Continuity, IVT, EVT, and Uniform Continuity: Examples and Counterexamples

Take $\mathbb{R}$ with its usual metric $d(x,y) = |x-y|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded), identify $\mathbb{Z}$ with its canonical copy inside $\mathbb{R}$ (The integers as equivalence classes of pairs of naturals, The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals), and let

$\psi(x) \;:=\; d(x, \mathbb{Z}) \;=\; \inf\{\, |x - n| \ : \ n \in \mathbb{Z} \,\}$

be the distance from $x$ to the nonempty set $\mathbb{Z}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). Then:

1. $\psi$ is $1$-Lipschitz on $\mathbb{R}$: $|\psi(x) - \psi(y)| \le |x-y|$ for all real $x, y$. Consequently $\psi$ is uniformly continuous on $\mathbb{R}$ (Uniform continuity of $f : A \to \mathbb{R}$: one $\delta$ serving every pair of points of $A$) and continuous on $\mathbb{R}$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).
2. The infimum is attained, and computed. Writing $m := \lfloor x \rfloor$ for the integer part of $x$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \le x < m + 1$) and $t := x - m$, so that $0 \le t < 1$, $\psi(x) \;=\; \min\{\, t,\ 1-t \,\} \;=\; \min\bigl\{\, |x-m|,\ |x-(m+1)| \,\bigr\} ,$ so $\psi(x) = |x - n|$ for $n = m$ or $n = m+1$.
3. Range. $0 \le \psi(x) \le 1/2$ for every real $x$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).
4. Zero set. $\psi(x) = 0$ if and only if $x \in \mathbb{Z}$.

Why this example is here. It is the standard uniformly continuous function of this track that is not defined by a formula in the field operations, and it is obtained from the metric machinery rather than rebuilt: claim 1 is $|d(x,A) - d(y,A)| \le d(x,y)$, so the distance to a fixed nonempty set is $1$-Lipschitz applied to $A := \mathbb{Z}$ in the metric space $\mathbb{R}$, transported to a statement about a real function by Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$, 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, clause 3. Only claims 2 to 4, which compute the value, need an argument of their own.

The same function is computed elsewhere, and nothing here depends on that. The trigonometry-free oscillator $\psi(x) = \inf_{n \in \mathbb{Z}} |x - n|$ is well defined and attained at a nearest integer, takes values in $[0, 1/2]$, vanishes exactly on $\mathbb{Z}$, equals $1/2$ at half-integers, and is $1$-periodic introduces $\psi$ on the companion page of The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ and proves the same computation together with $1$-periodicity and the value at half-integers. That item lives on an examples page, which is a leaf of the dependency graph, so no item may rest on it; the verification below is therefore self-contained, and the duplication is deliberate rather than an oversight.

• No completeness of $\mathbb{R}$ is spent on the infimum here. Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space produces $d(x,\mathbb{Z})$ from the greatest-lower-bound property, but step 3.1 does not need that route: it exhibits an element of the set of distances that is also a lower bound, which is the definition of the infimum read directly (Greatest lower bound (infimum)). Completeness does enter once, through Integer part: for every real $x$ there is exactly one integer $m$ with $m \le x < m + 1$, whose existence half is the Archimedean property.

• Why "nearest integer" is a theorem and not a phrase. The words presuppose that a nearest integer exists, and that is exactly what steps 2.2 and 2.3 establish. When $t = 1/2$ there are two nearest integers, $m$ and $m+1$, and the formula $\min\{t,1-t\}$ is indifferent to which is taken, so nothing is selected.

• What this contributes to the hierarchy. $\psi$ is Lipschitz, hence uniformly continuous (Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$, 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, clause 6), and it is bounded and not monotone; so it is a uniformly continuous function that is neither a polynomial nor eventually constant, and it is the natural domain-wide example to set beside $x \mapsto 1/x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1/(k+2)$ and $1/(k+3)$ defeating every $\delta$, where uniform continuity fails.

Refuted claim: every function $\mathbb{R} \to \mathbb{R}$ is continuous at at least one point (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

The witness is the Dirichlet function, the indicator of the rationals: writing $\mathbb{Q}$ for the canonical copy of the rationals inside $\mathbb{R}$ (The rationals embed densely in the reals),

$\mathbf{1}_{\mathbb{Q}} : \mathbb{R} \to \mathbb{R}, \qquad \mathbf{1}_{\mathbb{Q}}(x) := \begin{cases} 1 & \text{if } x \in \mathbb{Q},\\ 0 & \text{if } x \notin \mathbb{Q}. \end{cases}$

It is continuous at no point of $\mathbb{R}$. The mechanism is that both $\mathbb{Q}$ and its complement are dense (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable), so every neighbourhood of every real contains a point of each, and the two values differ by $1$.

The argument is choice free. Density is used in the form "every neighbourhood of every point meets the set", which is The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points applied to a closure equal to $\mathbb{R}$; no sequence is built, so neither A point lies in the closure of $A \subseteq \mathbb{R}$ iff some sequence in $A$ converges to it, so a subset of $\mathbb{R}$ is closed iff it is sequentially closed nor $f$ is continuous at $c \in A$ if and only if $f(x_k) \to f(c)$ for every sequence in $A$ converging to $c$, the converse direction costing countable choice is invoked, and the countable choice those two spend is not spent here.

• Why $\varepsilon_0 = 1/2$ and not $1$. Any $\varepsilon_0 \le 1$ works, since the discrepancy produced is exactly $1$. Taking $1/2$ leaves the inequality strict and makes it visible that the failure is not a boundary effect.

