This recent question led to some discussion on hemicompactness. A topological space $X$ is said to be hemicompact if there is an increasing sequence of compacta $K_1\subseteq\dots \subseteq K_n\subseteq \dots$ such that every compact $K\subseteq X$ lies in some $K_n$.

This is always the case when $X$ is $\sigma$-compact and weakly locally compact (every point has a compact neighborhood), and the converse is true under the additional assumption of first countability (since a hemicompact space is trivially $\sigma$-compact, the nontrivial part of the converse is the deduction of weak local compactness).

On the other hand, here it was shown that we cannot relax the first countability assumption to sequentialness. The counterexample is the Arens space, which is sequential (though not Fréchet-Urysohn) and hemicompact, yet not weakly locally compact. It should be noted that the Arens space has rather good separation properties ($T_6$).

A natural question becomes, given that sequentialness is too weak, can we relax first countability to the Fréchet-Urysohn property?

The short answer is no, at least not without further assumptions. I'll describe a counterexample next, then conclude with the titular question:


Let $X=\mathbb R\cup \infty$, where $\mathbb R$ has the Euclidean topology and neighborhoods of $\infty$ are those of the form $\{\infty\}\cup\mathbb R\backslash D$, where $D\subset \mathbb R$ is discrete and closed (equivalently, $D$ has no limit points in $\mathbb R$).

Lemma. $K\subseteq X$ is compact if and only if either $K\subseteq \mathbb R$ is Euclidean compact, or $\infty\in K$ and $K\cap \mathbb R$ is bounded.

If $K\subseteq \mathbb R$, then since $\mathbb R$ has the Euclidean topology $K$ is compact if and only if $K$ is Euclidean compact. On the other hand, if $\infty\in K$, and $K\cap \mathbb R\subseteq [-M,M]$, then every open cover of $K$ has a member $U\ni \infty$, whereby $\mathbb R\backslash U$ is closed and discrete, so that $K\backslash U\subseteq [-M,M]\backslash U$ is finite.

Finally, if $K\cap \mathbb R$ is unbounded, then let $x_n\in K\cap \mathbb R$ be a sequence eventually leaving every bounded subset of $\mathbb R$. Then the sets $D_n=\{x_k\mid k\geq n\}$ are discrete and closed, and so the sets $U_n= \{\infty\}\cup\mathbb R\backslash D_n$ form an open cover of $K$ with no finite subcover.

Corollary. $X$ is hemicompact, Fréchet-Urysohn, and not weakly locally compact.

Proof. Every neighborhood of $\infty$ intersects $\mathbb R$ in an unbounded set, so there are no compact neighborhoods of $\infty$, hence $X$ is not weakly locally compact. On the other hand, the sets $K_n:=[-n,n]\cup\{\infty\}$ are an increasing sequence of compacta in $X$, and every compact $K\subset X$ lies in some $K_n$, so $X$ is hemicompact.

Finally, to verify the Fréchet-Urysohn property, since every point in $\mathbb R$ has a countable basis, it suffices to consider $S\subseteq \mathbb R$ with $\infty\in \overline{S}$. In this case, $S$ must have some limit point $x\in \mathbb R$ (otherwise $\{\infty\}\cup \mathbb R\backslash S$ is a neighborhood of $\infty$). But then if $s_n\in S$ is a (distinct) sequence with $s_n\to x\in\mathbb R$, then $s_n$ eventually leaves every closed discrete subset of $\mathbb R$, whereby $s_n\to\infty$ as well.


Now, we could let the matter rest there, but compared to the Arens space, the space $X$ constructed above has rather poor separation properties: it is $T_1$, but just barely, in the sense that it is not even $US$ (convergent sequences do not have unique limits, as seen from the proof above).

Is there a Hausdorff counterexample, i.e., is there a Fréchet-Urysohn hemicompact Hausdorff space that fails to be locally compact*? More generally, if a space is hemicompact and Fréchet-Urysohn, might any separation axioms, anywhere from $US$ through to $T_6$**, guarantee weak local compactness?

