Continuity, IVT, EVT, and Uniform Continuity

Throughout, $\mathbb{R}$ is the complete ordered field with its order and absolute value (Complete ordered field (least-upper-bound property), Basic properties of the absolute value), and neighbourhoods are those of The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$.

Let $A\subseteq \mathbb{R}$, let $f:A\to \mathbb{R}$ and let $c\in A$. Then $f$ is continuous at $c$ when

$(\mathrm{\forall }\epsilon >0)(\mathrm{\exists }\delta >0)(\mathrm{\forall }x\in A)[\mathrm{\mid }x-c\mathrm{\mid }<\delta ⟹\mathrm{\mid }f(x)-f(c)\mathrm{\mid }<\epsilon ],$

with $\epsilon$ and $\delta$ ranging over the positive reals. In the language of neighbourhoods: for every real $\epsilon >0$ there is a real $\delta >0$ with

$f(A\cap N_{\delta }(c))\subseteq N_{\epsilon }(f(c))\mathrm{.}$

$f$ is continuous on $A$ when it is continuous at every point of $A$.

The point $c$ is required to lie in $A$, and the condition is unpunctured. Both differ from The $\epsilon$-$\delta$ limit ${\lim }_{x\to c}f(x)=L$ of $f:A\to \mathbb{R}$ at a limit point $c$ of $A$, and deliberately. There the quantifier runs over $0<\mathrm{\mid }x-c\mathrm{\mid }<\delta$, which removes $x=c$; here $x=c$ is allowed, and at $x=c$ the implication reads $\mathrm{\mid }f(c)-f(c)\mathrm{\mid }=0<\epsilon$, which is automatic. So allowing $x=c$ costs nothing, and it is what lets the definition be stated at every point of $A$, including the points where no limit exists.

Three clauses, and all three are part of the definition.

1. At a limit point. Suppose $c\in A$ is a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$). Then $f$ is continuous at $c$ if and only if the limit of $f$ at $c$ exists and ${\lim }_{x\to c}f(x)=f(c)$ (The $\epsilon$-$\delta$ limit ${\lim }_{x\to c}f(x)=L$ of $f:A\to \mathbb{R}$ at a limit point $c$ of $A$). Indeed, for a given $\epsilon >0$ a $\delta$ witnessing continuity witnesses the limit condition, because the limit condition quantifies over a subset of the points continuity quantifies over; and conversely a $\delta$ witnessing ${\lim }_{x\to c}f(x)=f(c)$ witnesses continuity, because the one point it omits, $x=c$, satisfies $\mathrm{\mid }f(c)-f(c)\mathrm{\mid }<\epsilon$ anyway.

2. At an isolated point. Suppose $c\in A$ is an isolated point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), so that $N_{\eta }(c)\cap A=\{c\}$ for some real $\eta >0$. Then every $f:A\to \mathbb{R}$ is continuous at $c$: take $\delta :=\eta$, so that the only $x\in A$ with $\mathrm{\mid }x-c\mathrm{\mid }<\delta$ is $c$ itself, and $\mathrm{\mid }f(c)-f(c)\mathrm{\mid }=0<\epsilon$.

3. On a set. Continuity on $A$ is continuity at each point of $A$, and nothing more. It is not a condition relating $f$ to points outside $A$.

Every point of $A$ is either a limit point of $A$ or an isolated point of $A$, and never both (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), so clauses 1 and 2 between them describe continuity at every point of $A$.

This is not the raw $\epsilon$-$\delta$ formula of FALSE: a function has at most one limit at every point of its domain, isolated points included. That item records what goes wrong when the punctured formula of The $\epsilon$-$\delta$ limit ${\lim }_{x\to c}f(x)=L$ of $f:A\to \mathbb{R}$ at a limit point $c$ of $A$ is written down at an arbitrary point of the domain: at an isolated point it is satisfied vacuously by every real $L$ at once, so it defines nothing, and this library therefore leaves ${\lim }_{x\to c}f(x)$ undefined at an isolated point. Continuity at an isolated point is a different matter: the formula above is not vacuous — it is a genuine condition on $f(c)$, satisfied because $f(c)$ is the only value being compared with itself — and it names a single, well-defined property. The limit is undefined there; the continuity is defined, and is automatic. Clause 1 is the only place where the two notions meet, and it is stated only where the limit exists as a notion.

