Negation of uniform convergence Suppose $f_{n}$ is a sequence of functions which does not convergence uniformly to $f$. Does this mean that there exists an $\varepsilon_{0} > 0$, an $x_{0}$, and a sequence of integers $n_{k} \rightarrow \infty$ such that $|f_{n_{k}}(x_{0}) - f(x_{0})| \geq \varepsilon_{0}$?
 A: If you know how to negate logical formulas with quantifiers, you can do this more or less mechanically.
Definition of uniform convergence can be written like this:
$$(\forall \varepsilon>0) (\exists n_0) (\forall x\in S) (\forall n>n_0) (|f_n(x)-f(x)|<\varepsilon) $$
Negation of uniform convergence
$$(\exists \varepsilon>0) (\forall n_0) (\exists x\in S) (\exists n>n_0) (|f_n(x)-f(x)|\ge\varepsilon) $$
Pointwise convergence is defined as follows
$$(\forall \varepsilon>0) (\forall x\in S) (\exists n_0) (\forall n>n_0) (|f_n(x)-f(x)|<\varepsilon) $$
negation
$$(\exists \varepsilon>0) (\exists x\in S) (\forall n_0) (\exists n>n_0) (|f_n(x)-f(x)|\ge\varepsilon)$$
If you look closely at the negation of pointwise convergence, it is equivalent to condition from your question. Indeed, we have existence of $\varepsilon>0$ and existence of a point, which we many denote $x_0$, such that $|f_n(x_0)-f(x_0)|\ge\varepsilon$ happens for infinitely many $n$'s.
So any function which converges pointwise but not uniformly is a counterexample to the claim in your post.