(* Note that in the Hausdorff case, weak local compactness and local compactness coincide - the latter term indicating the existence of a basis of compact neighborhoods at each point.)

(** See here for an in depth description of a chain of properties between $T_2$ and $T_1$, with the weakest being the aforementioned $US$.)

  • 1
    $\begingroup$ Thanks for posting this question. I had been trying to generalize topology.pi-base.org/theorems/T000279 for a long time without success. And now the answers here explain why! $\endgroup$
    – PatrickR
    Jan 15 at 7:03

2 Answers 2


The sequential fan with $\omega$ many spines is a counter-example:

It is defined as follows:
Let $X = \omega \times (\omega+1)$ with the product topology, where both factors have the order topology, i.e., the first factor is discrete, the second one is a convergent sequence.
Identify all points $(n, \omega)$ to a point called $\infty$.
The sequential fan with $\omega$ many spines is the quotient space $S = S(\omega) = (\omega \times \omega) \cup \{\infty\}$.

Each singleton except $\{\infty\}$ is closed and open in $S$, hence $S$ is T2, regular and thus even perfectly normal (since it is countable).
Neighborhoods of $\infty$ are of the form $\{\infty\} \cup \bigcup_{n \in \omega} (\{n\} \times [m_n, \omega))$.

Obviously, each $K_n := ([0, n] \times \omega) \cup \{\infty\}$ is compact (since it is the image of a compact set in $X$).

Moreover, by the above it is easy to see:

  1. Each compact subset of $S$ is contained in some $K_n$. Hence $S$ is hemicompact, but not locally compact.
  2. If $\infty \in \overline{B} \setminus B$ for some $B \subset S$, then there is a $n$ such that $(\{n\} \times \omega) \cap B$ is infinite. Hence $S$ is Fréchet-Urysohn.
  • $\begingroup$ Thank you for this, very nice! It seems your idea also generalizes to a lot of examples - more or less any quotient collapsing a closed noncompact set will do (with one or two details). It was a little too long for a comment to explain so I posted a follow-up answer detailing. $\endgroup$
    – M W
    Jan 15 at 4:40
  • $\begingroup$ fyi now in pi-base: topology.pi-base.org/spaces/S000131 $\endgroup$
    – PatrickR
    Jan 25 at 1:33

As a sort of postscript to Ulli's nice answer, we can generalize the ideas from that answer to obtain a large class of counter-examples as follows:

Let $X$ be a topological space, let $S\subset X$ be nonempty, and for $x_1,x_2\in X$ let $x_1\sim_S x_2$ if either $x_1=x_2$ or $x_1,x_2\in S$. Let $Y=X/\sim_S$ be the quotient space, with $\pi\colon X\to Y$ the quotient map.

Proposition. If $X$ is $T_6$, Fréchet-Urysohn, $\sigma$-compact, and locally compact, and $S$ is closed, with $\partial S$ noncompact, then $Y$ is $T_6$, Fréchet-Urysohn, hemicompact, and not locally compact.

Example. If $S\subset \mathbb R^n$ is closed, unbounded, and has unbounded complement, then $\mathbb R^n/\sim_S$ is $T_6$, Fréchet-Urysohn, hemicompact, and not locally compact. For example, the infinite loop space $\mathbb R/\sim_{\mathbb Z}$, or the plane with a collapsed line $\mathbb R^2/\sim_L$, are examples that satisfy these properties.

This provides a rich source of counter-examples that can have many nice properties (connectedness, local connectedness, contractability, etc.), illustrating the difficulties with replacing the first countability condition to obtain local compactness from hemicompactness, even alongside the Fréchet-Urysohn property and good separation.

Proof of Proposition.

The proposition follows immediately from the ensuing three lemmas. Denote by $\infty$ the single element of $\pi(S)$ in $Y$.

Lemma 1. If $X$ is Fréchet-Urysohn, then $Y$ is as well.


[For our purposes we could simplify this a decent amount by assuming $S$ is closed, but it is nice to know this lemma is true regardless, so we prove it in general.]

