# Question About the Failure of Uniform Convergence

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.

• When negate the statement, turn every $\exists$ to $\forall$ and vice versa. In the definition, it is $\forall \varepsilon > 0$, hence in the negated one it should be $\exists \varepsilon_0 > 0$.
– xbh
Commented Oct 1, 2019 at 4:50

$$f_n \to f$$ 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 $$|f_{n_N}(x_N)-f(x_N)| \geq s$$

• Yeah, sorry, that was a typo: I know $s$ should merely need to exist. But Bartle's lemma uses subsequences -- that's what I'm having trouble with. Where does that come from?
– Rain
Commented Oct 1, 2019 at 5:07
• @Rain i edited my answer. Commented Oct 1, 2019 at 5:12
• Ah, that makes sense -- the subsequence is lurking behind $N$! Thank you.
– Rain
Commented Oct 1, 2019 at 5:14
• You are welcome. Commented Oct 1, 2019 at 5:18