Continuity, Rigorously

Informally, a function is continuous if you can draw its graph "without lifting the pen". To prove theorems we need that intuition pinned to an inequality. The working definition is that the limit equals the value, and the ε–δ definition of a limit spells that out with no appeal to motion at all.

A function f is continuous at a when, for every target tolerance \varepsilon > 0, there is an input tolerance \delta > 0 such that

|x - a| < \delta \;\;\Longrightarrow\;\; |f(x) - f(a)| < \varepsilon.

Read it as a game: an adversary names how close to f(a) the output must land (\varepsilon); you must answer with a window around a (\delta) so narrow that every input inside it is mapped inside the adversary's band. Continuity means you can always win — no matter how small the \varepsilon.

Centrepiece: f(x) = x^2 is continuous at every a

Fix a point a and a challenge \varepsilon > 0. We must produce a \delta that works. We build it line by line.

Step 1 — write down the quantity to control. We need |f(x) - f(a)| small, so factor it:

|x^2 - a^2| = |x - a|\,|x + a|.

Step 2 — the awkward factor is |x + a|. It is not constant, so first tether x near a. Insist from the outset that \delta \le 1; then |x - a| < \delta \le 1, and by the triangle inequality

|x + a| = |(x - a) + 2a| \le |x - a| + 2|a| < 1 + 2|a|.

Step 3 — bound the product. Combining Steps 1 and 2, whenever |x - a| < \delta \le 1,

|x^2 - a^2| = |x - a|\,|x + a| < \delta \,(1 + 2|a|).

Step 4 — choose \delta to beat \varepsilon. We want that right-hand side \le \varepsilon, so pick

\delta = \min\!\left(1,\; \frac{\varepsilon}{1 + 2|a|}\right).

Step 5 — verify. With this \delta, if |x - a| < \delta then both clauses hold: \delta \le 1 validates Step 2, and \delta \le \varepsilon / (1 + 2|a|) gives

|x^2 - a^2| < \delta\,(1 + 2|a|) \le \frac{\varepsilon}{1 + 2|a|}\,(1 + 2|a|) = \varepsilon.

We met an arbitrary \varepsilon with a concrete \delta, so x^2 is continuous at every a. Notice the catch worth remembering: the \delta we needed shrinks as |a| grows. That dependence is the seed of the next idea.

Pointwise versus uniform

Above, \delta depended on both \varepsilon and the point a. That is ordinary (pointwise) continuity. A stronger demand is that one \delta work for all points at once.

f is uniformly continuous on a set S when, for every \varepsilon > 0, there is a single \delta > 0 (depending on \varepsilon only) with

|x - y| < \delta \;\;\Longrightarrow\;\; |f(x) - f(y)| < \varepsilon \qquad \text{for all } x, y \in S.

Why x^2 is NOT uniformly continuous on \mathbb{R}. Take two points a fixed gap h apart, x and x + h:

|(x + h)^2 - x^2| = |2hx + h^2| = h\,|2x + h|.

However small the gap h > 0, the output gap h|2x + h| \to \infty as x \to \infty. So no single \delta can keep the outputs within (say) \varepsilon = 1 everywhere: out at large x the parabola is too steep. Uniform continuity fails precisely because the \delta from the proof above had to keep shrinking with |a|.

The rescue is to stay on a closed, bounded interval. There x^2 can't run off to steep regions, and uniform continuity is automatic.

Let f be a real function. Then:

ε–δ continuity at a: for every \varepsilon > 0 there is \delta > 0 with |x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon. This is equivalent to \lim_{x \to a} f(x) = f(a).
Uniform continuity on S: for every \varepsilon > 0 there is one \delta > 0 with |x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon for all x, y \in S. Uniform continuity \Rightarrow continuity, but not conversely (x^2 on \mathbb{R} is a counterexample).
Heine–Cantor: if f is continuous on a closed bounded interval [a, b], then it is uniformly continuous there.

Closed bounded intervals make continuity extraordinarily well-behaved. Two cornerstone theorems hold there — and both fail without the hypotheses.

Extreme Value Theorem. If f is continuous on [a, b], it attains a maximum and a minimum:

\exists\, c, d \in [a, b] : \quad f(c) \le f(x) \le f(d) \quad \text{for all } x \in [a, b].

The closed interval is essential. On the open interval (0, 1] the continuous function 1/x has no maximum — it shoots to +\infty near the missing endpoint 0. Drop continuity instead and a single jump can step over every candidate maximum.

Intermediate Value Theorem. If f is continuous on [a, b] and y lies between f(a) and f(b), then f(c) = y for some c \in [a, b] — a continuous graph cannot get from one side of a level to the other without crossing it. This is what guarantees a root of x^2 - 2 in [1, 2], hence the very existence of \sqrt{2} as a real number. Remove continuity and the graph can leap across y without ever equalling it.

See the box close in

Here is the game made visual on y = x^2 at the point a = 1. The horizontal band is the \varepsilon-tolerance around f(a) = 1; the vertical strip is the \delta-window the proof produces. Pull the slider down: as \varepsilon shrinks, the \delta-strip narrows just enough that the parabola never escapes the band inside it.