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Is this space $F[R]$ with the Pixley-Roy topology paracompact?

In general, when the space $F[X]$ is paracompact for general topological space?


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

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The paracompactness of Pixley-Roy Hyperspaces seems to have been solved by Przymusiński (Normality and paracompactness of Pixley-Roy hyperspaces, Fund. Math. 113, pp.201-219). Therein he gives the following theorem:

Theorem: The following are equivalent for a Pixley-Roy hyperspace $\mathcal{F}[X]$:

  1. $\mathcal{F}[X]$ is paracompact;
  2. $\mathcal{F}[X]^n$ is paracompact for all $n \in \mathbb{N}$;
  3. $\mathcal{F}[X]$ is collectionwise Hausdorff;
  4. There is a neighbourhood assignment $F \mapsto V_F$ for the finite nonempty subsets of $X$ such that given finite, nonempty $F ,H \subseteq X$ the inclusions $F \subseteq V_H$ and $H \subseteq V_F$ imply $F \cap H \neq \emptyset$.

It follows from this that $\mathcal{F}[\mathbb{R}]$ is not paracompact. Suppose that $V \mapsto V_F$ is any neighbourhood assignment for the finite nonempty $F \subseteq \mathbb{R}$. Note that given any $x \in \mathbb{R}$ we may assume without loss of generality that $V_{\{x\}}$ is of the form $( x - \varepsilon_x , x + \varepsilon_x )$ for some $\varepsilon_x > 0$. From now on, I will write $V_x$ for $V_{\{x\}}$. The non-paracompactness of $\mathcal{F}[\mathbb{R}]$ is then an immediate consequence of the following claim:

Claim: There are distinct $x, y \in \mathbb{R}$ such that $y \in V_{x}$ and $x \in V_{y}$.

Proof. We construct sequences $\{ x_i \}_{i \in \mathbb{N}}$ and $\{ \delta_i \}_{i \in \mathbb{N}}$ so that

  • $x_0 = 0$;
  • $\delta_i = \varepsilon_{x_i}$;
  • $x_{i+1} = x_i + (-1)^i \frac{\delta_i}{2}$

Note that for $j > i$ we clearly have that $x_j \in V_{x_i}$. If for any $j > i$ we also have $x_i \in V_{x_j}$ we are done. So assume that this never happens.

Given any $i \in \mathbb{N}$, we have $x_i \notin V_{x_{i+1}}$. It thus follows that $$| x_{i+1} - x_i | > \varepsilon_{x_{i+1}} = \delta_{i+1} = 2 | x_{i+2} - x_{i+1} |,$$ and so the sequence $\{ x_i \}_{i \in \mathbb{N}}$ converges to some $y \in \mathbb{R}$.

It can be shown that $y \in V_{x_i}$ for all $i \in \mathbb{N}$. (If $i$ is even, then for all $j > i+1$ we have $x_i < x_j < x_{i+1}$ and therefore $x_i \leq y \leq x_{i+1}$; similarly for $i$ odd.) As $x_i \rightarrow y$, then $| y - x_i | < \varepsilon_y$ for all large enough $i$, and therefore $x_i \in V_y$ for all such $i$. $\Box$

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