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Ur Ya'ar
Ur Ya'ar
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The theorem should state that the separative quotient of $B_{0}/{\sim}$ is isomorphic to $B_{\kappa}^+$.

Proof. Recall that the separative quotient of a poset $P$ is the unique (up to isomorphism) separative $Q$ such that there is an order preserving $h:P\to Q$ such that $x$ is compatible with $y$ iff $h(x)$ is compatible with $h(y)$ (see Jech pg. 205). Since $B_{\kappa}$ is separative as a Boolean algebra, we want to provide such $h:B/{\sim}\to B_{\kappa}^{+}$.

Let $b\in B_{0}^{+}$. For every $\alpha<\kappa$ let $b_{\alpha}=\inf \{ d\in B_{\alpha}\mid d\geq b \} $, and let $\bar{b}=\sup\{b_{\alpha}\mid\alpha<\kappa\}$. Note that $\{b_{\alpha}\mid\alpha<\kappa\}$ is an ascending sequence, so in fact for every $\beta$, $\bar{b}=\sup\{b_{\alpha}\mid\beta\leq\alpha<\kappa\}$, and this is an element of $B_{\beta}$, so all-in-all $\bar{b}\in B_{\kappa}$. Now if $b'\sim b$ then for all large enough $\alpha$, $b_{\alpha}=b_{\alpha}'$ so $\bar{b}=\bar{b'}$. So the function $h([b])=\bar{b}$ is well defined, and since in particular $\bar{b}\geq b>0$, it is into $B_{\kappa}^{+}$. It is order preserving since if $[b]\leq[c]$ then for all large enough $\alpha$, we have $b_\alpha \leq c_\alpha $ so also $\bar{b}\leq\bar{c}$.

Let $[b],[c]\in B_{0}/{\sim}$. We want to show they are compatible iff $\bar{b}$ and $\bar{c}$ are compatible.

  • If $[b],[c]$ are compatible, $[d]\leq[b],[c]$, then for all large enough $\alpha$ $d_{\alpha}\leq b_{\alpha},c_{\alpha}$ so $\bar{d}\leq\bar{b},\bar{c}$.
  • If $[b],[c]$ are incompatible, it means that there is an unbounded $I\subseteq\kappa$ such that for $\alpha\in I$, $b_{\alpha}$ and $c_{\alpha}$ are incompatible. But this also implies that for every $\alpha,\beta\in I$ $b_{\alpha}$ and $c_{\beta}$ are incompatible (if $\alpha<\beta$ and there is e.g. $d\leq b_{\alpha},c_{\beta}$ then since $b_{\alpha}\leq c_{\beta}$ we get $d\leq b_{\beta},c_{\beta}$), so, as we are in a complete Boolean algebra, $$ \overline{b}\cdot\bar{c}=\sum_{\alpha\in I}b_{\alpha}\cdot\sum_{\beta\in I}c_{\beta}=\sum_{\alpha,\beta}b_{\alpha}\cdot c_{\beta}=0 $$ i.e. $\bar{b},\overline{c}$ are incompatible. $\square$

Regarding the specific case, my advisor pointed out to me that if we are considering tails of a full support product, say of length $\kappa$, then the quotient poset will be $\kappa$-closed.

Proof sketch. Let $P=\prod_{\alpha<\kappa}Q_{\alpha}$ be a full support product of posets, and for $\alpha<\kappa$ let $P_{\alpha}=\prod_{\alpha\leq\xi<\kappa}Q_{\xi}$. Let $\left\langle p_{i}\mid i<\kappa\right\rangle $ be sequence such that $i<j$ implies $[p_{i}]>[p_{j}]$. Then to construct $q$ such that $[q]$ is a lower bound, we diagonalize - let $q(\xi)=p_0(\xi)$ until the coordinate witnessing $[p_0]>[p_1]$, then $q(\xi)=p_1(\xi)$ until $p_2$ is smaller, and so on. I'll spare you the indexing monstrosity. $\square$

The theorem should state that the separative quotient of $B_{0}/{\sim}$ is isomorphic to $B_{\kappa}^+$.

Proof. Recall that the separative quotient of a poset $P$ is the unique (up to isomorphism) separative $Q$ such that there is an order preserving $h:P\to Q$ such that $x$ is compatible with $y$ iff $h(x)$ is compatible with $h(y)$ (see Jech pg. 205). Since $B_{\kappa}$ is separative as a Boolean algebra, we want to provide such $h:B/{\sim}\to B_{\kappa}^{+}$.

Let $b\in B_{0}^{+}$. For every $\alpha<\kappa$ let $b_{\alpha}=\inf \{ d\in B_{\alpha}\mid d\geq b \} $, and let $\bar{b}=\sup\{b_{\alpha}\mid\alpha<\kappa\}$. Note that $\{b_{\alpha}\mid\alpha<\kappa\}$ is an ascending sequence, so in fact for every $\beta$, $\bar{b}=\sup\{b_{\alpha}\mid\beta\leq\alpha<\kappa\}$, and this is an element of $B_{\beta}$, so all-in-all $\bar{b}\in B_{\kappa}$. Now if $b'\sim b$ then for all large enough $\alpha$, $b_{\alpha}=b_{\alpha}'$ so $\bar{b}=\bar{b'}$. So the function $h([b])=\bar{b}$ is well defined, and since in particular $\bar{b}\geq b>0$, it is into $B_{\kappa}^{+}$. It is order preserving since if $[b]\leq[c]$ then for all large enough $\alpha$, we have $b_\alpha \leq c_\alpha $ so also $\bar{b}\leq\bar{c}$.

Let $[b],[c]\in B_{0}/{\sim}$. We want to show they are compatible iff $\bar{b}$ and $\bar{c}$ are compatible.

