Topology of $\mathbb{R}$

Throughout, $\mathbb{R}$ is the complete ordered field (Complete ordered field (least-upper-bound property), Ordered field) with its order (Order on the reals) and its absolute value (Absolute value in an ordered field).

Let $x\in \mathbb{R}$ and let $\epsilon \in \mathbb{R}$ with $\epsilon >0$. The $\epsilon$-neighbourhood of $x$ is

$N_{\epsilon }(x):=\{y\in \mathbb{R}:\mathrm{\mid }y-x\mathrm{\mid }<\epsilon \},$

and the punctured $\epsilon$-neighbourhood of $x$ is

$N_{\epsilon }^{\ast }(x):=N_{\epsilon }(x)\setminus \{x\}=\{y\in \mathbb{R}:0<\mathrm{\mid }y-x\mathrm{\mid }<\epsilon \}\mathrm{.}$

The two descriptions of $N_{\epsilon }^{\ast }(x)$ agree because $\mathrm{\mid }y-x\mathrm{\mid }=0$ holds exactly when $y=x$ (Basic properties of the absolute value).

A neighbourhood is an open interval. For every $x$ and every $\epsilon >0$,

$N_{\epsilon }(x)=(x-\epsilon ,x+\epsilon ),$

the interval of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length. Indeed Basic properties of the absolute value gives, for $\epsilon >0$, the equivalence $\mathrm{\mid }y-x\mathrm{\mid }<\epsilon ⟺-\epsilon <y-x<\epsilon$, and adding $x$ throughout turns the right-hand side into $x-\epsilon <y<x+\epsilon$ (Ordered field).

The centre lies in its own neighbourhoods. $x\in N_{\epsilon }(x)$, since $\mathrm{\mid }x-x\mathrm{\mid }=\mathrm{\mid }0\mathrm{\mid }=0<\epsilon$ (Basic properties of the absolute value).

Punctured neighbourhoods are never empty. The element $y:=x+\epsilon \mathrm{/}2$ satisfies $\mathrm{\mid }y-x\mathrm{\mid }=\epsilon \mathrm{/}2$, which is $>0$ and $<\epsilon$, so $y\in N_{\epsilon }^{\ast }(x)$ (Basic properties of the absolute value, Ordered field).

Monotonicity in the radius. If $0<\delta \le \epsilon$ then $N_{\delta }(x)\subseteq N_{\epsilon }(x)$, because $\mathrm{\mid }y-x\mathrm{\mid }<\delta \le \epsilon$ (Ordered field).

Nesting at an interior point. If $y\in N_{\epsilon }(x)$ and $0<\delta \le \epsilon -\mathrm{\mid }y-x\mathrm{\mid }$, then

$N_{\delta }(y)\subseteq N_{\epsilon }(x)\mathrm{.}$

Indeed for $z\in N_{\delta }(y)$ the triangle inequality (The triangle inequality) gives $\mathrm{\mid }z-x\mathrm{\mid }=\mathrm{\mid }(z-y)+(y-x)\mathrm{\mid }\le \mathrm{\mid }z-y\mathrm{\mid }+\mathrm{\mid }y-x\mathrm{\mid }<\delta +\mathrm{\mid }y-x\mathrm{\mid }\le \epsilon$. Note that $\epsilon -\mathrm{\mid }y-x\mathrm{\mid }>0$ precisely because $y\in N_{\epsilon }(x)$, so such a $\delta$ always exists.

• The radius is a real number, not a rational. Nothing on this page tests a condition against rational radii only. That convention belongs to Limits and Cauchy sequences of reals, where the quantifier is over rational $\epsilon$ and the passage between the rational and the real form is the sanctioned remark of Sequences of reals: bounded, eventually, frequently, tails, subsequences. Here $\epsilon$ ranges over the positive reals throughout, and every statement above is proved for an arbitrary positive real.

• Why the punctured version is separated out. A limit point of a set is a point every punctured neighbourhood of which meets the set (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), and deleting the centre is exactly what stops a point of the set from qualifying automatically. The unpunctured condition defines the weaker notion of an adherent point, and the difference between the two is precisely an isolated point.

