Continuity

In this lesson you’ll learn the definition of continuity at a point and on a set, the sequential characterization, the three types of discontinuity, and the theorem that continuous functions pull back open sets to open sets. This is where limits become a property of functions rather than of points.

$f$ is continuous at $a$ if

$$\forall \varepsilon>0 \; \exists\delta>0 \; \text{such that} \; |x - a| < \delta \implies |f(x) - f(a)| < \varepsilon$$

Compare the limit definition. Two changes, and both matter.

• The condition is $|x-a| < \delta$, not $0 < |x-a| < \delta$. The point $a$ is included now.
• The target is $f(a)$, not some separate $L$.

So continuity at $a$ packages three requirements:

1. $f(a)$ is defined.
2. $\lim_{x\to a}f(x)$ exists.
3. They are equal.

$$\lim_{x\to a} f(x) = f(a)$$

That one equation is the whole definition, and each of the three parts can fail independently.

Removable. The limit exists but the value is wrong or missing. $f(x) = \frac{x^2-4}{x-2}$ has limit 4 at $x=2$ and no value there; defining $f(2)=4$ repairs it. Hence “removable.”

Jump. The one-sided limits both exist and differ, so no value at the point can help. $\frac{|x|}{x}$ at 0 is the model, and so is any step function. Common in physical models with switching behaviour.

Essential. No limit exists at all. $\sin\frac1x$ at 0. Nothing can be redefined to fix it, because there is nothing to converge to.

$f$ is continuous at $a$ $\iff$ for every sequence $x_n \to a$, we have $f(x_n) \to f(a)$.

Notice $x_n = a$ is now allowed, unlike the limit version. In words: continuous functions commute with limits.

$$\lim_{n\to\infty} f(x_n) = f\left(\lim_{n\to\infty}x_n\right)$$

That is what “continuous” buys you, and it is why continuity is the hypothesis whenever you want to move a function inside a limit.

Same practical benefit as before: to disprove continuity, find one sequence $x_n \to a$ with $f(x_n) \not\to f(a)$.

If $f, g$ are continuous at $a$ then so are $f+g$, $cf$, $fg$, and $f/g$ when $g(a) \ne 0$. These follow from the sequential characterization plus the algebraic limit theorems - no new epsilon-delta work.

Composition deserves its own statement:

If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f \circ g$ is continuous at $a$.

Proof (sequential). Let $x_n \to a$. Continuity of $g$ gives $g(x_n) \to g(a)$. Continuity of $f$ at $g(a)$ gives $f(g(x_n)) \to f(g(a))$. ∎

Two lines, versus a genuinely fiddly epsilon-delta argument with two nested tolerances. This is the sequential criterion paying off.

Consequences: polynomials are continuous everywhere; rational functions are continuous where the denominator is nonzero; $\sin$, $\cos$, $\exp$ are continuous everywhere; $\sqrt x$ is continuous on $[0,\infty)$; and any composition of these is continuous on its natural domain.

$f: \mathbb{R}\to\mathbb{R}$ is continuous $\iff$ $f^{-1}(V)$ is open for every open $V$.

Preimages of open sets are open. This is the definition used in general topology, where there is no $\varepsilon$ to speak of.

Note it is about preimages. Images of open sets need not be open: $f(x) = x^2$ maps the open $(-1,1)$ to $[0,1)$, which is not open. And a constant function maps every open set to a single point.

This characterization is what made the “continuous image of a compact set is compact” proof in the last lesson work, and it is why compactness and continuity interact so cleanly.

Define

$$f(x) = \begin{cases} x & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q}\end{cases}$$

This is continuous at exactly one point, $x = 0$, and discontinuous everywhere else.

At 0: $|f(x)| \le |x|$ always, so $\delta = \varepsilon$ works. At any $a \ne 0$: sequences of rationals approaching $a$ give images approaching $a$, while sequences of irrationals give images approaching 0, and $a \ne 0$.

The point of the example: continuity is a pointwise condition, and the set where a function is continuous can be extremely strange. Intuition about curves you can draw is not a guide.

Example 1: Show $f(x) = 3x+1$ is continuous everywhere.

Solution. Fix $a$. Given $\varepsilon>0$, take $\delta = \frac{\varepsilon}{3}$. If $|x-a|<\delta$ then

$$|f(x)-f(a)| = |3x+1-3a-1| = 3|x-a| < 3\delta = \varepsilon$$

∎

Here $\delta$ depends on $\varepsilon$ only, not on $a$. That extra property is uniform continuity, coming in three lessons.

Example 2: Classify the discontinuity of $f(x) = \frac{x^2-9}{x-3}$ at $x=3$.

