# Try to have an intuitional impress of a special topological space

Let $R$ be Real line and let $F[R]$ be $\{x\subset R:\text{is finite}\}$ with Pixley-Roy topology.

Definition of Pixley-Roy topology is this: Basic neighborhoods of $F\in F[R]$ are the sets $$[F,V]=\{H\in F[R]; F\subseteq H\subseteq V\}$$ for open sets $V\supseteq F$, see e.g. here.

I'm not very clear to the topology of this space. I'm try to have an intuitional impress of the space. Could someone draw a picture for me such that help me to familar with this special toplogical space?

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Consider the very rough sketch below:

                p        s       q  t          x            r   u
---------(-|--------|-------|--|----)-----|------(-----|---|---)------
U                                W


Let $V=U\cup W$, the union of the two open intervals shown in the sketch, and let $F=\{p,q,r\}$. Then $[F,V]$ is the collection of all finite subsets $H$ of $\Bbb R$ such that $F\subseteq H$ and $H\subseteq V$. For example, let $H=\{p,q,r,s,t,u\}$; then $F\subseteq H\subseteq V$, so $H\in[F,V]$. On the other hand, $K=\{p,q,s,t,u\}$ is not in $[F,V]$, because $F\nsubseteq K$ (since $r\notin K$), and $G=\{p,q,s,t,u,x\}\notin[F,V]$ because $G\nsubseteq V$ (since $x\notin V$).

Added: It’s not easy to give a good global picture of $X=\mathscr{F}[\Bbb R]$, but perhaps these observations will help. For $n\in\Bbb Z^+$ let $X_n=\{F\in\mathscr{F}[\Bbb R]:|F|=n\}$.

1. Each $X_n$ is a discrete subspace of $X$. Suppose that $F\in X_n$, and let $V$ be an open set in $\Bbb R$. If $F\nsubseteq V$, then of course $[F,V]=\varnothing$, so assume that $F\subseteq V$. If $G\in[F,V]\cap X_n$, then $F\subseteq G\subseteq V$; but $|F|=|G|=n$, so $F\subseteq G$ iff $F=G$, and therefore $[F,V]=\{F\}$.

2. For each $n\in\Bbb Z^+$ let $Y_n=\bigcup_{k\le n}X_k$, the set of points of $X$ that contain at most $n$ points of $\Bbb R$; then $Y_n$ is closed in $X$. For suppose that $F\in X\setminus Y_n$; then $|F|>n$, and $[F,\Bbb R]\cap Y_n\varnothing$, since every member of $[F,\Bbb R]$ contains $F$ and therefore has more than $n$ points of $\Bbb R$.

3. For each $n\in\Bbb Z^+$, $X_n$ is a dense set of isolated points of $Y_n$. That the points of $X_n$ are isolated in $Y_n$ follows from (1) and (2). To see that they are dense in $Y_n$, let $F\in Y_n$ be arbitrary, and let $V$ be any open subset of $\Bbb R$ containing $F$. Clearly $|F|\le n$. If $|F|=n$, $F\in[F,V]\cap X_n$. If $|F|<n$, let $G$ be a set of $n-|F|$ points of $V\setminus F$; clearly $F\cup G\in[F,V]\cap X_n$.

4. An easy extension of (3) is that the Cantor-Bendixson rank of $Y_n$ is $n$, and that in fact $Y_n^{(k)}=Y_{n-k}$ for $k\le n$, where $Y_0=\varnothing$.

5. Let $[F,V]$ and $[G,W]$ be basic open sets in $X$. Then $$[F,V]\cap[G,W]=[F\cup G,V\cap W]\;,$$ which is empty iff $F\cup G\nsubseteq V\cap W$. $[F,V]\cup[G,W]$ doesn’t have a really simple description.

6. It’s not too hard to use (5) and the second countability of $\Bbb R$ to show that $X$ is ccc: every collection of pairwise disjoint non-empty subsets of $X$ is countable.

7. $X$ is metacompact. To see this, let $\mathscr{U}$ be an open cover of $X$. Without loss of generality assume that the members of $\mathscr{U}$ are basic open sets. For each $F\in X$ there is a $U_F=[G_F,V_F]\in\mathscr{U}$ such that $F\in U_F$. Let $R_F=[F,V_F]$; $G_F\subseteq F$, so $[F,V_F]\subseteq U_F$, and $\mathscr{R}=\{R_F:F\in X\}$ is an open refinement of $\mathscr{U}$. Let $G\in X$ be arbitrary; if $G\in R_F\in\mathscr{R}$, then $F\subseteq G$. $G$ has only finitely many subsets, so $G$ belongs to at most finitely many members of $\mathscr{R}$. Thus, $\mathscr{R}$ is a point-finite open refinement of $\mathscr{U}$, and $X$ is metacompact. (In fact $X$ is hereditarily metacompact, by a similar argument.)

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Thanks Brain for the local struct picture. Could you show me more on the global struct of the space? – Paul Jul 14 '12 at 3:54