• Nesting is the workhorse. Almost every openness verification on this page has the shape "given $y$ in the set, shrink the radius by the distance already travelled", which is the nesting property above. It is recorded here once so that no later proof has to redo the triangle inequality in passing.

Let $U,F\subseteq \mathbb{R}$, with neighbourhoods as in The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$.

• $U$ is open when for every $x\in U$ there is a real $\epsilon >0$ with $N_{\epsilon }(x)\subseteq U$.
• $F$ is closed when its complement $\mathbb{R}\setminus F$ is open.
• A set is clopen when it is both open and closed.

The whole of the topology of $\mathbb{R}$ developed on this page rests on this one definition: closedness is defined as openness of the complement, and every other description of a closed set on this page is a theorem (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, 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).

$\mathrm{\varnothing }$ and $\mathbb{R}$ are clopen. The condition defining openness quantifies over the elements of the set, so it holds vacuously for $\mathrm{\varnothing }$; and for $x\in \mathbb{R}$ one has $N_1(x)\subseteq \mathbb{R}$, so $\mathbb{R}$ is open. Since each of the two is the complement of the other, each is also closed.

Every neighbourhood is open. Let $y\in N_{\epsilon }(x)$ and put $\delta :=\epsilon -\mathrm{\mid }y-x\mathrm{\mid }$, which is $>0$ because $y\in N_{\epsilon }(x)$. The nesting property of The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$ gives $N_{\delta }(y)\subseteq N_{\epsilon }(x)$. So every point of $N_{\epsilon }(x)$ has a neighbourhood inside it.

The four open forms of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length are open sets. Let $a,b\in \mathbb{R}$.

• $(a,b)$: for $x$ with $a<x<b$, both $x-a>0$ and $b-x>0$, so $\delta :=\min \{x-a,b-x\}$ is a positive real (the minimum of a two-element set of reals exists, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set). If $\mathrm{\mid }y-x\mathrm{\mid }<\delta$ then $y>x-\delta \ge x-(x-a)=a$ and $y<x+\delta \le x+(b-x)=b$, so $y\in (a,b)$; hence $N_{\delta }(x)\subseteq (a,b)$.
• $(a,\mathrm{\infty })$: for $x>a$ take $\delta :=x-a>0$; then $\mathrm{\mid }y-x\mathrm{\mid }<\delta$ gives $y>x-\delta =a$.
• $(-\mathrm{\infty },b)$: for $x<b$ take $\delta :=b-x>0$; then $\mathrm{\mid }y-x\mathrm{\mid }<\delta$ gives $y<x+\delta =b$.
• $(-\mathrm{\infty },\mathrm{\infty })=\mathbb{R}$: already treated above.

The four closed forms of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length are closed sets. In each case the complement is shown open directly.

• $[a,b]$: if $x\mathrm{\notin }[a,b]$ then $x<a$ or $x>b$ by trichotomy (Ordered field). If $x<a$, take $\delta :=a-x>0$; every $y\in N_{\delta }(x)$ has $y<x+\delta =a$, hence $y\mathrm{\notin }[a,b]$. If $x>b$, take $\delta :=x-b>0$; every $y\in N_{\delta }(x)$ has $y>x-\delta =b$, hence $y\mathrm{\notin }[a,b]$. So $\mathbb{R}\setminus [a,b]$ is open.
• $[a,\mathrm{\infty })$: its complement is $(-\mathrm{\infty },a)$, which is open by the previous paragraph.
• $(-\mathrm{\infty },b]$: its complement is $(b,\mathrm{\infty })$, which is open.
• $(-\mathrm{\infty },\mathrm{\infty })=\mathbb{R}$: its complement is $\mathrm{\varnothing }$, which is open.

• Open and closed are not opposites, and not exhaustive. A set may be neither: the half-open interval $[0,1)$ is neither open nor closed (FALSE: every subset of $\mathbb{R}$ is either open or closed). A set may be both: $\mathrm{\varnothing }$ and $\mathbb{R}$ are clopen. The words are inherited from the interval terminology of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, and the agreement between the two usages is exactly the two lists verified above: an interval called open there is an open set here, and an interval called closed there is a closed set here.