Suppose $A\subset Y$ with $y\in \overline{A}$. We consider a few cases.

If $y\notin \overline{\{\infty\}}$, then $y$ has a neighborhood $V$ avoiding $\infty$, and then the restriction $\pi|_{\pi^{-1}(V)}$ is a homeomorphism, so by Fréchet-Urysohn on $X$ we have a sequence in $A$ converging to $y$.

If $y\in \overline{\{\infty\}}$ and $\infty\notin \overline{A}$, then $\overline{\pi^{-1}(A)}\subseteq \pi^{-1}(\overline{A})\subseteq X\backslash S$, so $\overline{\pi^{-1}(A)}$ is a saturated closed set. It follows that $\pi(\overline{\pi^{-1}(A)})\supseteq \overline{A}$, so if $y=\pi(x)$ we have $\overline{\pi^{-1}(A)}= \pi^{-1}(\overline{A})\ni x$. By Fréchet-Urysohn there is a sequence $x_n\in \pi^{-1}(A)$ converging to $x$, whereby $\pi(x_n)$ is a sequence in $A$ converging to $y$.

Lastly, if $y\in \overline{\{\infty\}}$ and $\infty\in \overline{A}$, then we must have $\overline{\pi^{-1}(A)} \cap S\neq \emptyset$, since otherwise $\overline{\pi^{-1}(A)}$ is a saturated closed set avoiding $S$, whereby $\pi(\overline{\pi^{-1}(A)})$ is a closed superset of $A$ avoiding $\infty$, contradicting that $\infty\in \overline{A}$.

Therefore we let $s\in \overline{\pi^{-1}(A)} \cap S$. Since $X$ is Fréchet-Urysohn we have a sequence $x_n\to s$ with $x_n\in \pi^{-1}(A)$, so by continuity the sequence $\pi(x_n)$ is a sequence in $A$ converging to $\pi(s)=\infty$. Finally, since $y\in \overline{\{\infty\}}$, $\pi(x_n)\to y$ as well.

Lemma 2. If $X$ is perfectly normal and $S$ is closed, then $Y$ is perfectly normal as well.


Let $C\subseteq Y$ be closed. If $\infty\in C$, then since $\pi^{-1}(C)$ is closed and $X$ is perfectly normal, there is a continuous function $f\colon X\to \mathbb R$ with $f^{-1}(0)=\pi^{-1}(C)\supseteq S$. Then since $f$ is constant on fibers of $\pi$, it factors as $f=g\circ \pi$ for some continuous $g\colon Y\to \mathbb R$ ($g$ is continuous since $\pi$ is a quotient map), and $g^{-1}(0)=C$.

If $\infty \notin C$, then by an alternate characterization of perfect normality there is a continuous function $f\colon X\to \mathbb R$ with $f^{-1}(0)=\pi^{-1}(C)$ and $f^{-1}(1)=\pi^{-1}(\infty)=S$, so $f$ is again constant on fibers and factors as before to give a continuous $g$ with $g^{-1}(0)=C$.

[thanks to PatrickR for pointing out we don't need $S$ closed for the first part of the next lemma.]

Lemma 3. If $X$ is $T_1$ and exhaustible by compact sets (equivalently, $X$ is $\sigma$-compact and weakly locally compact), then $Y$ is hemicompact. If in addition $S$ is closed, and $\partial S$ is not contained in $\overline{K}$ for any compact $K\subset X$, then $Y$ is not weakly locally compact.


Let $\{K_n\}$ be an exhaustion of compacta in $X$, i.e., $K_n\subseteq \operatorname{int}(K_{n+1})$ for all $n$ and $\bigcup_n K_n=X$.

We claim every compact subset $L\subseteq Y$ lies in $\pi(K_n)$ for some $n$. To prove this, suppose otherwise and choose a sequence $y_n\in L\backslash \pi(K_n)$. With no loss of generality each $K_n$ intersects $S$, so $y_n\neq \infty$, and so the sequence $x_n=\pi^{-1}(y_n)\in X\backslash S$ is well-defined and intersects each $K_n$ finitely. Let $U_n=X\backslash \bigcup_{k=n}^\infty \{x_k\}$.

