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Collapsing the set of naturals inside $R$ to a point gives a quotient of $R$ that is not locally compact at the collapsed point

Statement refuted

Refuted: that a continuous image of a locally compact space is locally compact (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space). Local compactness is not preserved by continuous maps, and it is not even preserved by quotient maps.

Witness. Write $ι : N \to R$ for the canonical natural (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field) and put $N : = {ι (n) : n \in N}$ , the set of naturals inside $R$ . Let

$Y : = (R ∖ N) \cup {N}, q : R \to Y, q (t) : = t (t \notin N), q (t) : = N (t \in N),$

and give $Y$ the quotient topology of $q$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection). Then $q$ is a continuous surjection, $R$ is locally compact, and $Y$ is not locally compact: the point $* : = N$ of $Y$ has no compact neighbourhood.

Facts & Assumptions

Given: $R$ with its usual topology, the set $N = {ι (n) : n \in N}$ , the set $Y$ , the surjection $q$ , and the quotient topology on $Y$ .

[L1]

A subset $V \subseteq Y$ is open exactly when $q^{- 1} [V]$ is open in $R$ ; $q$ is continuous; and a set $G \subseteq R$ with $G \cap N = \emptyset$ satisfies $q^{- 1} [q [G]] = G$ , while a set $G \supseteq N$ satisfies $q^{- 1} [q [G]] = G$ as well (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection, For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map, Continuity of a map of topological spaces at a point and globally).

[L2]

$U \subseteq R$ is open exactly when every $x \in U$ has a real $r > 0$ with $(x - r, x + r) \subseteq U$ ; $R$ is metrizable and is locally compact, the set ${t : ∣ t - p ∣ \leq 1}$ being a compact neighbourhood of $p$ (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, 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, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide, Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space).

[L3]

The canonical-natural map $ι$ is strictly increasing, hence injective, and satisfies $ι (n + 1) = ι (n) + 1$ (Canonical naturals are positive and strictly increasing, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, The natural numbers $N$ (von Neumann)), so distinct naturals have distinct canonical naturals and the members of $N$ are spaced at distance at least $1$ ; and for every real $t$ there is a natural $n$ with $t < ι (n)$ (Every complete ordered field is Archimedean, Complete ordered field (least-upper-bound property)).

[L4]

A subset $A$ is a compact subset when the subspace it carries is compact; a closed subset of a compact space is a compact subset of it; a space in which every singleton is open is discrete, and an infinite discrete space is not compact, its singletons covering it with no finite subcover (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L5]

$A$ is a neighbourhood of $x$ when some open set lies between them (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

Counterexample

technique · direct

1.1

$q$ is a continuous surjection by [L1] and $R$ is locally compact by [L2], so any failure of local compactness in $Y$ refutes the claim. Suppose $K \subseteq Y$ is a compact neighbourhood of $*$ , and fix an open $O$ of $Y$ with $* \in O \subseteq K$ ; then $G : = q^{- 1} [O]$ is open in $R$ and contains $N$ .

L1L2L5construct

2.1

For $n \in N$ put $ρ_{n} : = \frac{1}{2} \sup {r \in R : 0 < r < \frac{1}{2} and (ι (n) - r, ι (n) + r) \subseteq G}$ , a supremum of a nonempty set of reals bounded above by $\frac{1}{2}$ , so $0 < ρ_{n} < \frac{1}{2}$ and $(ι (n) - ρ_{n}, ι (n) + ρ_{n}) \subseteq G$ ; nothing is selected, the supremum being determined by $n$ and $G$ . Put $x_{n} : = ι (n) + ρ_{n}$ , so that $ι (n) < x_{n} < ι (n) + \frac{1}{2}$ and $x_{n} \in G$ .

L2L3step 1.1construct

3.1

No $x_{n}$ lies in $N$ , since $ι (n) < x_{n} < ι (n) + 1 = ι (n + 1)$ and the members of $N$ are the $ι (m)$ ; and $x_{n} \neq x_{m}$ for $n \neq m$ , the two lying in disjoint intervals $(ι (n), ι (n) + \frac{1}{2})$ and $(ι (m), ι (m) + \frac{1}{2})$ . So $P : = {x_{n} : n \in N}$ is an infinite subset of $R ∖ N$ and $q$ is injective on it.