• A clopen set is a disconnection waiting to happen. If $A$ is clopen and both $A$ and $\mathbb{R}\setminus A$ are nonempty, then each of the two is its own closure, so the two are separated in the sense of Separated sets, disconnection, and connected subset of $\mathbb{R}$ and $\mathbb{R}=A\cup (\mathbb{R}\setminus A)$ is a disconnection. Since $\mathbb{R}$ is order-convex it is connected (A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval), so no such $A$ exists: $\mathrm{\varnothing }$ and $\mathbb{R}$ are the only clopen subsets of $\mathbb{R}$.

• The half-open forms are the ones the two lists omit, and deliberately so: $[a,b)$ and $(a,b]$ with $a<b$ are neither open nor closed as subsets of $\mathbb{R}$.

• The radius depends on the point. Openness asks for some $\epsilon$ at each point, and that $\epsilon$ may shrink to nothing as the point approaches the edge of the set, as the computation for $(a,b)$ shows: there $\delta =\min \{x-a,b-x\}$ tends to $0$ as $x$ tends to either endpoint. Asking instead for a single $\epsilon$ that works simultaneously at every point of the set is a strictly stronger condition, and it is not what is defined here; nothing on this page uses it.

Let open and closed subsets of $\mathbb{R}$ be as in Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen.

1. Arbitrary unions of open sets are open. If $\mathcal{U}$ is any family of open subsets of $\mathbb{R}$, then $\bigcup \mathcal{U}$ is open.
2. Finite intersections of open sets are open. If $n\in \mathbb{N}$ and $U_0,\dots ,U_n$ are open, then $U_0\cap \cdots \cap U_n$ is open.
3. Arbitrary intersections of closed sets are closed. If $\mathcal{F}$ is a nonempty family of closed subsets of $\mathbb{R}$, then $\bigcap \mathcal{F}$ is closed.
4. Finite unions of closed sets are closed. If $n\in \mathbb{N}$ and $F_0,\dots ,F_n$ are closed, then $F_0\cup \cdots \cup F_n$ is closed.

The word finite in claims 2 and 4 is not decoration: an arbitrary intersection of open sets need not be open, and dually an arbitrary union of closed sets need not be closed; the remarks below say where that is settled. Claim 3 asks $\mathcal{F}$ to be nonempty only so that $\bigcap \mathcal{F}$ is a subset of $\mathbb{R}$ without appeal to a convention about the empty intersection.

• Completeness plays no part. Nothing above uses the least-upper-bound property, or even the Archimedean property: the only facts about $\mathbb{R}$ the proof touches are the definition of a neighbourhood, its monotonicity in the radius, and the comparison of two positive radii. What needs completeness is not the algebra of open sets but the theorems about compactness that come later.

• Why finiteness cannot be dropped in claim 2. The minimum taken in step 2.1 is a minimum of finitely many positive radii, and it is positive precisely because it is one of them (Every nonempty finite set of reals has a maximum and a minimum). An infinite family of positive radii has an infimum that may be $0$, and then no positive $\delta$ survives. That is exactly what happens for the shrinking intervals of FALSE: an arbitrary intersection of open subsets of $\mathbb{R}$ is open, whose named witness is ${\bigcap }_k(-1\mathrm{/}k,1\mathrm{/}k)=\{0\}$ is not open ↗.

• The four claims are a rewriting of two. Claims 3 and 4 are claims 1 and 2 read through complementation, and closedness is defined by complementation (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen), so no separate argument about closed sets is possible or needed.

Let $A\subseteq \mathbb{R}$, with open and closed sets as in Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen.

• The interior of $A$ is the union of all open subsets of $A$: $A^{\circ }:=\bigcup \{U\subseteq \mathbb{R}:U\text{ open and }U\subseteq A\}\mathrm{.}$
• The closure of $A$ is the intersection of all closed supersets of $A$: $\bar{A}:=\bigcap \{F\subseteq \mathbb{R}:F\text{ closed and }A\subseteq F\}\mathrm{.}$
• The boundary of $A$ is $\mathrm{\partial }A:=\bar{A}\setminus A^{\circ }$.
• The exterior of $A$ is $\mathrm{ext}A:=(\mathbb{R}\setminus A{)}^{\circ }$.

