Topology of $\mathbb{R}$: Examples and Counterexamples

Take

$U \;:=\; (-1,0) \;\cup\; (0,1) \;\cup\; (2,\infty) .$

This is an open subset of $\mathbb{R}$, and its order components in the sense of Every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, namely its order components are exactly the three intervals $(-1,0)$, $(0,1)$ and $(2,\infty)$: they are pairwise disjoint, their union is $U$, and there are three of them, a finite and hence at most countable family. The example is chosen so that one component is unbounded and two are bounded, and so that the two bounded ones are separated by a single missing point, $0$, rather than by a gap of positive length.

• The count is the number of components, not the number of points. There are three components here, while each of them is an uncountable set (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable). The at most countable family of Every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, namely its order components is a family of intervals, and a finite family is one instance of it.

• What keeps two components apart may be a single missing point. $A$ and $B$ are kept apart by $0 \notin U$ alone, and no gap of positive length is required, although $B$ and $C$ happen to have one. This is why the components are defined by an equivalence relation on $U$ and not by measuring distances between the pieces.

• Reading the decomposition off the formula is legitimate here only because the three pieces were checked to be the classes. A presentation of an open set as a union of open intervals is not automatically its decomposition into components: $(0,2) = (0,1) \cup (0,2)$ writes an open set as a union of open intervals that are neither disjoint nor components.

Write $\mathbb{Q}_{\mathbb{R}}$ for the copy of $\mathbb{Q}$ inside $\mathbb{R}$ (The rationals embed densely in the reals). Then

$\overline{\mathbb{Q}_{\mathbb{R}}} = \mathbb{R}, \qquad (\mathbb{Q}_{\mathbb{R}})^{\circ} = \varnothing, \qquad \partial \mathbb{Q}_{\mathbb{R}} = \mathbb{R},$

with closure, interior and boundary as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$. So the rationals are as large as possible for the closure operator and as small as possible for the interior operator at once, and their boundary is everything.

• The same computation applies verbatim to the irrationals. $X$ is dense by [L2] and its complement $\mathbb{Q}_{\mathbb{R}}$ is dense too, so $\overline{X} = \mathbb{R}$, $X^{\circ} = \varnothing$ and $\partial X = \mathbb{R}$. Two complementary sets can therefore both have boundary everything, which is what the density of each of them forces.

• Empty interior is not smallness in any counting sense. $X$ has empty interior and is uncountable (The irrationals are uncountable), while $\mathbb{Q}_{\mathbb{R}}$ has empty interior and is countable ($\mathbb{Q}$ is countably infinite). The interior measures whether the set contains an interval, and nothing else.

• Where the density of the irrationals comes from. It is proved in Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable by counting: an interval is uncountable and the rationals are not, so an interval cannot consist of rationals alone. No explicit irrational is needed for this computation, though one is available (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$, FALSE: some rational number squares to 2).

Put

$B \;:=\; \{\, 1/k : k \in \mathbb{N},\ k \ge 1 \,\} \qquad \text{and} \qquad A \;:=\; B \cup \{0\} .$

The index runs over $k \ge 1$ because $1/0$ is undefined. Then $A$ is compact (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded) and $B$ is not closed (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). The single point $0$ is the whole difference: it is a limit point of $B$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$) that $B$ omits, and adjoining it turns a non-closed bounded set into a compact one.

• $B$ is bounded and not compact, and it fails to be closed by a single point. Adjoining $0$ is what the verification above shows to be enough: $A = B \cup \{0\}$ is closed. The same computation, run at a point $x \notin A$ lying between two consecutive reciprocals, is the one that isolates each $1/k$ from the rest of $B$.

• $A$ is compact and has an isolated point. Every $1/k$ is isolated in $A$, so $A$ is not perfect (Perfect subset of $\mathbb{R}$: closed with no isolated points); compactness and perfectness are independent properties, and this is a compact set that is countable, which is possible exactly because it is not perfect (Every nonempty perfect subset of $\mathbb{R}$ is uncountable).

• The index range matters. The set is indexed from $k = 1$; $1/0$ does not exist. Since $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), a set written $\{1/k : k \in \mathbb{N}\}$ without a restriction would be ill formed, and the same care is needed at the threshold $K = n-1$ used in the convergence arguments on the parent page.

Let $a, b \in \mathbb{R}$ with $a < b$. Then the closed interval $E := [a,b]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) is perfect (Perfect subset of $\mathbb{R}$: closed with no isolated points), and therefore uncountable by Every nonempty perfect subset of $\mathbb{R}$ is uncountable.

This is a second proof of the uncountability of a nondegenerate interval. The first, Every nondegenerate interval of $\mathbb{R}$ is uncountable, runs a trisection argument directly against an assumed enumeration; the route here checks two purely local properties, closedness and the absence of isolated points, and lets the perfect-set theorem do the counting.

• Nondegeneracy is exactly what is needed. For $a = b$ the set $[a,a] = \{a\}$ is closed, its single point is isolated, and it is finite; the argument of step 1.2 breaks precisely there, since neither $x < b$ nor $x > a$ holds. This matches the hypothesis $a < b$ of Every nondegenerate interval of $\mathbb{R}$ is uncountable.

• The open interval supplies only half of the definition. The computation of step 1.2 applies verbatim inside $(a,b)$ and shows it has no isolated points, but $(a,b)$ is not closed, so it is not perfect. Perfectness needs both halves, which is why the example is stated for the closed interval.

