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Let $f_n$ be a sequence of functions continuous on a set $T\subset \mathbb{C}$ (where $\mathbb{C}$ is the set of all complex numbers) and converging uniformly to $f$ on $T$. Then $f$ is also continuous on $T$.

The proof given in the book states:

To prove that $f$ is continuous at a point $z_0$ of $T$, we must show that for any $\epsilon >0$ there is a $\delta >0$ such that if $z$ belongs to $T$ and $|z_0-z|<\delta$, then $|f(z_0)-f(z)|<\epsilon$ ...

I don't understand the beginning of this proof, can anyone help elaborate?

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What is $C$? Is it $\bf C$, the set of all complex numbers? – Gerry Myerson Mar 4 '13 at 22:22
@GerryMyerson Yes sir it is. Going to edit it. – Q.matin Mar 4 '13 at 22:27
The beginning of the proof as quoted by you here i smerely stating the definition of "$f$ is continuous at $z_0$". The proof itself starts later (and is probably a standard $\frac\epsilon3$ proof). – Hagen von Eitzen Mar 4 '13 at 23:04
@HagenvonEitzen Yes, it is the standard $\frac{\epsilon}{3}$ proof but I dont understand the definition. – Q.matin Mar 4 '13 at 23:27
Thats okay. I will have to be bold next time and just submit an answer =p – Frudrururu Mar 6 '13 at 3:12
up vote 1 down vote accepted

Comment-answer by Wishingwell:

We want to show that $f$ is continuous at every point in $T$ so if we pick an arbitrary point of $T$ and show that $f$ is continuous there since the point was arbitrary and could have been any point in $T$ it will follow that $f$ will be continuous on $T$. So we let $\epsilon >0$ and then we just need the $\delta >0$. You'll notice $$|f(z_0)−f(z)|=|f(z_0)+f_n(z)−f_n(z)+f_n(z_0)−f_n(z_0)−f(z)|$$ and so $$|f(z_0)−f(z)|\le |f(z_0)−f_n(z)|+|f_n(z)−f_n(z_0)|+|f_n(z_0)−f(z)|$$ and the book I am sure has what you need etc. Continuity of $f_n$ gives the middle part, and uniform convergence will give the first and last parts of the sum, i.e., these things will let us make the quantities smaller than $\epsilon/3$.

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