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I'm posting the problem 2.14 and 2.15 of the book "Set theory" of J.L. Bell. These problem are proposed after the forcing relation chapter and I'm new in this kind of stuff, so I have some little problems. I think these exercise are pretty easy since I have some hint, but my solution are not so precise I think, so I hope someone could help me. The notation used is the one of the book.

Let $k$ be an infinite cardinal, $\lambda$ any cardinal, and let $\lambda ^k$ and $\hat{\lambda} ^\hat{k}$ denote respectively the set of function from $k$ to $\lambda$ and the set of function from $\hat{k}$ to $\hat{\lambda}$ in $V^{(B)}$.

$B$ is said to be $(k,\lambda)-distributive$ if for any double sequence $\{ b_{\alpha , \beta} : (\alpha , \beta) \in k \times \lambda \}\subseteq B$ we have:

$$ \bigwedge_{\alpha < k} \bigvee_{\beta < \lambda}b_{\alpha , \beta} =\bigvee_{f \in \lambda^k}\bigwedge_{\alpha < k} b_{\alpha , f(\alpha)} $$

Show the following are equivalent:

i)$B$ is $(k,\lambda)-distributive$

ii) $V^{(B)}\models \hat{\lambda} ^\hat{k} = \hat{(\lambda^k)}$

Hint: use fact that $[[ h \in \hat{\lambda} ^\hat{k} ]]= \bigwedge_{\alpha < k} \bigvee_{\beta < \lambda}b_{\alpha , \beta} $ and $[[ h \in \hat{(\lambda^k)} ]]= \bigvee_{f \in \lambda^k}\bigwedge_{\alpha < k} b_{\alpha , f(\alpha)} $ where $b_{\alpha , \beta}=[[h(\hat{\alpha})=\hat{\beta}]]$

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1 Answer 1

$(i)\Rightarrow(ii)$: We have that, for arbitrary $h\,\varepsilon\,V^{(B)}$, $A\,=\,\|h\,\varepsilon\,\hat{\lambda}^{\hat{\kappa}}\|\,=\,\bigvee_{f\,\varepsilon\,\lambda^{\kappa}}\bigwedge_{\alpha\,<\,\kappa}b_{\alpha f(\alpha)}$, and $B\,=\,\|h\,\varepsilon\,\hat{(\lambda^{\kappa})}\|\,=\,\bigwedge_{\alpha\,<\,\kappa}\bigvee_{\beta<\lambda}b_{\alpha\,\beta}$, where $b_{\alpha\beta}\,=\,\|h(\hat{\alpha})\,=\,\hat{\beta}\|$. By $(i)$, $A\,=B$, and so, since $h$ was arbitrary, $V^{(B)}\models\,\hat{\lambda}^{\hat{\kappa}}\,=\,\hat{(\lambda^{\kappa})}$

$(ii)\Rightarrow(i)$: Define $f\,\varepsilon\,V^{(B)}$ by $\operatorname{dom}(f)\,=\,\operatorname{dom}(\hat{\kappa})\times \operatorname{dom}(\hat{\lambda})$ and $f(\langle\hat{\alpha},\hat{\beta}\rangle)\,=\,b_{\alpha \beta}$, for all $\langle\alpha,\beta\rangle\,\varepsilon\,\kappa \times \lambda$. Then, $\|f(\hat{\alpha})\,=\,\hat{\beta}\|\,=\,\|\langle\hat{\alpha},\hat{\beta}\rangle\,\varepsilon\,f\|\,=\,b_{\alpha \beta}.$ So, by $(ii)$, $\|f\,\varepsilon\,\hat{\lambda}^{\hat{\kappa}}\|\,=\,\|f\,\varepsilon\,\hat{(\lambda^{\kappa})}\|$, i.e., $\bigwedge_{\alpha < \kappa}\bigvee_{\beta < \lambda}b_{\alpha \beta}\,=\,\bigvee_{g\,\varepsilon\,\lambda^{\kappa}}\bigwedge_{\alpha < \kappa}b_{\alpha g(\alpha)}$, i.e. $A\,=\,B$, as desired.

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1  
You might find this useful: it is more standard (and easier) to use "\in"($\in$) rather than "\varepsilon" in tex markup. –  rschwieb Mar 4 '13 at 14:08
    
Thanks. That is useful. –  Ben Eva Mar 5 '13 at 2:49

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