# The necessary and sufficient condition for no uniform convergence

Let < $$f_n$$> be a functional sequence on $$A\subset \mathbb{R}$$ and let $$f:A\rightarrow\mathbb{R}$$ then < $$f_n$$>is not uniform convergence iif there exists $$\epsilon_0$$ such that there exists the subsequnce of $$$$, $$$$ and a sequance $$$$ for all $$k\in \mathbb{N}$$ $$|f_{n_k}(x_k)-f (x_k)|\geq \epsilon_0$$ I don't know how to prove this from the defintion of uniform convergence.

Actually the definition of uniform convergence is

Let $$$$ be a functional sequence on $$A\subset\mathbb {R}$$ and $$f:A\rightarrow \mathbb {R}$$ then for all $$\epsilon>0$$ there exist $$K (\epsilon)\in \mathbb {N}$$ such that $$n \geq K (\epsilon ), x\in A$$ imply $$|f_n(x)-f (x)|<\epsilon$$ Then we say < $$f_n$$> uniform convergent to $$f$$ on $$A$$

• You are to find the negation of the statement of uniform convergence. Do you know how to negate sentences in first-order logic? Aug 12, 2019 at 2:02

Uniform Convergence $$f_n \rightarrow f, \text{ uniformly if } \forall \epsilon>0, \exists N, \forall n > N, \forall x\in A, \ |f_n(x) - f(x)| <\epsilon$$ Not Uniform Convergence $$f_n \not\rightarrow f, \text{ uniformely if } \exists \epsilon >0, \forall N>0, \exists n>N, \exists x\in A, |f_n(x) - f(x)| \geq\epsilon$$
In words: Suppose $$f_n\not\rightarrow f$$ uniformly, then there exists $$\epsilon_0 >0$$ such that for each $$N\in \mathbb{N}$$ there exists some $$n_N>N$$ and $$x_N\in A$$ such that $$|f_{n_N}(x_N) - f(x_N)| \geq\epsilon_0$$