Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions

Let $X$ be a set and, for each $k \in \mathbb{N}$, let $f_k : X \to \mathbb{R}$ be a real-valued function (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$). Let $f : X \to \mathbb{R}$.

The sequence $(f_k)$ converges pointwise to $f$ on $X$ when, for every $x \in X$, the real sequence $(f_k(x))$ converges to $f(x)$ (Limits and Cauchy sequences of reals). Thus the index after which $|f_k(x)-f(x)|<\varepsilon$ may depend on both $\varepsilon$ and $x$.

The sequence $(f_k)$ converges uniformly to $f$ on $X$ when

$(\forall \varepsilon>0)(\exists N\in\mathbb{N})(\forall k\ge N)(\forall x\in X)\quad |f_k(x)-f(x)|<\varepsilon,$

where $\varepsilon$ ranges over the positive reals. Here one index $N$ serves every point of $X$.

The sequence $(f_k)$ is uniformly Cauchy on $X$ when

$(\forall \varepsilon>0)(\exists N\in\mathbb{N})(\forall m,n\ge N)(\forall x\in X)\quad |f_m(x)-f_n(x)|<\varepsilon.$

For each of the three notions above, restricting the error to positive rationals gives an equivalent condition. The real-error condition immediately implies the rational-error condition. Conversely, given a real $\eta>0$, choose $n\ge1$ with $1/n<\eta$ by For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$; the condition for the positive rational $1/n$ implies the condition for $\eta$. The real-error form is used because it makes the uniform quantifiers transparent.