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Let $X$ be a $T_1$ space and let $F[X]$ be $\{x\subset X:\text{is finite}\}$ with Pixley-Roy topology.

If $X$ is not discrete, how to proof $F[X]$ is not a Baire space?

Thanks ahead:)


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.


I don't know in the theorem 2.2 why each $Z \cap F_n[X]$ is closed, nowhere dense subspace of $Z$?

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I've added the definition of Pixley-Roy topology to your questions, since this topology is (probably) not very well-known. –  Martin Sleziak Jun 21 '12 at 12:32
    
BTW you could check Theorem 2.2 in the paper I linked there. topology.auburn.edu/tp/reprints/v03/tp03111s.pdf –  Martin Sleziak Jun 21 '12 at 12:33
    
O, thank you:) Why each $Z \cap F_n[X]$ is closed, nowhere dense subspace of $Z$? –  Paul Jun 21 '12 at 12:42

1 Answer 1

up vote 2 down vote accepted

(explaining the theorem 2.2 in the paper linked by Martin Sleziak)

Let $X$ not be discrete, and $p$ be a limit point.

Put $F_n[X]=\lbrace A\subseteq X\vert \lvert A\rvert\leq n\rbrace$, $Z=[\lbrace p\rbrace,X]$

If $F[Z]$ were Baire, so would $Z$. But $Z=\bigcup_n (Z\cap F_n[X])$, and each of $Z\cap F_n[Z]$ is nowhere dense and closed:

  • it is closed because its complement $\bigcup_{\lvert A\rvert >n} (Z\cap[A,X])$ is open.
  • it is nowhere dense, because any nonempty basic open subset of $Z$ is of the form $[A,U]$ for $p\in A\subseteq U$, so in particular $U$ is infinite (as a neighbourhood of $p$), so it has an $n+1$-element subset $B$. Then $[A\cup B,U]$ is a subset of $[A,U]$ disjoint from $F_n[X]\cap Z$
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Yeer, thaks tomasz; I got it:) –  Paul Jun 21 '12 at 13:22
    
I made a slight mistake here: not every open subset of $Z$ is of the form, but every BASIC subset is (so in particular any open subset contains one like this). –  tomasz Jun 21 '12 at 13:31

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