Where the distinction disappears. If $A$ is an open subset of $\mathbb{R}$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen), then every $c\in A$ has some $N_{\eta }(c)\subseteq A$, and a punctured neighbourhood is never empty (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$), so every point of $A$ is a limit point of $A$ and clause 1 covers the whole of $A$. The same holds when $A$ is a nondegenerate interval (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Isolated points are what force clause 2 to exist at all, and they occur as soon as $A$ is allowed to be an arbitrary subset of $\mathbb{R}$, as in $A=\{0\}\cup [1,2]$.

• Continuity is local. If $\eta >0$ and $g:A\to \mathbb{R}$ agrees with $f$ on $A\cap N_{\eta }(c)$, then $f$ is continuous at $c$ if and only if $g$ is: any $\delta$ may be replaced by $\min \{\delta ,\eta \}$, after which the condition only ever evaluates the two functions where they agree. So continuity at $c$ sees only an arbitrarily small neighbourhood of $c$, exactly as the limit does (The limit at $c$ depends only on the restriction of $f$ to a punctured neighbourhood of $c$, and passes to any subset of the domain having $c$ as a limit point).

• Continuity passes to subsets of the domain. If $B\subseteq A$ and $c\in B$, then continuity of $f$ at $c$ gives continuity of the restriction $f{\mathrm{\mid }}_B$ at $c$, with the same $\delta$: the condition on $f{\mathrm{\mid }}_B$ quantifies over fewer points. The converse fails, and the standard witness is the indicator of $\mathbb{Q}$ restricted to $\mathbb{Q}$, which is constant and hence continuous, while the indicator itself is continuous nowhere (The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ ↗).

• The radius is a real number. As in The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$, $\epsilon$ and $\delta$ range over the positive reals here. Restricting either quantifier to the positive rationals defines the same relation, by the passage recorded in The $\epsilon$-$\delta$ limit ${\lim }_{x\to c}f(x)=L$ of $f:A\to \mathbb{R}$ at a limit point $c$ of $A$: below every positive real lies a positive rational (The rationals embed densely in the reals), and a real $\delta$ may be shrunk to a rational one below it.

• The word continuous is used for two things in this library, and they agree. Continuity of a map between metric spaces, at a point and globally, in the $\epsilon$-$\delta$ form defines continuity of a map between metric spaces, and $A\subseteq \mathbb{R}$ carries the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$. The two notions coincide, and that is proved, not assumed: 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 is the dictionary, and it is stated on this page precisely so that no later item has to guess.

Let $A\subseteq \mathbb{R}$, let $f:A\to \mathbb{R}$ and let $c\in A$. The following are equivalent.

1. $f$ is continuous at $c$ (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).
2. For every sequence $(x_k{)}_{k\in \mathbb{N}}$ with $x_k\in A$ for every $k$ and $x_k\to c$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals), the sequence $(f(x_k){)}_{k\in \mathbb{N}}$ converges to $f(c)$.

The sequences here are not required to avoid $c$, which is the one difference from Heine criterion: ${\lim }_{x\to c}f(x)=L$ iff $f(x_k)\to L$ for every sequence in $A\setminus \{c\}$ converging to $c$ and is exactly what makes the criterion available at an isolated point of $A$, where no limit exists.

