$x\mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, the pairs $k+1$ and $k+1+1\mathrm{/}(k+1)$ defeating every $\delta$

Refuted claim: the function $f:\mathbb{R}\to \mathbb{R}$, $f(x):=x^2$ (Integer powers $a^m$), is uniformly continuous on $\mathbb{R}$ (Uniform continuity of $f:A\to \mathbb{R}$: one $\delta$ serving every pair of points of $A$).

$f$ is continuous on $\mathbb{R}$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point) and $\mathbb{R}$ is a closed subset of itself, so this is the complement of $x\mapsto 1\mathrm{/}x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1\mathrm{/}(k+2)$ and $1\mathrm{/}(k+3)$ defeating every $\delta$: there the domain was bounded and not closed, here it is closed and not bounded, and uniform continuity fails in both cases. Neither half of compactness suffices on its own, and Heine-Cantor in $\mathbb{R}$: a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$-natively from sequential compactness needs both (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded).

The refutation exhibits, for every $\delta >0$, a pair of reals closer than $\delta$ whose squares differ by more than $2$. The pairs are

$a_k:=k+1,b_k:=k+1+\frac{1}{k+1}(k\in \mathbb{N}),$

and the shift by $1$ is not cosmetic: $\mathbb{N}$ contains $0$ here (Sequences of reals: bounded, eventually, frequently, tails, subsequences is $0$-indexed), so the reciprocal $1\mathrm{/}k$ would be undefined at the first index.

• The mechanism is the growing slope, not a singularity. The increment $b_k-a_k$ is chosen to be the reciprocal of the point, so the product $(b_k-a_k)(b_k+a_k)$ stays above $2$ however small the increment becomes. Nothing blows up: $f$ is a polynomial, bounded on every bounded set, and the failure is entirely about how far out one is allowed to look.

• On every bounded interval it is uniformly continuous. On $[-M,M]$ one has $\mathrm{\mid }{x}^2-y^2\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }\mathrm{\mid }x+y\mathrm{\mid }\le 2M\mathrm{\mid }x-y\mathrm{\mid }$, so $f$ is Lipschitz there, hence uniformly continuous (Dictionary: for $A\subseteq \mathbb{R}$ with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$, continuity and uniform continuity of $f:A\to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace, clause 6). That is also what Heine-Cantor in $\mathbb{R}$: a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$-natively from sequential compactness gives, since $[-M,M]$ is compact.

• The consequence for products. The identity is uniformly continuous on $\mathbb{R}$ and its square is not, so the product of two uniformly continuous functions need not be uniformly continuous; that is The identity is uniformly continuous on $\mathbb{R}$ and its square is not, so uniform continuity is not preserved by products, which is this item read once more.