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[parent] the limit of a uniformly convergent sequence of continuous functions is continuous (Theorem)

Theorem. The limit of a uniformly convergent sequence of continuous functions is continuous.

Proof. Suppose $ f_n \rightarrow f$ uniformly and each $ f_n$ is continuous. Then given any $ \epsilon>0$, there exists $ N$ such that $ n>N$ implies $ \vert f(x)-f_{n}(x)\vert < \frac{\epsilon}{3}$ for all $ x$. Pick an arbitrary $ n$ larger than $ N$. Since $ f_n$ is continuous, given any point $ x_0$, there exists $ \delta>0$ such that $ 0<\vert x-x_0\vert<\delta$ implies $ \vert f_n(x) - f_n(x_0)\vert<\frac{\epsilon}{3}$. Therefore, given any $ x_0$ and $ \epsilon>0$, there exists $ \delta>0$ such that

$\displaystyle 0<\vert x-x_0\vert<\delta\Rightarrow \vert f(x)-f(x_0)\vert \leq ... ...)\vert + \vert f_n(x) - f_n(x_0)\vert + \vert f_n(x_0) - f(x_0)\vert < \epsilon$
Therefore, $ f$ is continuous.



"the limit of a uniformly convergent sequence of continuous functions is continuous" is owned by neapol1s.
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See Also: limit function of sequence


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Cross-references: point, implies, continuous
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This is version 6 of the limit of a uniformly convergent sequence of continuous functions is continuous, born on 2005-06-25, modified 2008-05-15.
Object id is 7191, canonical name is LimitOfAUniformlyConvergentSequenceOfContinuousFunctionsIsContinuous.
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Classification:
AMS MSC: 40A30 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences of functions)
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Some suggestions by goldberg on 2007-09-07 08:18:49
I think it should be emphasized that $\delta$ depends on $n$. Furthermore the assumption $n>N$ is not mentioned in the conclusion. Since $n$ is an arbitrary number greater than $N$, why not take $n=N+1$ and write $\delta=\delta_{N+1}$?
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