Let $C \subset \mathbb{R}$ be a compact set. Let $f$ be a function $\mathbb{R} \rightarrow C$ which is continuous in $x_0$, and an $\varepsilon$-interval around it, $x_n$ a sequence of numbers s.t. $\lim_{n}{x_n}=x_0$, and $f_n: \mathbb{R} \rightarrow C$ a sequence of functions s.t. $\forall x \in \mathbb{R} \ \lim_{n}{f_n(x)}=f(x)$. Note: $f_n$ are not necessarily continuous in $x_0$.
I can prove that
$$ \lim_{n \to \infty}{f_n(x_n)}=f(x_0) $$
using continuity + chaining:
$ |f_n(x_n) - f(x_0)| \leq |f_n(x_n) - f(x_n) + f(x_n) - f(x_0)| \overset{Trianlge}{\leq} |f_n(x_n) - f(x_n)| + |f(x_n) - f(x_0)| \leq \varepsilon_1 + \varepsilon_2$
where $\varepsilon_1$ is due to point-wise uniform convergence and $\varepsilon_2$ due to continuity of $f$ around $x_0$. I'm specifically omitting a lot of details here as I'm not interested in the correctness of the proof itself. E.g., I still have to figure out uniform convergence.
I have a few questions regarding $\lim_{n \to \infty}{f_n(x_n)}$
- Can it be computed as $\lim_{m \to \infty}\lim_{n \to \infty}{f_n(x_m)}$? Note $f_n$ is not necessarily continuous, but $\lim_n(f_n)$ is continous in an epsilon ball around $x_0$.
- Is the above question related to the Moore-Osgood Theorem? How does the uniform convergence in the latter relate to my setup?
- answered in the comments: my proof would not work without it
- What are the conditions for $\lim_{n \to \infty}\lim_{m \to \infty}{f_n(x_m)} = \lim_{m \to \infty}\lim_{n \to \infty}{f_n(x_m)} = \lim_{n \to \infty}{f_n(x_n)}$ in general?