If we have a function sequence $\left\{ f_{n} \right\}$ and it uniformly convergences to a $f$ on every compact subset of $\mathbb{C}$. Can I say $f_{n}$ is uniformly convergent to $f$ on $\mathbb{C}$?
Since $\mathbb{C}$ has no infinity, I think that for any $z \in \mathbb{C}$, $z$ is not infinity and then there should be a large enough disk $\overline{D(0, R)}$ such that $z \in \overline{D(0, R)}$, and $f_{n}$ is uniformly convergent to $f$ on $\overline{D(0, R)}$
But this argument makes me unrigorously feel that $\mathbb{C}$ is bounded which is clearly false. I'm pretty confused.
Also if the statement above is false. Is there a theorem that can ensure that if $f_n$ is uniformly convergent to $f$ on every compact set of $\mathbb{C}$, then $f_n$ is uniformly convergent to $f$ on $\mathbb{C}$?
Any help on this? Thanks.
Consider the sequence $\{f_n\}$ defined for $n=1,2,\ldots$ by $f_n(z):=z/n$.
Suppose $K \subset \mathbb{C}$ is compact.
Since every compact subset of $\mathbb{C}$ is closed and bounded, there is a positive number $M$ so that $|z|\leq M$ for every $z \in K$. So $\sup_{z \in K} |f_n(z)| \leq M/n$ for every $n \in \mathbb{N}$. This shows that $f_n \to 0$ uniformly on $K$.
Assume $f_n$ converges uniformly on $\mathbb{C}$. Since $f_n \to 0$ pointwise, we must have that $f_n \to 0$ uniformly on $\mathbb{C}$. So there is a positive integer $N$ such that $|f_n(z)|=|z|/n<1$ whenever $n \geq N$. But $f_N(2N)=2 \geq1$. This is a contradiction, and so our assumption must be false.