LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The $K$ -topology on $R$ , generated by the open intervals together with their complements of $K = {1 / (n + 1) : n \in N}$ , is $T_{1}$ and Hausdorff but not regular

Statement

Write $ι$ for the canonical natural of $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field), so that $1 / (n + 1)$ abbreviates the inverse of $ι (n + 1)$ , and put

$K : = {1 / (n + 1) : n \in N} \subseteq R, B_{K} : = {(a, b) : a < b} \cup {(a, b) ∖ K : a < b},$

the bounded open intervals of $R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) together with those same intervals with $K$ removed. Then:

$B_{K}$ is a basis for a unique topology $T_{K}$ on $R$ (A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis), the $K$ -topology, and $T_{K}$ is finer than the usual topology of $R$ (The absolute value makes $R$ 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, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).
$(R, T_{K})$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and $T_{1}$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).
$K$ is closed in $T_{K}$ .
$(R, T_{K})$ is not regular (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly): the point $0$ and the closed set $K$ have no disjoint open neighbourhoods.

Facts & Assumptions

Given: $R$ with its order, its usual metric $d_{R} (s, t) = ∣ s - t ∣$ and its usual topology; the set $K$ and the family $B_{K}$ above; reals $a, b, c, d$ and naturals $m, n$ . Throughout $1 / (n + 1)$ is the inverse of the canonical natural $ι (n + 1)$ .

[A1]

$(a, b) = {t \in R : a < t < b}$ , and for $a < b$ the midpoint satisfies $a < (a + b) / 2 < b$ , so $(a, b) \neq \emptyset$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L1]

A family satisfying (B1) and (B2) is a basis for exactly one topology, namely the family of sets each of whose points lies in a member inside the set (A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis, Basis and subbasis for a topology, and the topology generated by a family of sets).

[L2]

$U \subseteq R$ is open in the usual topology exactly when every $x \in U$ has $r > 0$ with $(x - r, x + r) \subseteq U$ (The absolute value makes $R$ 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 3, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[L3]

For every real $ε > 0$ there is a natural $k \geq 1$ with $1 / k < ε$ ; every nonzero natural is a successor; and for every real $x$ there is a natural $k \geq 1$ with $x < k$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every nonzero natural number is a successor, Every complete ordered field is Archimedean).

[L4]

For $n \geq 1$ the canonical natural $ι (n)$ is positive and $n \mapsto ι (n)$ is strictly increasing on the naturals $\geq 1$ (Canonical naturals are positive and strictly increasing); and $0 < u < v$ implies $0 < 1 / v < 1 / u$ (Inverses of positives are positive, and reciprocation reverses order).

[L5]

A two-element set of reals has a maximum and a minimum, each of which is one of the two elements (Maximum and minimum of a set, Every nonempty finite set of reals has a maximum and a minimum).

[L6]

A space is Hausdorff when distinct points have disjoint open neighbourhoods; every Hausdorff space is $T_{1}$ ; a space is regular when a point and a closed set not containing it have disjoint open neighbourhoods (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Every Urysohn space is Hausdorff, every Hausdorff space is $T_{1}$ and hence $T_{0}$ , and every regular $T_{1}$ space is Urysohn, Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly, $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).

[L7]

A set is closed exactly when its complement is open, and an arbitrary union of open sets is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1

(B1): every $x \in R$ lies in $(x - 1, x + 1) \in B_{K}$ , so $B_{K}$ covers $R$ .

1.2

(B2): for $a < b$ and $c < d$ the intersection $(a, b) \cap (c, d)$ is $(\max {a, c}, \min {b, d})$ by [L5] and [A1], which is a member of $B_{K}$ when $\max {a, c} < \min {b, d}$ and is empty otherwise; removing $K$ from one or from both factors intersects the same interval with the complement of $K$ , giving a member of $B_{K}$ or the empty set; and in the empty case (B2) is vacuous.

A1L5

2.1

By steps 1.1 and 1.2 and [L1] the family $B_{K}$ is a basis for a unique topology $T_{K}$ on $R$ , which is claim 1's first half.

step 1.1step 1.2L1

3.1

$T_{K}$ is finer than the usual topology: if $U$ is usually open and $x \in U$ , then [L2] gives $r > 0$ with $x \in (x - r, x + r) \subseteq U$ , and $(x - r, x + r) \in B_{K}$ , so $U \in T_{K}$ by [L1]; this completes claim 1.

step 2.1L1L2

3.2

Let $x \neq y$ in $R$ ; by [L5] we may assume $x < y$ , and put $r : = (y - x) / 2 > 0$ , so that $x + r = (x + y) / 2 = y - r$ . The sets $(x - r, x + r)$ and $(y - r, y + r)$ lie in $B_{K}$ , hence are open, contain $x$ and $y$ respectively, and are disjoint, a common point being both $< x + r$ and $> y - r = x + r$ . So $(R, T_{K})$ is Hausdorff, and $T_{1}$ by [L6]; this is claim 2.

step 2.1A1L5L6

3.3

$R ∖ K = ⋃_{k \geq 1} ((- k, k) ∖ K)$ , the union being over naturals $k \geq 1$ : each term is a member of $B_{K}$ and misses $K$ , and conversely a point $x \notin K$ has $∣ x ∣ < k$ for some natural $k \geq 1$ by [L3], so $x$ lies in the term of index $k$ . Hence $R ∖ K$ is open by [L7] and $K$ is closed, which is claim 3.

