Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for a (Hausdorff or better) space $X$ and a subset $A$ of $X$ that is relatively countably compact (every sequence from $A$ has an accumulation point in $X$) such that the closure of $A$ is not countably compact. It is known that in many "nice" spaces such examples do not exist (a classical case being normed spaces in their weak topology).

Edit: for $T_4$ spaces this cannot happen, as the closure of a relatively countably compact subset is pseudocompact (Suppose A is relatively countably compact. If f from cl(A) to R is unbounded then f|A is unbounded as well, as A is dense in cl(A). So we can find {x_n: n in N} in A such that |f(x)| >= n. Let p in cl(A) be an accumulation point of this set, by A being relatively countably compact. Then continuity at f implies that |f(x_n)| <= |f(p)| + 1, for all but finitely many n. This contradicts the choice of the x_n, contradiction.) So cl(A) is pseudocompact, and hence in a normal space, countably compact. This explains the properties of the example given below.

share|improve this question
Thanks for the fun question, Henno; I enjoyed it very much. –  Joel David Hamkins Jul 6 '10 at 23:26

3 Answers 3

up vote 6 down vote accepted

Let $X$ be the space $(\omega+1)\times(\omega_1+1)-\{(\omega,\omega_1)\}$, putting the transfinite order topology on each coordinate and the product topology on the whole space.

 (0,omega_1)  (1,omega_1)  (2,omega_1)   --->     O
      :            :            :                 
      :            :            :                 :
      :            :            :                 :
      :            :            :                 :
 (0,alpha)    (1,alpha)    (2,alpha)     --->   (omega,alpha)
      :            :            :                 :
      :            :            :                 :
   (0,1)        (1,1)        (2,1)       --->   (omega,1)
   (0,0)        (1,0)        (2,0)       --->   (omega,0)

Thus, the space consists of $\omega_1+1$ many copies of a (horizontally) convergent sequence, stacked on top of each other, but with the limit of the final sequence omitted (at O in the diagram). Horizontally, the $\alpha$-th row has $(n,\alpha)$ converging to $(\omega,\alpha)$ as $n\to\omega$. Vertically, the $n$-th column has a transfinite sequence $(n,\alpha)$ converging to $(n,\omega_1)$ as $\alpha\to\omega_1$. This is a Hausdorff space, and comparatively nice in several ways.

Let $A=(\omega+1)\times\omega_1\subset X$ be the lower portion of this diagram, without the top row. Any countable sequence from $A$ is bounded below a countable ordinal and has accumulation points in $X$, and in fact in $A$ itself: either the sequence contains infinitely many points from one coordinate, in which case it accumulates on the limit supremum in that coordinate, or it contains points from infinitely many coordinates, which will accumulate on a point of the form $(\omega,\alpha)$ at the right-most column in $X$. In particular, $A$ already contains all its limit points of sequences in $A$. Nevertheless, the closure of $A$ in $X$ is all of $X$, and this includes the top row of points of the form $(n,\omega_1)$. But this sequence has no accumulation points in $X$, as desired.

share|improve this answer
Ah, the Tychonoff plank. I tried to define a similar example on $\omega \times \omega$, but failed... Is it essential that such a space be non-normal, as this is? –  Henno Brandsma Jul 6 '10 at 4:29
Adde: we cannot have a countably compact $A$ with non-countably compact closure for a non-normal space, like here, because cl(A) would then be pseudocompact (having a dense countably compact subset) and hence (by normality) countably compact. But, can we still have a relatively countably compact $A$ in a $T_4$ space $X$ with non countably compact closure? –  Henno Brandsma Jul 6 '10 at 14:30
You may want to replace `non-normal' by normal in the first line. –  KP Hart Jul 8 '10 at 9:42

To fix Pietro's idea: take a Maximal Almost Disjoint family on $\omega$, i.e., a family $\mathcal{A}$ of infinite subsets such that any two distinct elements have a finite intersection and such that it is maximal with respect to this property. Topologize $\omega\cup\mathcal{A}$ by making each $n\in\omega$ isolated and have $\bigl\lbrace\lbrace A\rbrace\cup(A\setminus F): F$ finite$\rbrace$ as a local base at $A\in\mathcal{A}$. By maximality $\omega$ is relatively countably compact; its closure is the whole space, which is not countably compact. (This is generally know as the/a Moore Mrowka space, a $\psi$-space, or $\Psi$-space ...)

share|improve this answer
Ah, the other classical Tychonov non-normal space. Figures... Thanks. –  Henno Brandsma Jul 8 '10 at 20:41

I just invented this "uglification of $\mathbb{N}$" for you. Shouldn't it work, at least, the name will be justified.

Let $A:=\mathbb{N}$ and let $B$ be the quotient set of $\mathcal{P}(\mathbb{N})$ modulo the equivalence relation $$S\sim S'\Leftrightarrow |S\Delta S'|< \aleph_0.$$ Say that a subset $\Omega$ of the disjoint union $X:=A\sqcup B$ is open if and only if $$\forall \omega\in \Omega\cap B\\ \exists\\, S\in \omega\\ : \\ S\subset \Omega .$$ You can check that that actually defines a Hausdorff topology on $X$. Moreover, $X$ is not countably compact: take e.g. a countable partition $\{S_k\}_k$ of $\mathbb{N}$ into infinite sets, and consider the sequence $\{\omega_k\}_k$ corresponding to their classes in $B$. Finally, any countable subset $S$ of $A$ has an accumulation point, namely its class.

share|improve this answer
Is it really Hausdorff? If $S\subset T\subset\mathbb{N}$ but $S$ is not equivalent to $T$, then the equivalence classes $[S]$ and $[T]$ are distinct elements of $B$, but every open neighborhood of $[S]$ has all but finitely many elements of $S$, co-finitely many of which will be in any given open neighborhood of $[T]$. So it doesn't seem you can separate these two points in $X$. Have I made a mistake? –  Joel David Hamkins Jul 6 '10 at 2:24
you're right. I think it can be fixed taking a subspace though... I'll see! –  Pietro Majer Jul 6 '10 at 6:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.