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Assume that $2 \leq |J| \leq \aleph_{0} $. Let $\mathbb{P}=\operatorname{Fn}(I,J)$

$\mathbb{P}=\operatorname{Fn}(I,J)$ is $\sigma$-centered iff $|I| \leq \mathcal{c}$ where $\mathcal{c}=2^{\aleph_{0}}$

a suggestion to show this please.

Added: $\operatorname{Fn}(I,J)$ is the set of finite partial functions from $I$ to $J$, with the order $p\preceq q$ iff $p\supseteq q$. A poset $\langle\Bbb P,\preceq\rangle$ is centred if for each finite $F\subseteq\Bbb P$ there is a $q\in\Bbb P$ such that $q\preceq p$ for each $p\in F$. A poset is $\sigma$-centred if it is the union of countably many centred subposets.

Brian M. Scott
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Arthur
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  • Could you define your symbols? What is the difference between $Fn(I,J)$ and $J^I$? What is $c$, do you mean ${\frak c}=2^{\aleph_0}$, the cardinality of the reals? – Mario Carneiro Feb 10 '15 at 20:12
  • Nothing appears to relate $I$ and $J$ in this description. You could have $\Bbb P=Fn(I,J)$ for any $I$ and $|J|\le\aleph_0$, and in particular both $|I|>\frak c$ and $|I|\le\frak c$ are possibilities. Also, I don't see any posets in the question, even though it is mentioned in the title. – Mario Carneiro Feb 10 '15 at 20:15
  • hello Mariano Carneiro I want to show that $\mathbb{P}$ is $\sigma$ centered iff $ |I| \leq 2^{\aleph_{0}}$ I thought that thinking can be solved $J^{I}$. – Arthur Feb 10 '15 at 20:21
  • I still don't know what $\Bbb P$ is or $Fn(I,J)$ or $J^I$ until you tell me. Why does the definition of $\Bbb P$ being $\sigma$-centered not involve $\Bbb P$ but uses $I$ directly? Does the definition make an assumption about the form of $\Bbb P$? You really need to define your notation before it is possible to give a proof of anything. – Mario Carneiro Feb 10 '15 at 20:26
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    Do you know how to prove that the product of continuum many separable space (e.g. your space $J^I$) is separable? – bof Feb 10 '15 at 20:38
  • hello bof, the idea of the proof is similar to what you mean – Arthur Feb 10 '15 at 20:46

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You can use the Hewitt-Marczewski-Pondiczery theorem, which says that if $\kappa$ is an infinite cardinal, $\mathscr{X}$ is a family of at most $2^\kappa$ spaces, each of density at most $\kappa$, then $\prod\mathscr{X}$ has density at most $\kappa$. Here we give $J$ the discrete topology and take $\kappa=\omega$ to conclude that ${^JI}$, viewed as the product of $|J|$ copies of $I$, is a separable space. Let $D$ be a countable dense subset of ${^JI}$.

For each $f\in\Bbb P$ the set $B(f)=\{x\in{^JI}:f\subseteq x\}$ is a basic open set in ${^JI}$, so it contains some member of $D$. For each $x\in D$ let

$$\Bbb P_x=\{f\in\Bbb P:x\in B(f)\}=\{f\in\Bbb P:f\subseteq x\}\;;$$

then $\Bbb P=\bigcup_{x\in D}\Bbb P_x$, and it suffices to show that each $\Bbb P_x$ is centred, which is straightforward.

There’s a sketch of a proof of the $\kappa=\omega$ case of the Hewitt-Marczewski-Pondiczery theorem here; Hewitt’s original paper can be found here; and there’s a proof here at Ask a Topologist by Henno Brandsma.

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Brian M. Scott
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