step 2.1L3L7

4.1

Suppose $U, V \in T_{K}$ are disjoint with $0 \in U$ and $K \subseteq V$ ; note $0 \notin K$ , since every element of $K$ is positive by [L4].

step 3.3L4assume-hyp

5.1

Under step 4.1: by [L1] there is $B \in B_{K}$ with $0 \in B \subseteq U$ , and $B$ is $(a, b)$ or $(a, b) ∖ K$ with $a < 0 < b$ .

step 4.1L1A1

6.1

Under step 4.1: by [L3] there is a natural $k \geq 1$ with $1 / k < b$ , and $k = n + 1$ for some $n \in N$ , so $1 / (n + 1) \in (a, b)$ , since $0 < 1 / (n + 1) < b$ by [L4] and $a < 0$ .

step 5.1L3L4

7.1

Under step 4.1: $B \neq (a, b)$ , for otherwise $1 / (n + 1) \in B \subseteq U$ by step 6.1 while $1 / (n + 1) \in K \subseteq V$ , contradicting $U \cap V = \emptyset$ ; hence $B = (a, b) ∖ K$ .

step 5.1step 6.1

7.2

Under step 4.1: $1 / (n + 1) \in K \subseteq V$ , so by [L1] there is $B^{'} \in B_{K}$ with $1 / (n + 1) \in B^{'} \subseteq V$ ; and $B^{'}$ is not of the form $(c, d) ∖ K$ , which contains no point of $K$ , so $B^{'} = (c, d)$ with $c < 1 / (n + 1) < d$ .

step 6.1L1

8.1

Under step 4.1: put $t : = \max {\max {a, c}, 1 / (n + 2)}$ by [L5]. Then $t < 1 / (n + 1)$ : indeed $a < 0 < 1 / (n + 1)$ and $c < 1 / (n + 1)$ , and $1 / (n + 2) < 1 / (n + 1)$ by [L4], since $ι (n + 1) < ι (n + 2)$ and both are positive.

step 6.1step 7.1step 7.2L4L5

9.1

Under step 4.1: the interval $(t, 1 / (n + 1))$ contains no element of $K$ . An element $1 / (m + 1)$ of it satisfies $1 / (m + 1) < 1 / (n + 1)$ , hence $ι (n + 1) < ι (m + 1)$ by [L4], hence $n + 1 < m + 1$ and so $m + 1 \geq n + 2$ , whence $1 / (m + 1) \leq 1 / (n + 2) \leq t$ by [L4] and step 8.1, contradicting $t < 1 / (m + 1)$ .

step 8.1L4

10.1

Under step 4.1: put $z : = (t + 1 / (n + 1)) / 2$ , so $t < z < 1 / (n + 1)$ by [A1] and step 8.1, and $z \notin K$ by step 9.1.

step 8.1step 9.1A1

11.1

Under step 4.1: $z \in (a, b) ∖ K = B \subseteq U$ , since $a \leq \max {a, c} \leq t < z$ and $z < 1 / (n + 1) < b$ by step 6.1 and step 8.1; and $z \in (c, d) = B^{'} \subseteq V$ , since $c \leq \max {a, c} \leq t < z$ and $z < 1 / (n + 1) < d$ by step 7.2.

step 6.1step 7.1step 7.2step 8.1step 10.1L5

12.1

Step 11.1 puts $z$ in $U \cap V$ , contradicting the disjointness assumed in step 4.1; so no such $U$ and $V$ exist, and by [L6] the space $(R, T_{K})$ is not regular, which is claim 4.

step 4.1step 11.1L6∎

Remarks

One pair suffices, and only one pair is claimed. Regularity is a statement about every point and every closed set missing it, so a single pair that cannot be separated refutes it; the pair exhibited is $(0, K)$ . Nothing above asserts that the space is regular at any other pair, and nothing needs it: what the lemma is for is the refutation of "Hausdorff implies regular", and that needs exactly one failure.
Why the gap $(t, 1 / (n + 1))$ is the right place to look. The basic neighbourhood of $0$ inside $U$ has had all of $K$ deleted, so it cannot be told apart from a usual interval except at the points of $K$ ; and any neighbourhood of the point $1 / (n + 1)$ of $K$ must be an ordinary interval, because the deleted basic sets miss $K$ altogether. Two such sets overlap in a nonempty interval, and the interval between consecutive members of $K$ supplies a point of the overlap that is not in $K$ . Writing "clearly some point of the overlap avoids $K$ " would be the gap that this argument exists to close.
The index shift is not cosmetic. $N$ contains $0$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field), so the set is written ${1 / (n + 1) : n \in N}$ and its largest element is $1$ ; writing ${1 / n : n \in N}$ would divide by zero.

Depends on

Used by

In the K-topology on ℝ the closed set K ∪ {0} carries a continuous two-valued function with no continuous extension Counterexample
The K-topology on ℝ is not uniformizable Counterexample
FALSE: every Hausdorff space is regular False statement

Dependency tree · two levels

59 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K-topology (Wikipedia) (standard reference, not scraped)
Regular space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §13 and §31 (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 13: Basis for a Topology (East Tennessee State University) (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 31: The Separation Axioms (East Tennessee State University) (standard reference, not scraped)

The KK-topology on RR, generated by the open intervals together with their complements of K={1/(n+1):n∈N}K={1/(n+1):n∈N}, is T1T1​ and Hausdorff but not regular