) implies negative answers to both (a) and (c). (Singletons are compact.) $\endgroup$ – Brian M. Scott Dec 6 '12 at 20:19

If it doesn't converge at a particular point, it certainly can't converge uniformly or in the compact-open topology (i.e. uniformly on compact sets).


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