The following problem was in a supplement to Rudin's "Principles of Mathematical Analysis"
Let $\{f_n\}_{n \in Z_+}$ be a sequence of complex valued and continuous functions on a compact metric space $(K, d)$ and suppose that it converges pointwise to a continuous limit function $f$. Now let $p$ be any limit point of $K$, if for every sequence $\{x_n\}_{n \in Z_+} \subseteq K$ that converges to $p$, $f_n(x_n)$ converges to $f(p)$, then $f_n$ converges uniformly on $K$.
My Problem
I have tried to prove this statement without success so far. Therefore I turned to the site in order to find clues on how to proceed. However, I noticed a possibly incorrect trend in the answers to similar questions, namely:
"We shall proceed by contradiction. Suppose that $f_n$ does not converge uniformly on $K$, then, by the negation of the definition: $$(\exists M > 0): (\forall n \in Z_+, \exists x_n \in K: |f(x_n) - f_{n}(x_n)| \geq M)$$ and the proof follows easily for there is a convergent subsequence $x_{k_n} \rightarrow p$ of $x_n$ (here the construction of the sequence itself is subtended) and $f_{k_{n}}(x_{k_n}) \rightarrow f(p)$"
However, an incorrect negation of the definition of uniform convergence is used in such proofs. Recall that $f_n$ converges uniformly to $f$ on $K$ if and only if: $$(\forall \epsilon > 0), (\exists N \in Z_+: \forall x \in K, \forall n \geq N, |f(x) - f_{n}(x)| < \epsilon)$$
therefore the correct negation of the definition of uniform convergence is:
$$(\exists M > 0): (\forall N \in Z_+, \exists x_N \in K \text{ and } \exists n_N \geq N: |f(x_N) - f_{n_N}(x_N)| \geq M)$$
For example, let $N=2$, then, if we subtend the construction of the sequences $\{x_N\}$ and $\{n_N\}$, we could have $n_N$ = 1000 and hence $|f(x_2) - f_{1000}(x_2)| \geq M$ insted of $|f(x_2) - f_{2}(x_2)| \geq M$. Moreover, I do not see an easy way to to extend $\{x_N\}$ so that there is correspondence with $\{n_N\}$ for when we extract the subsequence we need to be sure that there are infinitely many elements of the subsequence satisfying the negation above.
My Question
Now, I am not completely sure that what I have written is completely right, therefore I leave to anyone to prove me wrong. Furthermore, I would like to know how can the statement be proved.
As always, any comment or answer is welcome and let me know if I can explain myself clearer!
Your suspicions are correct, that negation is slightly incorrect, but it may not actually affect the rest of a proof that someone gave, I don't know.
Since uniform convergence says "eventually you are close in the supremum norm", the negation says "infinitely often you are not close in the supremum norm". i.e. The correct negation gives you a "bad" $M > 0$ for which it is the case that for infinitely many $n$ we have $\sup_{x \in K}|f(x) - f_n(x)| \geq M$. Then to unwind 'infinitely many' you say something like you have said: For every $N \in \mathbb{N}$, there exists a later $n_N > N$ such that....
I suppose since $K$ is compact you can now get a convergent subsequence of the $x_{n_N}$ to a limit point $p$ and then...