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A space $X$ is said almost metrizable if for every $x\in X$ there exists a compact set $F \subset X$ with a countable neighbourhood base in $X$ such that $x\in F$.

Is every almost metrizable space $q$-space? How about $k$-spaces?

A space $X$ is $q$-space if for every $x\in X$ there exists a sequence $\{U_n : n\in\omega\}$ of open neighbourhoods of $x$ in $X$ such that For every sequence $\xi=\{ x_n : n\in \omega\}$ of points in $X$ such that $x_n\in U_n$ for each $n$, there exists a limit point in $X$.

A Hausdorff space $X$ is a $k$-space if and only if for each $A\subset X$, the set $A$ is open in $X$ provided that the intersection of $A$ with any compact subspace $Z$ of the space $X$ is open in $Z$.

TXC
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  • Could you also add the definition for $k$-space? – Stefan Hamcke Jun 27 '13 at 17:47
  • @StefanH. A Hausdorff space $X$ is a $k$-space if and only if for each $A\subset X$, the set $A$ is open in $X$ provided that the intersection of $A$ with any compact subspace $Z$ of the space $X$ is open in $Z$. – TXC Jun 27 '13 at 18:22
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    "almost metrizable" spaces are more commonly known as spaces of point-countable type, already since at least 1971. See http://msp.org/pjm/1971/36-1/pjm-v36-n1-p15-s.pdf e.g. – Henno Brandsma Jun 27 '13 at 20:39

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Your almost metrizable spaces are called spaces of point-countable type (that's how I knew them); they were introduced by Arhangel'skij, see the first two references in this paper, where it is mentioned that they are $k$-spaces, as proved in the second reference.

I also think they are $q$-spaces: just use the local base $U_n$ for the compact set $K$ that contains $x$; make it decreasing WLOG. If we pick $x_n \in U_n$, I think we must get a cluster point on $K$ by compactness. BTW this paper in corollary 0.2 claims that this is proved in a Quintuple Quotient Quest by E. Michael (to which I have no access).

Added from comments To make the answer more self-contained, I'll add the proof as filled in by Alex Ravsky in the comments. My idea from above is indeed correct: suppose no point of $K$ is a cluster point of $\{x_n: n \in \omega\}$, then every point $x \in K$ has an open neighbourhood $O_x$ that only contains at most finitely many $x_n$. The open cover $\{O_x: x \in K\}$ has a finite subcover with union $O$. Then $O$ is a neighbourhood of $K$ that only contains finitely many $x_n$ as well, so from some index $N$ onwards, no $x_n$ with $n \ge N$ will be in $O$. But now we have a contradiction, as for some $k > N$ we will have that $U_k \subset O$ (as we have a decreasing neighbourhood base for $K$ in the $U_n$), and then $x_k \in U_k \subset O$, which cannot be.

The Quintuple Quotient Quest also indirectly shows that a space of pointwise countable type (as an open image of a paracompact $M$-space) is a $k$-space (as a quotient image of a paracompact $M$-space). But as said, this was already known to Arhangel'skij.

Henno Brandsma
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    Why compactness of $K$ implies that ${x_n:n\in \omega}$ has a cluster point in $X$? You can download A Quintuple Quotient Quest, here. – TXC Jul 01 '13 at 06:11
  • I'm not sure it works as I stated (hence the "I think", I'm not quite sure). The proof for point-countable type implies $q$-space should be easy, as an online reference just said it was immediate from the definition. The download link doesn't work for me. – Henno Brandsma Jul 01 '13 at 13:02
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    Assume that the sequence ${x_n}$ has no cluster point in $K$. Then each point $x\in K$ has an open neighborhood $O_x$ containing only finitely many members of the sequence ${x_n}$. Since the set $K$ is compact, the cover ${O_x:x\in K}$ of the set $K$ has a finite subcover $\mathcal O$. Then the set $O=\bigcup\mathcal O$ is an open neighborhood of the set $K$ containing only finitely many members of the sequence ${x_n}$. – Alex Ravsky Jul 07 '13 at 04:15
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    Choose a number $N$ such that $x_n\not\in O$ for each $n>N$. Since the family ${U_n}$ is a decreasing base at the set $K$ in $X$, there exists a number $n>N$ such that $U_n\subset O$. Then $x_n\in U_n\subset O$, a contradiction. – Alex Ravsky Jul 07 '13 at 04:15
  • @HennoBrandsma and AlexRavsky: I appreciate your help. – TXC Jul 12 '13 at 12:01
  • @AlexRavsky: The second paragraph states that "make it decreasing WLOG", how is this possible? – M.A. Sep 17 '13 at 12:08
  • @GeorgeL. If ${V_n}$ is any base then ${U_n}$ is a decreasing base, where $U_n=\bigcap_{i=1}^n V_i$ for each $n$. – Alex Ravsky Sep 17 '13 at 13:32
  • @AlexRavsky: Perhaps my question was unclear; by definition of almost metrizable space, for an open subset $W$ such that $K\subset W$ where $K$ is compact, there exists $n\in\omega$ such that $K\subset V_n\subset W$. now if we consider ${V_n:n\in\omega}$ decreasing as you say ($U_n=\bigcap_{i=1}^n V_i$) how can we find an element in ${U_n:n\in\omega}$ such that $K\subset U_n\subset W$? – M.A. Sep 17 '13 at 14:02
  • @GeorgeL. We can find $V_n$ such that $K\subset V_n\subset W$ because we assumed that ${V_n}$ is a neigborhood base of the set $K$. Then we also have $K\subset U_n=\bigcap_{i=1}^n V_i\subset V_n\subset W$. – Alex Ravsky Sep 17 '13 at 14:12