Both operators are well defined and deliver what their names claim. The family whose union defines $A^{\circ }$ always contains $\mathrm{\varnothing }$, and the family whose intersection defines $\bar{A}$ always contains $\mathbb{R}$, so the second family is nonempty and both expressions denote subsets of $\mathbb{R}$ without appeal to any convention about empty unions or intersections. Moreover:

• $A^{\circ }$ is open, being a union of open sets (Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets, claim 1), and $A^{\circ }\subseteq A$, since every set in the family is a subset of $A$. It is therefore the largest open subset of $A$: any open $U\subseteq A$ is a member of the family and so $U\subseteq A^{\circ }$.
• $\bar{A}$ is closed, being an intersection of a nonempty family of closed sets (Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets, claim 3), and $A\subseteq \bar{A}$, since every set in the family contains $A$. It is therefore the smallest closed superset of $A$: any closed $F\supseteq A$ is a member of the family and so $\bar{A}\subseteq F$.

Pointwise description of the interior. For $x\in \mathbb{R}$,

$x\in A^{\circ }⟺\text{there is }\epsilon >0\text{ with }{N}_{\epsilon }(x)\subseteq A\mathrm{.}$

If $x\in A^{\circ }$ then, $A^{\circ }$ being open and containing $x$, there is $\epsilon >0$ with $N_{\epsilon }(x)\subseteq A^{\circ }\subseteq A$. Conversely if $N_{\epsilon }(x)\subseteq A$ then $N_{\epsilon }(x)$ is an open subset of $A$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen) containing $x$, hence $x\in N_{\epsilon }(x)\subseteq A^{\circ }$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$).

The corresponding pointwise description of the closure is not a definitional matter and is proved separately, as 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.

• The four sets partition nothing by themselves, but three of them do. For every $A$ the three sets $A^{\circ }$, $\mathrm{\partial }A$ and $\mathrm{ext}A$ are pairwise disjoint with union $\mathbb{R}$. This is not proved here and is not used on this page; what is used is only the definitions above and the characterisations of 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.

• Interior and closure are dual. Complementation exchanges the two families above, since $U$ is open exactly when $\mathbb{R}\setminus U$ is closed, so $\mathbb{R}\setminus A^{\circ }=\overline{\mathbb{R}\setminus A}$ and $\mathbb{R}\setminus \bar{A}=(\mathbb{R}\setminus A{)}^{\circ }=\mathrm{ext}A$. The second identity is the reason the exterior is usually described as "the complement of the closure".

• $A$ is open exactly when $A=A^{\circ }$, and closed exactly when $A=\bar{A}$. For the first: $A^{\circ }\subseteq A$ always, and $A\subseteq A^{\circ }$ holds exactly when $A$ is one of the open subsets of $A$, that is, exactly when $A$ is open. The closure half is the same argument read the other way, and it is recorded as a claim of 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 because the rest of that theorem needs it.

• Boundary points may or may not belong to the set. $\mathrm{\partial }(0,1)=\mathrm{\partial }[0,1]=\{0,1\}$, and the two boundary points lie outside the first set and inside the second; the boundary sees only the way the set meets its complement, not which side the edge is assigned to.

Let $A\subseteq \mathbb{R}$ and $x\in \mathbb{R}$, with neighbourhoods as in The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$ and closure as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$.

• $x$ is an adherent point of $A$ when $N_{\epsilon }(x)\cap A\mathrm{\ne }\mathrm{\varnothing }$ for every real $\epsilon >0$.
• $x$ is a limit point (or accumulation point) of $A$ when $N_{\epsilon }^{\ast }(x)\cap A\mathrm{\ne }\mathrm{\varnothing }$ for every real $\epsilon >0$: every punctured neighbourhood of $x$ meets $A$.
• $x$ is an isolated point of $A$ when $x\in A$ and there is a real $\epsilon >0$ with $N_{\epsilon }(x)\cap A=\{x\}$.
• The derived set of $A$ is $A^{\mathrm{\prime }}:=\{x\in \mathbb{R}:x\text{ is a limit point of }A\}\mathrm{.}$
• $A$ is dense in $\mathbb{R}$ when $\bar{A}=\mathbb{R}$.