Solution. For $x \ne 3$, $f(x) = x+3$, so $\lim_{x\to3}f(x) = 6$. But $f(3)$ is undefined.

The limit exists, so the discontinuity is removable. Setting $f(3) = 6$ makes it continuous.

Example 3: Classify the discontinuity of the floor function at an integer.

Solution. At $x = 2$: $\lfloor x \rfloor = 1$ for $x \in [1,2)$ and $= 2$ for $x \in [2,3)$.

$$\lim_{x\to2^-}\lfloor x\rfloor = 1, \qquad \lim_{x\to2^+}\lfloor x\rfloor = 2$$

Both one-sided limits exist and differ, so this is a jump discontinuity of size 1. No redefinition helps.

The floor function is continuous at every non-integer, so its set of discontinuities is $\mathbb{Z}$ - infinite but isolated.

Example 4: Show $f(x) = \frac1x$ is continuous on $(0,\infty)$.

Solution. Fix $a > 0$. From the limits lesson,

$$\left|\frac1x-\frac1a\right| = \frac{|x-a|}{|ax|}$$

Take $\delta = \min\left(\frac a2, \frac{\varepsilon a^2}{2}\right)$. Then $|x-a|<\frac a2$ gives $x>\frac a2$, so $|ax| > \frac{a^2}{2}$, and

$$\left|\frac1x-\frac1a\right| < \frac{2|x-a|}{a^2} < \frac{2}{a^2}\cdot\frac{\varepsilon a^2}{2} = \varepsilon$$

∎

Continuous at every point of $(0,\infty)$. There is no discontinuity at 0 to classify, because 0 is not in the domain - ”$\frac1x$ is discontinuous at 0” is a common but incorrect phrasing.

Example 5: Continuity of a composition.

Show $h(x) = \sqrt{x^2+1}$ is continuous on $\mathbb{R}$.

Solution. Write $h = f\circ g$ with $g(x) = x^2+1$ and $f(u) = \sqrt u$.

$g$ is a polynomial, continuous everywhere. And $g(x) \ge 1 > 0$ for all $x$, so $g$ maps into $[1,\infty)$ where $f$ is continuous.

By the composition theorem, $h$ is continuous on all of $\mathbb{R}$. ∎

The check that $g$ lands where $f$ is continuous is the step people skip, and it is the step that fails for something like $\sqrt{x^2-1}$.

Example 6: Disprove continuity with one sequence.

Let $f(x) = 1$ for $x \ge 0$ and $f(x) = 0$ for $x<0$. Show $f$ is discontinuous at 0.

Solution. Take $x_n = -\frac1n \to 0$. Then $f(x_n) = 0$ for every $n$, so $f(x_n) \to 0$.

But $f(0) = 1 \ne 0$. By the sequential characterization, $f$ is not continuous at 0. ∎

One sequence suffices, since the characterization requires every sequence to work.

Example 7: Preimages, not images.

For $f(x) = x^2$, compare $f((-1,1))$ with $f^{-1}((-1,1))$.

Solution. The image is $f((-1,1)) = [0,1)$, which is not open - 0 has no room to wiggle inside it.

The preimage is $f^{-1}((-1,1)) = \{x : x^2 < 1\} = (-1,1)$, which is open.

Continuity guarantees the second, never the first. Getting the direction backwards is a standard error, and the topological definition is stated with preimages precisely because that is the version that is true.

Numerical root finding. Bisection requires continuity to guarantee a sign change encloses a root. Applied to a discontinuous function it can converge confidently to a jump where no root exists, which is a real failure mode in practice.

Physical modelling. Jump discontinuities model switching: a thermostat, a diode turning on, a phase transition. Recognizing which type of discontinuity a model has determines whether standard numerical solvers apply or whether the problem must be split into smooth pieces.

Computer graphics. Visible seams and banding are discontinuities in a shading function. Texture filtering and level-of-detail transitions are engineered to be continuous, and often to have continuous derivatives too, because the eye detects even a derivative jump as a crease.

Signal processing. Discontinuities in a signal have infinite frequency content, which is why a sharp edge produces ringing after band-limited reconstruction. The Gibbs phenomenon is a jump discontinuity’s signature.

Optimization. Most solvers assume continuity, and many assume more. A discontinuous objective breaks gradient methods and can break even derivative-free ones, since neighbouring evaluations no longer predict nearby behaviour.

Finance. Asset prices are usually modelled as continuous paths, but jump-diffusion models add discontinuities explicitly to capture crashes. The choice changes both the mathematics and the hedging strategy.