L3step 2.1

4.1

$P$ is closed in $R$ : a real $t$ lies in $[ι (m), ι (m) + 1)$ for exactly one natural $m$ when $t \geq 0$ by [L3], and in $(- \infty, 0)$ otherwise; the interval $(t - \frac{1}{4}, t + \frac{1}{4})$ meets at most one of the disjoint intervals $(ι (n), ι (n) + \frac{1}{2})$ , hence contains at most one member of $P$ , so no real is a limit point of $P$ outside $P$ and the complement of $P$ is open by [L2].

L2L3step 3.1

4.2

The subspace $q [P]$ is discrete: for each $n$ the interval $I_{n} : = (x_{n} - η_{n}, x_{n} + η_{n})$ with $η_{n} : = \frac{1}{2} \min {ρ_{n}, \frac{1}{2}}$ misses $N$ , since $ι (n) < x_{n} - η_{n}$ and $x_{n} + η_{n} < ι (n) + 1$ , so $q^{- 1} [q [I_{n}]] = I_{n}$ is open by [L1] and $q [I_{n}]$ is open in $Y$ ; and $q [I_{n}] \cap q [P] = {q (x_{n})}$ , because $I_{n}$ contains no $x_{m}$ with $m \neq n$ .

L1L3step 2.1step 3.1

5.1

$q [P]$ is closed in $Y$ : its preimage is $P$ by [L1], since $P$ misses $N$ , and the preimage of the complement of $q [P]$ is the complement of $P$ , which is open by step 4.1, so $q [P]$ is closed by [L1]. Moreover $q [P] \subseteq q [G] = O \subseteq K$ , so $q [P]$ is a closed subset of the compact $K$ and hence a compact subset of $Y$ by [L4].

L1L4step 1.1step 2.1step 4.1

6.1

So $q [P]$ is an infinite discrete compact space, which [L4] forbids: its singletons form an open cover with no finite subcover. Hence $*$ has no compact neighbourhood in $Y$ , the space $Y$ is not locally compact, and the claim that a continuous image of a locally compact space is locally compact is refuted.

L4step 1.1step 5.1step 4.2∎

Remarks

Collapsing a compact set would not work, and that is the point. If $E$ is a compact subset of a locally compact Hausdorff space, the quotient collapsing $E$ to a point is again locally compact at the collapsed point: claim 4 of In a locally compact Hausdorff space every point has a neighbourhood base of compact sets, every open subspace and every closed subspace is locally compact, every open set around a point contains an open set with compact closure inside it, and every compact set sits inside an open set with compact closure gives an open $V \supseteq E$ with $\overline{V}$ compact, the saturation of $V$ is $V$ itself, so its image is open and lies inside the compact image of $\overline{V}$ . So a convergent sequence together with its limit is the wrong set to collapse; what is needed is an infinite closed discrete set, and $N$ is the simplest one.

What the failure looks like. Every open set of $Y$ containing $*$ pulls back to an open set containing all of $N$ , hence containing an interval around each $ι (n)$ ; the points $x_{n}$ chosen just to the right of each $ι (n)$ then form a closed discrete infinite set inside it, and no compact set can contain such a thing. There is no way to make the neighbourhood small, because it must be large near infinitely many separated places at once.

Compactness behaves differently. A continuous image of a compact space is compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, claim 1); it is only the local condition that fails to survive, and it fails because "locally" is a statement about each point separately and a quotient can glue infinitely many points together.

Quotient maps are the natural place to look. A quotient map is a continuous surjection, so this also refutes the same claim for continuous surjections; and since $q$ here is a closed map with one non-singleton fibre, closedness of the map, even with a single non-singleton fibre, is not enough to restore the conclusion.

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Sources

Locally compact space (Wikipedia) (standard reference, not scraped)
Quotient space (topology) (Wikipedia) (standard reference, not scraped)
J. M. Møller, General Topology (standard reference, not scraped)
I. Khatchatourian, Compactifications (MAT327 notes) (standard reference, not scraped)

Collapsing the set of naturals inside RR to a point gives a quotient of RR that is not locally compact at the collapsed point