# Show the equivalence between two negations of uniformly convergence.

$$(f_n)$$ is uniformly convergent iif $$\exists f:D\subseteq \mathbb{R}\longrightarrow \mathbb{R}$$ such that $$\forall \epsilon > 0 \quad\exists N=N(\epsilon) \in \mathbb{N}$$ such that $$\forall n \geq N, \forall x\in D, |f_n(x)-f(x)|<\epsilon$$

So the negation is:

$$(f_n)$$ is not uniformly convergent iff $$\exists \epsilon_0 >0$$ such that $$\forall N\in\mathbb{N}, \exists k \in \mathbb{N}, k \geq N$$ and $$\exists x_k\in D$$ such that $$|f_k(x_k)-f(x_k)|\geq\epsilon_0$$

The negation with subsequences is:

$$(f_n)$$ is not uniformly convergent iff $$\exists \epsilon_0 >0, \exists (f_{n_k})$$ subsequence of $$(f_n)$$, $$\exists (x_k)$$ sequence in $$D$$ such that $$\forall k\in\mathbb{N}$$ $$|f_{n_k}(x_k)-f(x_k)|\geq\epsilon_0$$

Show the equivalence between the two negations.

I've been trying this proof without success. Any suggestions would be great!

Hint: Idea is to construct a subsequence by taking $$N=1,2,3,\cdots$$
From the first definition, for $$N=1,\exists k_1\gt 1: x_{k_1}\in D$$ etc.
Similarly, get $$x_{k_2},x_{k_3}\cdots$$ etc.
• @SofíaContreras: For the other way, note the "$\forall k\in \mathbb N$" part and think about $n_k\ge k$.