Ordinary continuity and uniform continuity

Two definitions that look alike

First, let us recall ordinary continuity of $f$ on an interval $I$, written with its quantifiers:

$$\forall x \in I,\; \forall \varepsilon > 0,\; \exists \delta > 0,\; \forall y \in I,\quad \lvert y - x \rvert \leq \delta \implies \lvert f(y) - f(x) \rvert \leq \varepsilon$$

Uniform continuity on $I$ moves a single quantifier, and that changes everything:

$$\forall \varepsilon > 0,\; \exists \delta > 0,\; \forall (x, y) \in I^2,\quad \lvert x - y \rvert \leq \delta \implies \lvert f(x) - f(y) \rvert \leq \varepsilon$$

Where is the difference?

Notion	Order of quantifiers	What $\delta$ depends on
Continuity on $I$	$\forall x,\; \forall \varepsilon,\; \exists \delta$	on $\varepsilon$ and on $x$
Uniform continuity on $I$	$\forall \varepsilon,\; \exists \delta,\; \forall x$	on $\varepsilon$ only

In the second case, one and the same $\delta$ must work everywhere on $I$: that is the meaning of the word "uniform". We immediately deduce the implication

$$f \text{ uniformly continuous on } I \implies f \text{ continuous on } I$$

and the converse is false in general, as the following example shows.

Example: $x \mapsto x^2$ is not uniformly continuous on $\mathbb{R}$

Let $f(x) = x^2$ and, for $n \geq 1$:

$$x_n = n \qquad \text{and} \qquad y_n = n + \frac{1}{n}$$

The two points get as close as we like, since $\lvert y_n - x_n \rvert = \dfrac{1}{n} \to 0$. Yet their images stay apart:

$$\lvert f(y_n) - f(x_n) \rvert = \left(n + \frac{1}{n}\right)^2 - n^2 = 2 + \frac{1}{n^2} \geq 2$$

With $\varepsilon = 1$, no $\delta > 0$ can work for all $x$ at once: it suffices to take $n$ such that $\dfrac{1}{n} \leq \delta$ to break the definition. The function is indeed continuous at every point, but far from the origin $\delta$ must be chosen smaller and smaller, because the slope $f'(x) = 2x$ blows up.

Classic pitfall

Never write "let $\delta$ be associated with $\varepsilon$ by the continuity of $f$" before having fixed $x$: this $\delta$ depends on $x$. If your proof then needs the same $\delta$ for another point, you are silently using uniform continuity — which is exactly what Heine's theorem will allow, but only on a closed bounded interval.