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I gave a proof here some time ago but was wondering about the following:

Suppose that $f_n:K\subseteq \Bbb{R}\to\Bbb{R}$ is continuous for each $n\in \Bbb{N}$, then $\{f_n\}$ is uniformly continuous. So, for every $\epsilon > 0$ and $n\in \Bbb{N}$, there exists $\delta_n>0$ such that $\forall\,x,y\in K, \;|x − y| < \delta_n,$ implies $$|f_n(x)-f_n(y)| < \epsilon,\forall\;n\in \Bbb{N}.$$ Taking $\delta=\min\{\delta_n:\;n\in \Bbb{N}\}$ (which might not be true), then we have the definition given here, that "for every $\epsilon > 0$, there exists $\delta>0$ such that $\forall\,x,y\in K, \;|x − y| < \delta,$ implies $$|f_n(x)-f_n(y)| < \epsilon,\forall\;n\in \Bbb{N}.$$ which implies that $\{f_n\}$ is equicontinuous.

QUESTION: I'm I right? If not, can you please provide a counter-example?

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  • $\begingroup$ What is the min of an infinite set of numbers? What is $\min\{1/n : n \in \mathbb{N}\}$? Your proof is flawed. $\endgroup$ – RRL Jan 5 at 8:04
  • $\begingroup$ @RRL: Thanks for that! But how do I get an example? $\endgroup$ – Omojola Micheal Jan 5 at 8:07
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To construct a sequence $(f_n)$ of continuous functions on a compact set that is not equicontinuous, select one that is pointwise but not uniformly convergent and with a discontinuous limit function.

An example is $f_n(x) = x^n$ with $K = [0,1]$. To see that these are not equicontinuous, suppose that there exists $\delta > 0$ such that if $|x - 1| < \delta$ then $|f_n(x) - f_n(1)| =|x^n - 1| < 1/4$ for every $n$. However, since $x^n \to 0$ when $1- \delta < x < 1$, with $x$ fixed we can find $n$ such that $|x^n - 1| > 1/4$ a contradiction.

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  • $\begingroup$ So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right? $\endgroup$ – Omojola Micheal Jan 5 at 8:25
  • $\begingroup$ If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family $\{f_n\}$ may not be equicontinuous. $\endgroup$ – RRL Jan 5 at 8:27
  • $\begingroup$ So, is it right to say that $f_n:K\subseteq \Bbb{R}\to\Bbb{R}$ is continuous for each $n\in \Bbb{N}$, then $\{f_n\}$ is uniformly continuous. So, for every $\epsilon > 0$ and $n\in \Bbb{N}$, there exists $\delta_n>0$ such that $\forall\,x,y\in K, \;|x − y| < \delta_n,$ implies $$|f_n(x)-f_n(y)| < \epsilon,\forall\;n\in \Bbb{N}?$$ $\endgroup$ – Omojola Micheal Jan 5 at 8:27
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    $\begingroup$ "If each function fn is continuous on compact K then it is uniformly continuous, but the family {fn} may not be equicontinuous." Okay, I get you now! $\endgroup$ – Omojola Micheal Jan 5 at 8:28
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    $\begingroup$ Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof! $\endgroup$ – Omojola Micheal Jan 5 at 8:30

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