Now each $x\in U_n$ lies in $\operatorname{int}(K_m)$ for some $m$, so $x$ has a neighborhood intersecting at most finitely many members of $X\backslash U_n$, hence by $T_1$, it has a neighborhood contained in $U_n$, whereby each $U_n$ is open. Since $S\subset U_n$, $U_n$ is saturated as well, so $\{\pi(U_n)\mid n\in \mathbb N\}$ is an open cover of $L$ with no finite subcover, contradicting compactness.

Therefore the sets $\pi(K_n)$ witness that $Y$ is hemicompact. To prove the final statement, we observe that if $L$ is a compact neighborhood of $\infty$, then by what precedes we have $L\subseteq \pi(K_n)$ for some $n$, whereby

$$\pi^{-1}(L)\subseteq K_n\cup S.\tag{1}$$

But choosing $s\in \partial S\backslash \overline{K_n}$, we see that since $L$ is a neighborhood of $\infty$, $\pi^{-1}(L)\backslash \overline{K_n}$ is a neighborhood of $s$. Since $s$ is a boundary point of $S$, $\pi^{-1}(L)\backslash \overline{K_n}$ intersects $X\backslash S$ as well, contradicting (1). Thus there is no compact neighborhood of $\infty$, so $Y$ is not weakly locally compact.


Thanks to user558840 in the comments for directing my attention to the following two references:

  1. Theorem 2 of K. Morita, On closed mappings. Proc. Japan Acad. 32(1956), 538–543 appears to mostly generalize the above proposition, minus the conclusion of hemicompactness for the quotient (though that could conceivably be buried in the paper somewhere as well).

  2. This much more recent paper also seems quite relevant, though I haven't yet taken the time to look through it:

Lazar AJ, Somerset DWB. Pure quotients and Morita’s theorem for $k_{\omega }$ -spaces. Canadian Mathematical Bulletin. 2022;65(3):582-597. doi:10.4153/S0008439521000515

On another note, the $T_1$ assumption can probably be removed from Lemma 3 by a method similar to the one used in this answer, though I haven't carefully written that down, and since the proposition assumes $T_6$ anyway, I'll leave the details of that generalization to the interested reader.

  • 1
    $\begingroup$ Very nice results. In Lemma 3, the fact that $S$ is closed is not used anywhere for the first sentence (i.e. $X$ $T_1$ and exhaustible by compacts implies $Y$ hemicompact), right? And for the second sentence, $S$ closed is only used so that $\partial S\subseteq S$, so we can assert that $s\in\pi^{-1}(L)$? $\endgroup$
    – PatrickR
    Jan 16 at 4:54
  • 1
    $\begingroup$ @Ulli and M W: these examples are very useful. I'll make sure to add at least the "sequential fan" and the space $\mathbb R/{\sim}_{\mathbb Z}$ = "bouquet of circles" = "wedge sum of countably many circles" to pi-base, hopefully soon $\endgroup$
    – PatrickR
    Jan 16 at 5:31
  • 1
    $\begingroup$ I think that the Proposition above is a special case of Theorem 2 of K. Morita, On closed mappings. Proc. Japan Acad. 32(1956), 538–543. projecteuclid.org/journals/proceedings-of-the-japan-academy/… More recently it was shown that closed quotients of locally compact Frechet-Urysohn are basically the only source of hemicompact Hausdorff Frechet-Urysohn spaces: cambridge.org/core/journals/canadian-mathematical-bulletin/… $\endgroup$
    – user558840
    Jan 17 at 21:22
  • 1
    $\begingroup$ I think that their hemicompactness is Thm 9.5 of this paper by Steenrod (1967): projecteuclid.org/journals/michigan-mathematical-journal/… $\endgroup$
    – user558840
    Jan 18 at 23:34
  • 1
    $\begingroup$ fyi now in pi-base: topology.pi-base.org/spaces/S000139 $\endgroup$
    – PatrickR
    Jan 31 at 1:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .