# mrowka space or psi space

The construction of Mrowka space ( psi space) is not clear for me. Because of this I could not see why it is first countable, locally compact, and Hausdorff.

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What in particular about the construction is not clear to you? What construction are you following? – Potato Aug 7 '13 at 3:53
I'll add link to the blog about Mrowka space at Dan Ma's Topology Blog. – Martin Sleziak Aug 8 '13 at 8:07

Let us recap the construction, also to fix notation. We have $\omega$ (the natural numbers) and a collection $\mathcal{A}$ of infinite subsets of $\omega$ that is almost disjoint: for every $A,B \in \mathcal{A}, A \neq B$ we have that $A \cap B$ is finite.

We define a space $X = \omega \cup \{x_A: A \in \mathcal{A} \}$ (so we add to $\omega$ a collection of new points, one for each member of the almost disjoint family).

We define a topology on this by defining a neighbourhood base for each point: each $n \in \omega$ has neighbourhood base $\{ \{n \} \}$; it is an isolated point (like in the usual discrete topology on $\omega$). Every $x_A$ (for a fixed $A \in \mathcal{A}$) has a neighbourhood base consisting of all sets of the form $B(A,F) = \{x_A\} \cup (A \setminus F)$, where $F \subset A$ is finite. We can also define the open sets directly as all sets $O$ such that: $\forall_{A \in \mathcal{A}}: x_A \in O \rightarrow A \setminus O \mbox{ finite.}$

Note that all subsets of $\omega$ are still open (as usual) and that every $B(A,F)$ intersects $X$ only in $\omega$, and does not contain other added points. So the set $X \setminus \omega$ (all extra points $x_A$) forms a closed and discrete subset of $X$. This shows that if $\mathcal{A}$ is uncountable, then $X$ is not countably compact. Also, no $\{x_A\}$ is open by itself, because all $A$ are infinite, so all basic open sets of $x_A$ intersect $\omega$ non-trivially. So $\omega \subset X$ is dense (so $X$ is separable).

To see that $X$ is Hausdorff, we need to separate a pair of distinct points $x \neq y$ from $X$. If $x,y \in \omega$ the sets $\{x\}$ and $\{y\}$ will do, of course and if, say $x \in \omega$ and $y = x_A$ for some $A \in \mathcal{A}$, then $\{x\}$ and $B(A,\{x\})$ are open, pairwise disjoint neighbourhoods of $x$ and $y$ resp. As the final case, if $x = x_A$, $y = x_B$, for $A \neq B, A,B \in \mathcal{A}$, we know that $F = A \cap B$ is finite and then $x \in B(A,F)$, $y \in B(B, F)$ and these sets are clearly disjoint (we removed the possible intersection $F$). So here we (finally) use that $\mathcal{A}$ is almost disjoint.

That $X$ is first countable: trivial at isolated points, and there are only countably many finite subsets of $\omega$, so the other local bases as described above, are all countable already.

And $X$ is locally compact, because every set $A \cup \{x_A \}$ ($=B(A,\emptyset)$), and even every set $B(A,F)$, is homeomorphic to a convergent sequence (enumerate $A$ as $\{x_n: n \in \omega \}$, then $x_n \rightarrow x_A$ as $n \rightarrow \infty$: every neighbourhood of the limit contains a tail of the sequence. And convergent sequences are compact spaces. Also, this shows that $X$ is zero-dimensional (as all basic elements are clopen being compact in a Hausdorff space) and thus completely regular (Tychonoff, or $T_{3\frac{1}{2}}$).

All spaces of this form are often called $\Psi$-like spaces (or Mrówka spaces). If we choose $\mathcal{A}$ to be a maximal (by inclusion) almost disjoint family of subsets (which are always uncountable and can be as large as the continuum, then $X$ can be shown to be pseudocompact (all real-valued continuous functions defined on it are bounded), and as $T_4$ pseudocompact spaces are countably compact, we conclude that for maximal almost disjoint familes $\mathcal{A}$ this $X$ is then a completely regular, non-normal Hausdorff space that shows that pseudocompactness does not necessarily imply countable compactness for non-$T_4$ spaces.

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thanks so much. If possible, could you give example for added new – user89330 Aug 8 '13 at 3:33
thanks so much. If possible, could you give example for added new points $x_A$, because ı could not imagine. And also, the singletion $\{x_A\}$ is open if the compement of $\omega$ discrete and closed, as you said.If each singletion $\{x_A\}$ is open, the set complement of $\omega$ must be open. So, the complement of $\omega$ can be clopen?If every B(A,F) intersects X only in ω, how it can be said each singletion $\{x_A\}$ is open? I think there was a wrong here. why not $x_A \in (B(A,F) \cap X)$? – user89330 Aug 8 '13 at 3:50
There is no "example" of the added new points; we just add one point to $\omega$ for each $A \in \mathcal{A}$ that wasn't there before. In a way, we could have the subset $A$ itself be the new point, but that would get a bit confusing, IMHO, so I use indices. – Henno Brandsma Aug 8 '13 at 6:46
No, I edited my post to show that $\omega$ is not closed (it is discrete, of course), but dense in $X$. Of course by definition $B(A,F) \cap X = B(A,F)$ and it contains $x_A$, but $B(A,F) \cap \omega \subset A \subset \omega$ and the intersection is infinite, – Henno Brandsma Aug 8 '13 at 6:51
thanks again.It was very useful for me. – user89330 Aug 12 '13 at 8:00