I'm working through Bartle's Elements of Real Analysis, and I'm having trouble parsing Lemma 13.5, on the failure of uniform convergence. The lemma says,
"A sequence $(f_n)$ does not converge uniformly on $D_0$ to $f$ if and only if for some $\varepsilon_0 > 0$ there is a subsequence $(f_{n_k})$ of $(f_n)$ and a sequence $(x_k)$ in $D_0$ such that
\begin{align*}
|f_{n_k}(x_k) – f(x_k)| \geq \varepsilon_0 \text{ for } k \in \mathrm{N}."
\end{align*}
Bartle goes on to say how this can be proved simply by negating the definition of uniform convergence. But I don't understand how the two subsequences, $(f_{n_k})$ and $(x_k)$, arise simply by negating the definition. When I negate the definition, I get something like "there exists an $x \in D_0$ such that, for every $\varepsilon_0 > 0$ and every natural number $K$, there is a corresponding natural number $n \geq K$ for which
\begin{align*}
|f_{n}(x) – f(x)| \geq \varepsilon_0."
\end{align*}
Any clarification would be appreciated.
Best Answer
$f_n \to f$ uniformly if $\forall \epsilon>0$ exists $N \in \Bbb{N}$ such that $|f_n(x)-f(x)|<\epsilon ,\forall n \in \Bbb{N},\forall x \in D_0$
The negation is:
Exists $s>0$ such that: $\forall N \in \Bbb{N}$ exists $n \geq N$ and $x \in D_0$ such that $|f_n(x)-f(x)| \geq s$
So for $N=1,2,3.... $ exist $n_N \geq N$ and $x_N \in D_0$ such that $|f_{n_N}(x_N)-f(x_N)| \geq s$