Let us denote $ \ V$ to be the set of all real valued functions $ \ f: \mathbb{R} \to \mathbb{R} $. Consider the two Topologies $ \tau_1 , \ \tau_2 $ on $ \ V$ as follows:
$\tau_1=$ induced by uniform convergence on compact sets
$\tau_2=$induced by local neighborhoods $$ N(f,[a,b], \epsilon)=\{g \ : \ |g(x)-f(x)<\epsilon \} $$
Then compare these two topologies $ \tau_1, \ \tau_2$.
Are they same?
Answer:
The topology $ \tau_1$ induced by uniform convergence while the topology. So it has base $ \{g: |g(x)-f(x)<\epsilon \} $
which is same as $ \tau_2$.
Thus $ \tau_1, \tau_2$ are same topology .
Am I right ?
Help me doing this
Best Answer
You are right.
(i). If a sequence $(f_n)_n$ converges uniformly to $f$ on every compact subset of $\Bbb R$, it will, a fortiori, converge uniformly to $f$ on every compact (i.e. closed bounded) interval.
(ii). If $(f_n)_n$ converges uniformly to $f$ on every compact interval, and if $C$ is any compact subset of $\Bbb R,$ then there exist $a,b\in \Bbb R$ with $[a,b]\supset C.$ Now $(f_n)_n$ converges uniformly to $f$ on $[a,b],$ so $(f_n)_n$ converges uniformly to $f$ on any subset of $[a,b].$ In particular, $(f_n)_n$ converges uniformly to $f$ on $C.$