• Restricting the domain repairs it completely. The restriction of $\mathbf{1}_{\mathbb{Q}}$ to $\mathbb{Q}$ is constantly $1$ and the restriction to the irrationals is constantly $0$; both are continuous. This is the standard warning that continuity is a property of the pair (function, domain), recorded in Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point: continuity passes to subsets of the domain, never up from them.

• A near miss worth naming. Multiplying by $x$ repairs continuity at exactly one point: $x \mapsto x\,\mathbf{1}_{\mathbb{Q}}(x)$ is continuous at $0$ and nowhere else, which is $x \cdot 1_{\mathbb{Q}}(x)$ is continuous at $0$ and at no other point. The same argument as above, applied to any function taking two distinct values densely, shows nowhere-continuity; in particular the function equal to $1$ on $\mathbb{Q}$ and $-1$ elsewhere is nowhere continuous while its absolute value is constant.

Let $\mathbf{1}_{\mathbb{Q}}$ be the indicator of the rationals (The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$) and put

$h : \mathbb{R} \to \mathbb{R}, \qquad h(x) := x \cdot \mathbf{1}_{\mathbb{Q}}(x),$

so $h(x) = x$ for rational $x$ and $h(x) = 0$ for irrational $x$. Then:

1. $h$ is continuous at $0$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point);
2. $h$ is not continuous at any $c \ne 0$.

So the set of points of continuity of a real function can be a single point. Together with The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$, where that set is empty, this shows how little the set of continuity points is constrained by the mere existence of the function.

The point of the example. Continuity at $c$ compares $h(x)$ with $h(c)$, and the two branches of $h$ agree only where $x = 0$. Multiplying the indicator by $x$ damps the jump: near $0$ both branches are small, so the discrepancy is at most $|x|$; away from $0$ the discrepancy is at least $|c|/2$ on every neighbourhood, because each branch is realised arbitrarily close to $c$.

• Where the damping factor does its work. The estimate $|h(x)| \le |x|$ of step 1.1 is the whole of claim 1, and it is available only because the two branches of $h$ agree at $0$. Replacing $x$ by any function vanishing at $0$ and continuous there gives the same conclusion at $0$; replacing it by a nonzero constant gives The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$ back.

• The choice of $\delta' = \min\{\delta, |c|/2\}$ is what makes the irrational case work. Without shrinking to $|c|/2$ the rational point $p$ near $c$ could be close to $0$, and then $|h(p) - h(c)| = |p|$ would be small; the shrinking keeps $p$ away from $0$ by at least $|c|/2$.

• This example is choice free, for the same reason as The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$: density is used only in the form "every neighbourhood meets the set", never to build a sequence.

Refuted claim: the function $f : (0,1) \to \mathbb{R}$, $f(x) := 1/x$, is uniformly continuous on $(0,1)$ (Uniform continuity of $f : A \to \mathbb{R}$: one $\delta$ serving every pair of points of $A$, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).

$f$ is continuous on $(0,1)$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point) and its domain is bounded, so this is the sharpest simple instance of FALSE: every continuous real function is uniformly continuous on its domain: neither continuity nor boundedness of the domain implies uniform continuity, and what Heine-Cantor in $\mathbb{R}$: a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$-natively from sequential compactness actually needs is compactness, which $(0,1)$ does not have because it is not closed (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded).

The refutation exhibits, for every $\delta > 0$, a pair of points of $(0,1)$ closer than $\delta$ whose $f$-values differ by exactly $1$. The pairs are

$x_k := \frac{1}{k+2}, \qquad y_k := \frac{1}{k+3} \qquad (k \in \mathbb{N}),$

and the shift by $2$ and $3$ is not cosmetic: $\mathbb{N}$ contains $0$ here (Sequences of reals: bounded, eventually, frequently, tails, subsequences is $0$-indexed), so $1/k$ is undefined at $k = 0$ and $1/(k+1)$ leaves $(0,1)$ at $k = 0$.

• What this witnesses in the regularity hierarchy. Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent, transported to real functions by Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$, 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, gives uniformly continuous $\Rightarrow$ continuous and asserts no converse. This item is the witness that the converse fails, and it is one of the two named in the remarks of Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$, 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; the other, On $[0,1]$ the function $x^{\beta}$ is $\beta$-Hölder and is $\alpha$-Hölder for no rational $\alpha > \beta$, so the Hölder classes are strictly nested, separates the Hölder classes below it.

• The failure is at the missing endpoint, and it is repaired by restoring it. On $[a,1]$ with $0 < a < 1$ the same formula is uniformly continuous, by Heine-Cantor in $\mathbb{R}$: a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$-natively from sequential compactness, since $[a,1]$ is closed and bounded. It is also repaired by an explicit estimate: on $[a,1]$ one has $|1/x - 1/x'| = |x - x'|/(xx') \le |x-x'|/a^{2}$, so $f$ is even Lipschitz there.

• A second reading of the same fact. By A uniformly continuous real function on a subset $D \subseteq \mathbb{R}$ extends uniquely to a uniformly continuous function on the closure of $D$, a uniformly continuous function on $(0,1)$ would extend continuously to $[0,1]$ and hence be bounded there (A continuous real function on a compact subset of $\mathbb{R}$ is bounded); $1/x$ is unbounded on $(0,1)$, so it cannot be uniformly continuous. That route is shorter but rests on more, and the computation above is the elementary one.