The two directions do not cost the same. The implication from 1 to 2 is proved in ZF: the sequence is handed to the proof and nothing is selected. The implication from 2 to 1 is obtained below from Heine criterion: ${\lim }_{x\to c}f(x)=L$ iff $f(x_k)\to L$ for every sequence in $A\setminus \{c\}$ converging to $c$, and therefore inherits the one use of the axiom of countable choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) made in that theorem's converse direction. What this library does and does not claim about that cost is recorded once, in The sequence-to-$\epsilon$ direction of the Heine criterion uses countable choice for $\mathbb{R}$, and where this library records that cost, and is not restated here.

Nothing else on this page is routed through this theorem. The algebra of continuous functions, composition, the intermediate value theorem, the extreme value theorem and Heine-Cantor are all proved from $\epsilon$ and $\delta$, or from compactness, exactly as the previous page organised itself. The choice-free direction 1 to 2 is used, in the intermediate value theorem and in Heine-Cantor, and each of those two items says which direction it uses.

• Why the sequences are allowed to hit $c$. Heine criterion: ${\lim }_{x\to c}f(x)=L$ iff $f(x_k)\to L$ for every sequence in $A\setminus \{c\}$ converging to $c$ must exclude $c$, because The $\epsilon$-$\delta$ limit ${\lim }_{x\to c}f(x)=L$ of $f:A\to \mathbb{R}$ at a limit point $c$ of $A$ says nothing about $f(c)$ and a sequence constantly equal to $c$ would test the wrong thing. Continuity does look at $f(c)$, so no exclusion is needed, and dropping it is what makes statement 2 meaningful at an isolated point, where the only sequences converging to $c$ are those eventually equal to $c$.

• What the choice cost is, and what it is not. The claim recorded here is that the proof given above of 2 implies 1 uses countable choice, through Heine criterion: ${\lim }_{x\to c}f(x)=L$ iff $f(x_k)\to L$ for every sequence in $A\setminus \{c\}$ converging to $c$. No claim is made that it is necessary; The sequence-to-$\epsilon$ direction of the Heine criterion uses countable choice for $\mathbb{R}$, and where this library records that cost states in full what this library does and does not assert, including Sierpiński's ZF theorem that a function continuous sequentially at every point of $\mathbb{R}$ is continuous, which shows the everywhere-statement and the pointwise statement behave differently.

• The negative use is the common one. To show that $f$ is not continuous at $c$ it suffices to exhibit one sequence in $A$ converging to $c$ whose image sequence does not converge to $f(c)$, and that uses only the choice-free direction. The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ ↗ on the companion page is proved without sequences at all, directly from density, which is cheaper still.

Let $A\subseteq \mathbb{R}$, let $f,g:A\to \mathbb{R}$, let $\alpha \in \mathbb{R}$ and let $c\in A$. Suppose $f$ and $g$ are continuous at $c$ (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). Then:

1. $f+g$, $\alpha f$ and $fg$ are continuous at $c$;
2. $\mathrm{\mid }f\mathrm{\mid }$, the function $x\mapsto \mathrm{\mid }f(x)\mathrm{\mid }$, is continuous at $c$;
3. $\max \{f,g\}$ and $\min \{f,g\}$, defined pointwise by $x\mapsto \max \{f(x),g(x)\}$ and $x\mapsto \min \{f(x),g(x)\}$ (Maximum and minimum of a set), are continuous at $c$;
4. if $g(c)\mathrm{\ne }0$ then, writing $A_0:=\{x\in A:g(x)\mathrm{\ne }0\}$, the point $c$ lies in $A_0$ and the quotient $(f\mathrm{/}g){\mathrm{\mid }}_{A_0}:A_0\to \mathbb{R}$, $x\mapsto f(x)\mathrm{/}g(x)$, is continuous at $c$ as a function on $A_0$.

Moreover, with no hypothesis at all:

5. every constant function $A\to \mathbb{R}$ and the identity $\mathrm{i}\mathrm{d}:A\to \mathbb{R}$, $x\mapsto x$, are continuous on $A$; hence so is $x\mapsto x^n$ for every $n\in \mathbb{N}$ (Integer powers $a^m$), and hence so is every polynomial function $x\mapsto a_0+a_{1}x+\cdots +a_n{x}^n$ with real coefficients.

