Continuous Functions

Part 2, Chapter 2: Single-Variable Real Analysis

Learning objectives

Define continuity at a point via $\lim_{x\to a}f(x)=f(a)$ and via $\epsilon\text{-}\delta$
Apply the Intermediate Value Theorem and Extreme Value Theorem to closed intervals
Distinguish removable, jump, and essential discontinuities
Justify why closed and bounded matters for EVT, and completeness for IVT

Continuity is the gateway property of analysis. A function being continuous on $[a,b]$ unlocks two enormous theorems, the Intermediate Value Theorem and the Extreme Value Theorem, from which countless concrete consequences follow. Why every odd polynomial has a real root, why every continuous loop in the plane crosses itself when wrapped around the origin, why optimisation problems on closed and bounded sets always have solutions: all are corollaries of continuity. The structure is so clean precisely because we built continuity on the $\epsilon\text{-}\delta$ machinery of §2.1.

The definition

A function $f$ is continuous at $a$ if $\lim_{x\to a}f(x)=f(a)$ xtoaf(x)=f(a). Equivalently, for every $\epsilon>0$ there exists $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$ . Notice the subtle change from the limit definition: we no longer exclude $x=a$ , and the limit must equal the function value, not just exist.

A function is continuous on a set $S$ if it is continuous at every point of $S$ . Polynomials, exponentials, sines and cosines, and absolute value are continuous everywhere. Rational functions are continuous wherever their denominator is non-zero. Continuity is preserved by addition, multiplication, composition, and (where defined) division.

Three kinds of discontinuity

If $f$ fails to be continuous at $a$ , the failure has a name. A removable discontinuity occurs when $\lim_{x\to a}f(x)$ xtoaf(x) exists but disagrees with $f(a)$ (or $f(a)$ is undefined); we can "remove" it by redefining $f(a)$ . A jump discontinuity occurs when both one-sided limits exist but disagree. An essential discontinuity occurs when at least one one-sided limit fails to exist altogether (e.g. $\sin(1/x)$ at $0$ ).

IVT and EVT

The Intermediate Value Theorem (IVT) says: if $f$ is continuous on $[a,b]$ and $y_0$ lies between $f(a)$ and $f(b)$ , then $f(c)=y_0$ for some $c\in(a,b)$ . The Extreme Value Theorem (EVT) says: if $f$ is continuous on a closed and bounded interval $[a,b]$ , then $f$ attains a maximum and a minimum.

Both theorems require completeness of $\mathbb{R}$ : IVT fails over $\mathbb{Q}$ ( $f(x)=x^2-2$ is continuous, changes sign on $[1,2]$ , but has no rational root). EVT fails on open intervals ( $f(x)=x$ on $(0,1)$ has supremum 1 but never attains it) and on unbounded ones ( $f(x)=x$ on $[0,\infty)$ has no maximum).

Use the grapher to plot $f(x)=x^3-x-1$ on $[1,2]$ . You will see $f(1)=-1<0$ and $f(2)=5>0$ . Since $f$ is continuous and changes sign, IVT guarantees a root in $(1,2)$ . Then try plotting $1/x$ on $[1,5]$ to confirm that EVT applies (the maximum is at $x=1$ , the minimum at $x=5$ ); plotting $1/x$ on $(0,1]$ shows EVT failing because the interval is not closed.

Where this shows up

Root-finding in numerical computing: the bisection method is a direct algorithmic implementation of IVT, bracket a continuous sign change and halve the interval. Every solver in MATLAB, NumPy, and SciPy that ships under names like fzero or brentq assumes continuity.

Signal processing & Fourier transforms: the inverse Fourier transform requires the signal to be continuous (or at least piecewise continuous with finite jumps). Discontinuities in audio appear as audible "clicks" because they correspond to infinite-bandwidth content the speaker cannot reproduce.

Optimisation in operations research: EVT guarantees that minimising a continuous cost function over a closed and bounded feasible set has a solution, the bedrock of linear programming, supply-chain logistics, and portfolio optimisation.

Pause and think: The function $f(x)=1/x$ is continuous on $(0,1]$ (every point in the domain is a point of continuity) but does not attain a maximum on $(0,1]$ . Which hypothesis of EVT is violated, and why does that matter?

Try it

Predict first: is $f(x)=(x^2-9)/(x-3)$ continuous at $x=3$ ? Compute the limit and compare to $f(3)$ (after deciding whether $f(3)$ is defined at all).

Use IVT to argue that $\cos x = x$ has a solution in $[0,1]$ . Compute the function $g(x)=\cos x - x$ at the endpoints and observe its sign.

Construct a function continuous on $(0,1)$ that does not attain its supremum. Which EVT hypothesis fails?

True or false: a composition $g\circ f$ is continuous whenever $f$ and $g$ are. Justify by chasing the $\epsilon\text{-}\delta$ definition.

A trap to watch for

"Continuous on $[a,b]$ " means continuous at every point, INCLUDING the endpoints, but at the endpoints, only the one-sided limit needs to match $f(a)$ . A function may be continuous on a closed interval and still discontinuous on the larger set obtained by extending the interval. The closed-interval hypothesis in IVT and EVT is not cosmetic: removing it breaks the conclusions, as the $1/x$ example shows.

What you now know

You can verify continuity, classify discontinuities, apply IVT to find roots, apply EVT to conclude existence of optima, and explain why closed and bounded intervals are special. Section §2.3 (Differentiation) builds the next layer: a function differentiable at $a$ is automatically continuous at $a$ , but the converse fails dramatically.

Mark section complete →

References

Garrity, T. (2002). All the Mathematics You Missed. Cambridge University Press, ch. 2.

Rudin, W. (1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill, ch. 4.

Abbott, S. (2015). Understanding Analysis (2nd ed.). Springer, ch. 4.4-4.5.

Tao, T. (2016). Analysis I (3rd ed.). Hindustan Book Agency, ch. 9.

Apostol, T. M. (1974). Mathematical Analysis (2nd ed.). Addison-Wesley, ch. 4.

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