A limit point is an adherent point, since $N_{\epsilon }^{\ast }(x)\subseteq N_{\epsilon }(x)$; and an element of $A$ is an adherent point of $A$, since $x\in N_{\epsilon }(x)\cap A$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$). So the adherent points of $A$ are exactly the points of $A\cup A^{\mathrm{\prime }}$, a statement proved as part of 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.

Limit point and isolated point are exact opposites inside $A$. For $x\in A$: $x$ is an isolated point of $A$ exactly when it is not a limit point of $A$. Indeed $N_{\epsilon }(x)\cap A=\{x\}$ says precisely that $N_{\epsilon }^{\ast }(x)\cap A=\mathrm{\varnothing }$, because $x$ itself always lies in $N_{\epsilon }(x)\cap A$ when $x\in A$; so the existence of an $\epsilon$ witnessing isolation is the negation of the condition defining a limit point. A point of $A$ is therefore either isolated in $A$ or a limit point of $A$, and never both.

A limit point need not belong to the set, and a point of the set need not be a limit point. Both possibilities occur, and the two examples that matter later are $0$, which is a limit point of $\{1\mathrm{/}k:k\ge 1\}$ without belonging to it, and $0$ again, which belongs to $\{0\}\cup [1,2]$ as an isolated point.

• Terminology: limit point here is about a set, never about a sequence. This library reserves subsequential limit for the sequential notion (Subsequential limit of a real sequence, and the subsequential limit set), and the two are genuinely different: the constant sequence $x_k=0$ has $0$ as a subsequential limit, while its set of values $\{0\}$ has no limit point at all. The distinction is the one Subsequential limit of a real sequence, and the subsequential limit set records under "Terminology", and it is respected throughout this page.

• Density is defined through the closure, not through intervals. Saying $\bar{A}=\mathbb{R}$ is equivalent to saying that every nonempty open subset of $\mathbb{R}$ meets $A$, and also to saying that every neighbourhood of every real meets $A$; the equivalences follow from 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 and are used in that form 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.

• The derived set need not be comparable with the set. It can be strictly larger, as for $\mathbb{Q}$: every punctured neighbourhood of any real contains a rational, since density supplies one strictly between $x$ and $x+\epsilon$ (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 the derived set of $\mathbb{Q}$ is all of $\mathbb{R}$. It can be strictly smaller, as for $\{0\}$, whose derived set is empty; and it can be neither, as for $\{0\}\cup (1,2)$, whose derived set is $[1,2]$, a set containing points outside the original and omitting the point $0$ of it. A closed set satisfying $A\subseteq A^{\mathrm{\prime }}$ is called perfect (Perfect subset of $\mathbb{R}$: closed with no isolated points).

Let $A\subseteq \mathbb{R}$, with closure $\bar{A}$ as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ and derived set $A^{\mathrm{\prime }}$ as in Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$. Write

$E:=\{x\in \mathbb{R}:N_{\epsilon }(x)\cap A\mathrm{\ne }\mathrm{\varnothing }\text{ for every real }\epsilon >0\}$

for the set of adherent points of $A$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$). Then:

1. $\bar{A}=E$.
2. $\bar{A}=A\cup A^{\mathrm{\prime }}$.
3. $\bar{A}$ is the smallest closed superset of $A$: it is closed, it contains $A$, and it is contained in every closed $F$ with $A\subseteq F$.
4. $A$ is closed if and only if $A=\bar{A}$, if and only if $A^{\mathrm{\prime }}\subseteq A$.

Claim 3 is the content of the definition of $\bar{A}$ and is restated here so that the four descriptions stand together; claims 1, 2 and 4 are the ones that carry work.

• Which claim does the work in practice. Claim 1 is the one used almost everywhere below: to show a point lies in $\bar{A}$ one exhibits, for each $\epsilon >0$, a point of $A$ within $\epsilon$ of it. Claim 2 is what separates the two ways a point can be adherent, by membership or by accumulation, and it is what makes the notion of an isolated point visible.

