Which results on this page use the order of $\mathbb{R}$ and therefore have no general-topological analogue

This page builds the topology of $\mathbb{R}$ out of the order and the absolute value alone: a neighbourhood is an interval (The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$), and open and closed are defined from neighbourhoods (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). Some of what follows uses nothing else about $\mathbb{R}$, and some of it is written in order vocabulary from beginning to end. This remark separates the two, so a reader knows which results are candidates for reuse elsewhere and which are not even statable elsewhere. It asserts nothing about topological spaces in general: they are developed later in this library, on the page of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison ↗ and Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not ↗, where the metric development becomes a special case, but no claim about them is made or needed here and nothing below rests on them.

Results that use only the definitions. Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets and 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 use openness, closedness, closure and the nesting property of neighbourhoods, together with the comparison of two positive radii, and nothing beyond that. Neither the least-upper-bound property nor the Archimedean property appears in either proof, and their statements mention no interval, no bound and no order, so those statements would still make sense wherever a notion of neighbourhood is available, however it arises.

Results that cannot be separated from the order of $\mathbb{R}$. Four results on this page depend on the order. For the first three the order is in what they say and not merely in how they are proved; for the fourth it is in the proof only, and the bullet says so.

• Every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, namely its order components says that an open set is a countable disjoint union of open intervals. An interval is defined by the order (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), the components are the classes of an equivalence relation defined by order-convexity, and the identification of a component as an interval is carried out with $\sup$ and $\inf$. Delete the order and there is no statement left to prove.
• A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval characterises connectedness by order-convexity. Connectedness itself is defined without the order (Separated sets, disconnection, and connected subset of $\mathbb{R}$), but the property it is being equated with is an order property, so the theorem is a bridge between an order notion and a topological one and exists only where both are present.
• Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact and A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded speak of closed bounded intervals and of bounded sets. Boundedness is an order notion (Lower bound, bounded below, bounded set), and the proof of the first is a bisection, which uses the midpoint and hence the field operations as well. The completeness of $\mathbb{R}$ enters 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 $0$.
• A subset of $\mathbb{R}$ is compact iff it is sequentially compact routes both implications through the previous item, whose backward half spends the least-upper-bound property, and its forward implication additionally uses Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence, which spends that property again; the backward implication does not use Bolzano-Weierstrass at all. Its statement mentions only compactness and sequences (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset), so unlike the three above it is statable without an order; the dependence lies entirely in the proof. This library provides no other proof, so nothing here licenses the equivalence outside $\mathbb{R}$.

Where the dependence on completeness is visible rather than merely present. FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness refutes, inside this library, the claim that closed and bounded implies compact in an arbitrary ordered field, and $\{q \in \mathbb{Q} : q \ge 0,\ q^2 < 2\}$ is closed and bounded in $\mathbb{Q}$ and is not compact ↗ names the witness in $\mathbb{Q}$. That is the sharpest statement this page makes about the limits of its own results: the Heine-Borel characterisation is not a formal consequence of the definitions, and it fails in the nearest ordered field that is not complete.

The metric topology of $\mathbb{R}$ is the topology of this page. This library does develop metric spaces (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and $\mathbb{R}$ under $d(x,y) = |x-y|$ is one of them (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). The two resulting notions of open subset of $\mathbb{R}$ are not merely equivalent but literally the same condition, and unfolding the definitions is the whole of the proof: claim 2 of that lemma gives $B(x,r) = (x-r,\ x+r)$, which is exactly the neighbourhood $N_r(x)$ of The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$, so "every point of $U$ admits a ball inside $U$" and "every point of $U$ admits a neighbourhood inside $U$" (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen) say the same thing word for word. This page still proves everything from the order directly, so that nothing here rests on the metric development; the identification is recorded so that a reader moving between the two pages knows they are looking at one topology and not two. Because the two collections of open sets are one collection, everything built from them is one notion as well. The interior, closure and boundary of Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ and of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space are the same three sets, since each side characterises them in both of the same two ways, pointwise by neighbourhoods and by extremality among the open subsets and the closed supersets (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 closure of a nonempty $A$ is $\{x : d(x,A) = 0\}$, equals $A$ together with its limit points, and is the smallest closed superset); limit point, isolated point, adherent point and dense are defined by literally the same condition on the two sides (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), once $N_\varepsilon(x)$ and $B(x,\varepsilon)$ are recognised as the same set; and 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 is the case $X = \mathbb{R}$ of A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, spending the axiom of countable choice in the same one of the two directions.

Bounded means the same here as it does in the metric development, and that agreement is not a topological fact. Lower bound, bounded below, bounded set calls $A \subseteq \mathbb{R}$ bounded when it has both a lower and an upper bound, while Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space calls it bounded when it is empty or lies inside some ball of $(\mathbb{R}, d_{\mathbb{R}})$. The two conditions hold of exactly the same subsets of $\mathbb{R}$. If $A \subseteq B(x_0,r) = (x_0 - r,\ x_0 + r)$ (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, claim 2, Open ball, closed ball and sphere in a metric space) then $x_0 - r$ and $x_0 + r$ bound $A$ below and above; conversely, if $\ell \le a \le u$ for every $a \in A$ and $A \ne \varnothing$, then $\ell \le u$ and $A \subseteq B(\ell,\ u - \ell + 1)$, the radius being at least $1$ and hence positive; and an empty $A$ is bounded under both definitions, vacuously under the first and by the explicit empty clause of the second. So the word bounded in A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded may be read in either sense without changing which sets the theorem names.

What may not be done is to read that theorem as a statement about the topology. Boundedness is a property of the metric and not of the topology it induces (FALSE: boundedness of a metric space is determined by its topology): the metric $\rho(u,v) = \min\{|u-v|,\ 1\}$ induces exactly the open sets of this page, and under it every subset of $\mathbb{R}$, including $\mathbb{R}$ itself, is bounded. Since the open sets are unchanged, compactness (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset) is unchanged too, and $\mathbb{R}$ is not compact, being unbounded in the order sense (A compact subset of $\mathbb{R}$ is closed and bounded); yet $\mathbb{R}$ is closed, and $\rho$-bounded. So "closed and bounded implies compact" is false for $\rho$ and true for $d_{\mathbb{R}}$, on one and the same topology. The identification in the previous paragraph is with the usual metric specifically, and Heine-Borel is a theorem about that metric and the order it comes from, not about the topology alone.

What is deliberately not claimed. Whether the results above have analogues in a setting carrying a topology with no order available at all, and whether compactness and sequential compactness agree there, are questions about general topological spaces. This library takes those questions up on a later page, where Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not ↗ identifies the metric development, this page's topology included, as a special case of the general one; but nothing on this page proves anything about general spaces, and the reader should take no assertion about them from here. What is claimed here is narrower and is checkable line by line against the proofs: in the four results listed above, the order of $\mathbb{R}$ is used, and in three of them it is used in the statement itself.