• Two proofs of one fact, sharing one ingredient. Both routes spend the completeness of $\mathbb{R}$ exactly once, Every nondegenerate interval of $\mathbb{R}$ is uncountable as a supremum and Every nonempty perfect subset of $\mathbb{R}$ is uncountable through A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$. They differ in everything else: the first trisects a given interval against a given enumeration, the second selects rational-endpoint intervals by least index. Neither is a corollary of the other.

Refuted claim: an arbitrary intersection of open subsets of $\mathbb{R}$ is open (FALSE: an arbitrary intersection of open subsets of $\mathbb{R}$ is open, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

The witness is $U_k := (-1/k,\ 1/k)$ for $k \ge 1$: each $U_k$ is a nonempty open interval, the family is nested, and

$\bigcap_{k \ge 1} \Big( -\frac{1}{k},\ \frac{1}{k} \Big) \;=\; \{0\},$

which is not open. The index runs over $k \ge 1$ because $1/0$ is undefined; read as a family indexed by $\mathbb{N}$ it is $j \mapsto (-1/(j+1),\ 1/(j+1))$. The refutation is carried out in full in FALSE: an arbitrary intersection of open subsets of $\mathbb{R}$ is open and is recorded here as the named counterexample. The comparison worth keeping in view is the finite case: any finite subfamily of $\{U_k\}$ has intersection the smallest of its members, which is open.

• Exactly one hypothesis of Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets is missing. That theorem asserts openness for finite intersections, and its proof takes the minimum of finitely many positive radii. Here the family is infinite and the radii at the surviving point $0$ are the numbers $1/k$, which have no positive lower bound (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$). So the true theorem is not contradicted; its finiteness hypothesis cannot be dropped.

• The Archimedean property is doing the work in step 2.1. In a non-Archimedean ordered field a positive infinitesimal lies in every $(-1/k, 1/k)$, and the intersection is then strictly larger than $\{0\}$. So this is a counterexample about $\mathbb{R}$, supplied by For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$, and not a formal consequence of openness alone.

• The intersection here is a single point, and that is not forced. An infinite intersection of open sets can be open, for instance when all members are equal. The claim refuted is a universal one, so one witness settles it.

Refuted claim: every subset of $\mathbb{R}$ is open or closed (FALSE: every subset of $\mathbb{R}$ is either open or closed, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

The witness is the half-open interval $E := [0,1)$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). It fails openness at its left endpoint $0$, which belongs to $E$ while every neighbourhood of $0$ reaches below $0$, and it fails closedness at $1$, which does not belong to $E$ while every neighbourhood of $1$ reaches into $E$. The refutation is carried out in full in FALSE: every subset of $\mathbb{R}$ is either open or closed and is recorded here as the named counterexample.

• The two failures are at different points and are independent. Openness fails only at $0$, since every $x$ with $0 < x < 1$ has a neighbourhood inside $E$; closedness fails only at $1$, since every point outside $E$ other than $1$ has a neighbourhood outside $E$. Repairing either failure separately gives a set that is open or closed but not both: $(0,1)$ is open, $[0,1]$ is closed.

• Nothing about the true results is contradicted. Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets says which combinations of open sets are open and which combinations of closed sets are closed; it says nothing about arbitrary sets, and "closed" was never the negation of "open" (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

• The mirror witness. $(0,1]$ is neither open nor closed for the same two reasons with the roles of the endpoints exchanged, and $\mathbb{Q}$ is a subset that is neither open nor closed with no endpoints at all ($\mathbb{Q}$ has closure $\mathbb{R}$, empty interior, and boundary $\mathbb{R}$).

Refuted claim: the bounded interval $(0,1)$ is compact (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset); equivalently, boundedness on its own is enough for compactness.

The witness is the family

$\mathcal{U} \;:=\; \{\, V_k : k \ge 1 \,\}, \qquad V_k := (1/k,\ 1),$

of open subsets of $\mathbb{R}$. The index runs over $k \ge 1$ because $1/0$ is undefined, and the first member is degenerate: $V_1 = (1,1) = \varnothing$, which is harmless, since a cover may contain empty members. The family covers $(0,1)$ and no finite subfamily does.

• The cover creeps up on the missing endpoint. Every member of $\mathcal{U}$ stops short of $0$, and the whole family reaches every point of $(0,1)$ only because the reciprocals $1/k$ get arbitrarily small (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$). A finite subfamily stops at the largest of its indices $K$ and therefore misses every point of $(0,1)$ that is at most $1/K$.

• The empty member is not a defect. $V_1 = (1,1)$ is empty because $1/1 = 1$; a family of open sets is a cover as long as its union contains the set, and an empty member contributes nothing either way. Writing the family from $k = 2$ instead would change nothing in the argument.

• The closed interval behaves differently, and that is the whole point. $[0,1]$ is compact by Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact, and the analogous family $\{(1/k, 1)\}$ is not even a cover of it, since it misses both $0$ and $1$. The endpoint that the cover above exploits is $0$, which $(0,1)$ omits and $[0,1]$ contains.

Refuted claim: in every ordered field a closed bounded set is compact, so the completeness hypothesis of the Heine-Borel characterisation is unnecessary (FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness).

The witness is the ordered field $\mathbb{Q}$ (The rationals as equivalence classes of pairs of integers, The rationals form a totally ordered field) together with

$S \;:=\; \{\, q \in \mathbb{Q} : q \ge 0 \text{ and } q^2 < 2 \,\} .$

The set $S$ is bounded, is closed in $\mathbb{Q}$, and is not compact in $\mathbb{Q}$, all with respect to the vocabulary of Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset transposed from $\mathbb{R}$ to $\mathbb{Q}$ exactly as set out in FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness, where the refutation is carried out in full. This item records the witness and says what makes it work.