• No special property of $\mathbb{R}$ is used. The argument uses the definitions of open, closed, neighbourhood and closure, and the order enters only through the nesting property of neighbourhoods; neither the least-upper-bound property nor the Archimedean property appears at any step. The results of this page that do use them are flagged in Which results on this page use the order of $\mathbb{R}$ and therefore have no general-topological analogue.

• The sequential form is a separate theorem and costs more. Replacing "every neighbourhood meets $A$" by "some sequence in $A$ converges to $x$" is 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, and the passage from the first to the second spends the axiom of countable choice, since it selects one point of $A$ from each of infinitely many neighbourhoods. The characterisation proved above is choice free.

Let $A\subseteq \mathbb{R}$ and $x\in \mathbb{R}$, with closure as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ and sequences and convergence as in Sequences of reals: bounded, eventually, frequently, tails, subsequences and Limits and Cauchy sequences of reals. Then

$x\in \bar{A}⟺\text{there is a sequence }(a_k{)}_{k\in \mathbb{N}}\text{ with }{a}_k\in A\text{ for every }k\text{ and }{a}_k\to x\mathrm{.}$

Consequently $A$ is closed if and only if it is sequentially closed: whenever a sequence with all its terms in $A$ converges, its limit lies in $A$.

The right-to-left direction is choice free; the left-to-right direction spends ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Producing a sequence from a point of the closure requires selecting one point of $A$ from each of the countably many sets $N_{1\mathrm{/}(k+1)}(x)\cap A$, and this library has no canonical rule for that selection, so the axiom of countable choice is invoked explicitly at step 2.2 and nowhere else.

• Where the choice is spent, and why it cannot be avoided here. Step 2.2 is the only appeal to The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$). A canonical selection would require a rule picking a distinguished element of an arbitrary nonempty subset of $\mathbb{R}$, and $\mathbb{R}$ carries no well-ordering that this library has constructed, so this library has no such rule to offer. Contrast Every subset of an at most countable set is at most countable and A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$, where the selection is from subsets of $\mathbb{N}$ and the least element is canonical.

• The choice is genuinely confined to one direction. Step 2.1 selects a single rational $q$ and a single index $K$ for one $\epsilon$ at a time, and finitely many selections need no choice principle. So "the limit of a convergent sequence in a closed set lies in the set" is a theorem of ZF, and only the production of a sequence out of a point of the closure is not.

• The indices start at $0$. Since $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), the shrinking radii are $1\mathrm{/}(k+1)$ and not $1\mathrm{/}k$; the latter is undefined at $k=0$. The threshold in step 3.1 is $K=n-1$ for the same reason, and $n\ge 1$ is exactly what makes $K$ a natural number.

Let $U\subseteq \mathbb{R}$ be open (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). For $x,y\in \mathbb{R}$ write

$H(x,y):=\{z\in \mathbb{R}:x\le z\le y\text{ or }y\le z\le x\}$

for the order-convex hull of the pair, and define a relation on $U$ by

$x\sim y:⟺H(x,y)\subseteq U\mathrm{.}$

Then $\sim$ is an equivalence relation on $U$. Its equivalence classes, called the order components of $U$, form a family $\mathcal{C}$ with the following properties:

1. the members of $\mathcal{C}$ are nonempty and pairwise disjoint, and $U=\bigcup \mathcal{C}$;
2. every member of $\mathcal{C}$ is an interval of one of the four open forms $(a,b)$, $(a,\mathrm{\infty })$, $(-\mathrm{\infty },b)$, $(-\mathrm{\infty },\mathrm{\infty })$ of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, and is an open set;
3. $\mathcal{C}$ is at most countable (Finite, countably infinite, countable, uncountable).

So every open subset of $\mathbb{R}$ is the union of an at most countable family of pairwise disjoint nonempty open intervals. For $U=\mathrm{\varnothing }$ the family $\mathcal{C}$ is empty and the union of the empty family is $\mathrm{\varnothing }$, so the statement holds in that case too.

No choice principle is used. The components are defined by an explicit equivalence relation, and the enumeration in claim 3 is obtained by sending a component to the least index of a rational lying in it, which is canonical by The well-ordering principle.