Convergence of sequence of function Let $\{ f_n \}$ be a sequence of continuous complex-valued functions on $R$ such that $f(x) = \lim_{n \to \infty } f_n(x)$ exists (as a complex number) for each $x \in R$, and let $\epsilon > 0$. 
Prove that there is a non-empty open subset $V$ of $R$ and a positive integer $N$ such that $|f_n(x) - f(x) | \leq \epsilon $ if $x \in V$ and $n \geq N$. 

I am a little confuse on this questions. It seems like what the question asked me to prove is just from the definition that $f(x) = \lim_{n \to \infty } f_n(x)$. 
If you can offer me some guidelines I would be greatly appreciated. 
 A: The question asks you to prove that there is a non-empty open set such that the convergence of $f_n$ to $f$ satisfies a condition similar to that of uniform convergence there. This is different from the pointwise convergence of $f_n$ to $f$: that $f_n \to f$ pointwise on $\mathbf R$ means that for every $x \in \mathbf R$ and every $\epsilon > 0$, there exists an integer $N$ - depending on $x$ and on $\epsilon$ - such that $|f_n(x) - f(x)| \le \epsilon$ for all $n \ge N$. A different $x$ may require a larger $N$. You are asked to show that there is a non-empty open set on which the choice of $N$ is independent of $x$.
I have managed to prove the statement by an application of the Baire category theorem, under the additional assumption that the limit function $f$ is continuous also.
I will assume acquaintance with the following (there is an abundance of resources on this topic):


*

*The definition of a nowhere dense set.

*The Baire category theorem: a non-empty complete metric space is not a countable union of nowhere dense sets.

*A closed subset of a metric space is nowhere dense if and only if it has empty interior.


As for the proof (assuming continuity of $f$!):


*

*Explain why we can assume, without loss of generality, that each $f_n$ is real-valued and that $f = 0$.


Suppose now that $\{ f_n \}$ is a sequence of continuous real functions on $\mathbf R$ converging pointwise to $0$, and let $\epsilon > 0$.


*Why is the set $f_n^{-1} \big( [-\epsilon, \epsilon] \big)$ closed for all $n$?


Define a sequence $\{ A_k \}_{k \in \mathbf N}$ of subsets of $\mathbf R$ by
$$A_k = \bigcap_{n=k}^{\infty} f_n^{-1} \big( [-\epsilon, \epsilon] \big).$$


*Why is each $A_k$ closed in $\mathbf R$?

*Explain why we have
$$\bigcup_{k=1}^{\infty} A_k = \mathbf R.$$


Since $\mathbf R$ is complete, it follows from $(4)$ and the Baire category theorem that at least one $A_k$ is not nowhere dense, i.e., there is an $N \in \mathbf N$ such that $A_N$ is not nowhere dense.


*Explain why $A_N$ contains a non-empty open set $V$.


If $x$ is any point of $V$, then $x$ also lies in $A_N$, and therefore in $f_n^{-1} \big( [-\epsilon, \epsilon] \big)$ for every $n \ge N$.


*Explain why the proof is finished.


Edit


*

*The completeness of $\mathbf R$ is now explicitly mentioned.

*Added the assumption in $(1)$ that each $f_n$ is real-valued.

*Reformulated the first paragraph stating what are are trying to achieve.