Consequently, if $f$ and $g$ are continuous on $A$ then so are $f+g$, $\alpha f$, $fg$, $\mathrm{\mid }f\mathrm{\mid }$, $\max \{f,g\}$ and $\min \{f,g\}$, and $(f\mathrm{/}g){\mathrm{\mid }}_{A_0}$ is continuous on $A_0$.

Claim 4 is stated on $A_0$ because $f\mathrm{/}g$ is not defined where $g$ vanishes, and $g$ may well vanish at points of $A$ far from $c$. The hypothesis is $g(c)\mathrm{\ne }0$, not "$g$ nowhere zero"; what it buys is that $c$ itself lies in the smaller domain, which is what makes continuity there mean anything.

Nothing here is proved through a sequence. Claims 1 and 4 are read off from Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero, which is itself proved from $\epsilon$ and $\delta$, and claims 2, 3 and 5 are proved directly below. So no choice principle is used anywhere in this item.

• Why the two-case shape, and why it is not an inconvenience. Continuity is defined at every point of the domain, including isolated points, where no limit exists (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). The algebra of limits therefore cannot be applied blindly; but at an isolated point every function is continuous, so the case is settled before it is opened. Claims 2 and 5 are proved directly from $\epsilon$ and $\delta$ and need no case distinction at all.

• Absolute value, maximum and minimum are not in Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero, and the reason is that they are not needed there. They are needed here: the extreme value theorem and the one-dimensional fixed point theorem both build auxiliary functions out of maxima, minima and differences, and Rudin 4.20, the sharp converse: on a noncompact $E\subseteq \mathbb{R}$ there is an unbounded continuous function and a bounded continuous function with no greatest value, and if $E$ is bounded there is a continuous function on $E$ that is not uniformly continuous builds its witnesses out of $\mathrm{\mid }x-x_0\mathrm{\mid }$ and quotients.

• The converse of claim 2 is false: $\mathrm{\mid }f\mathrm{\mid }$ may be continuous while $f$ is continuous nowhere. The function equal to $1$ on $\mathbb{Q}$ and to $-1$ elsewhere has constant absolute value; that it is nowhere continuous follows from the argument of The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ ↗ applied verbatim, since that argument uses only that the two values are distinct.

Let $A,B\subseteq \mathbb{R}$, let $g:A\to \mathbb{R}$ with $g(A)\subseteq B$, and let $f:B\to \mathbb{R}$, so that the composite $f\circ g:A\to \mathbb{R}$ is defined. Let $c\in A$. If $g$ is continuous at $c$ and $f$ is continuous at $g(c)\in B$, then $f\circ g$ is continuous at $c$ (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).

Consequently, if $g$ is continuous on $A$ and $f$ is continuous on $B$, then $f\circ g$ is continuous on $A$.

No side hypothesis is needed, and that is the whole point. The composition theorem for limits, Composition of limits holds under either hypothesis: $f$ is defined at $L$ with value $M$, or $g$ avoids $L$ on a punctured neighbourhood of $c$, must assume one of two extra conditions: either $L\in B$ with $f(L)=M$, or $g\mathrm{\ne }L$ on a punctured neighbourhood of $c$; with both dropped the statement is false, which is FALSE: ${\lim }_{x\to c}f(g(x))=M$ whenever ${\lim }_{x\to c}g=L$ and ${\lim }_{y\to L}f=M$. The first of those conditions is exactly continuity of $f$ at $L$ written out, so under the hypotheses above it holds automatically and nothing has to be assumed. The mechanism is visible in the proof: 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 quantifies over $\mathrm{\mid }y-b\mathrm{\mid }<\rho$ rather than over $0<\mathrm{\mid }y-b\mathrm{\mid }<\rho$, so the value $y=b$ that the limit version cannot control is precisely the one the continuity hypothesis does control.