  • If $[b],[c]$ are compatible, $[d]\leq[b],[c]$, then for all large enough $\alpha$ $d_{\alpha}\leq b_{\alpha},c_{\alpha}$ so $\bar{d}\leq\bar{b},\bar{c}$.
  • If $[b],[c]$ are incompatible, it means that there is an unbounded $I\subseteq\kappa$ such that for $\alpha\in I$, $b_{\alpha}$ and $c_{\alpha}$ are incompatible. But this also implies that for every $\alpha,\beta\in I$ $b_{\alpha}$ and $c_{\beta}$ are incompatible (if $\alpha<\beta$ and there is e.g. $d\leq b_{\alpha},c_{\beta}$ then since $b_{\alpha}\leq c_{\beta}$ we get $d\leq b_{\beta},c_{\beta}$), so, as we are in a complete Boolean algebra, $$ \overline{b}\cdot\bar{c}=\sum_{\alpha\in I}b_{\alpha}\cdot\sum_{\beta\in I}c_{\beta}=\sum_{\alpha,\beta}b_{\alpha}\cdot c_{\beta}=0 $$ i.e. $\bar{b},\overline{c}$ are incompatible.

The theorem should state that the separative quotient of $B_{0}/{\sim}$ is isomorphic to $B_{\kappa}^+$.

Proof. Recall that the separative quotient of a poset $P$ is the unique (up to isomorphism) separative $Q$ such that there is an order preserving $h:P\to Q$ such that $x$ is compatible with $y$ iff $h(x)$ is compatible with $h(y)$ (see Jech pg. 205). Since $B_{\kappa}$ is separative as a Boolean algebra, we want to provide such $h:B/{\sim}\to B_{\kappa}^{+}$.

Let $b\in B_{0}^{+}$. For every $\alpha<\kappa$ let $b_{\alpha}=\inf \{ d\in B_{\alpha}\mid d\geq b \} $, and let $\bar{b}=\sup\{b_{\alpha}\mid\alpha<\kappa\}$. Note that $\{b_{\alpha}\mid\alpha<\kappa\}$ is an ascending sequence, so in fact for every $\beta$, $\bar{b}=\sup\{b_{\alpha}\mid\beta\leq\alpha<\kappa\}$, and this is an element of $B_{\beta}$, so all-in-all $\bar{b}\in B_{\kappa}$. Now if $b'\sim b$ then for all large enough $\alpha$, $b_{\alpha}=b_{\alpha}'$ so $\bar{b}=\bar{b'}$. So the function $h([b])=\bar{b}$ is well defined, and since in particular $\bar{b}\geq b>0$, it is into $B_{\kappa}^{+}$. It is order preserving since if $[b]\leq[c]$ then for all large enough $\alpha$, we have $b_\alpha \leq c_\alpha $ so also $\bar{b}\leq\bar{c}$.

Let $[b],[c]\in B_{0}/{\sim}$. We want to show they are compatible iff $\bar{b}$ and $\bar{c}$ are compatible.

  • If $[b],[c]$ are compatible, $[d]\leq[b],[c]$, then for all large enough $\alpha$ $d_{\alpha}\leq b_{\alpha},c_{\alpha}$ so $\bar{d}\leq\bar{b},\bar{c}$.
  • If $[b],[c]$ are incompatible, it means that there is an unbounded $I\subseteq\kappa$ such that for $\alpha\in I$, $b_{\alpha}$ and $c_{\alpha}$ are incompatible. But this also implies that for every $\alpha,\beta\in I$ $b_{\alpha}$ and $c_{\beta}$ are incompatible (if $\alpha<\beta$ and there is e.g. $d\leq b_{\alpha},c_{\beta}$ then since $b_{\alpha}\leq c_{\beta}$ we get $d\leq b_{\beta},c_{\beta}$), so, as we are in a complete Boolean algebra, $$ \overline{b}\cdot\bar{c}=\sum_{\alpha\in I}b_{\alpha}\cdot\sum_{\beta\in I}c_{\beta}=\sum_{\alpha,\beta}b_{\alpha}\cdot c_{\beta}=0 $$ i.e. $\bar{b},\overline{c}$ are incompatible. $\square$

Regarding the specific case, my advisor pointed out to me that if we are considering tails of a full support product, say of length $\kappa$, then the quotient poset will be $\kappa$-closed.

Proof sketch. Let $P=\prod_{\alpha<\kappa}Q_{\alpha}$ be a full support product of posets, and for $\alpha<\kappa$ let $P_{\alpha}=\prod_{\alpha\leq\xi<\kappa}Q_{\xi}$. Let $\left\langle p_{i}\mid i<\kappa\right\rangle $ be sequence such that $i<j$ implies $[p_{i}]>[p_{j}]$. Then to construct $q$ such that $[q]$ is a lower bound, we diagonalize - let $q(\xi)=p_0(\xi)$ until the coordinate witnessing $[p_0]>[p_1]$, then $q(\xi)=p_1(\xi)$ until $p_2$ is smaller, and so on. I'll spare you the indexing monstrosity. $\square$

Source Link
Ur Ya'ar
Ur Ya'ar
  • 329
  • 1
  • 11

The theorem should state that the separative quotient of $B_{0}/{\sim}$ is isomorphic to $B_{\kappa}^+$.

Proof. Recall that the separative quotient of a poset $P$ is the unique (up to isomorphism) separative $Q$ such that there is an order preserving $h:P\to Q$ such that $x$ is compatible with $y$ iff $h(x)$ is compatible with $h(y)$ (see Jech pg. 205). Since $B_{\kappa}$ is separative as a Boolean algebra, we want to provide such $h:B/{\sim}\to B_{\kappa}^{+}$.

Let $b\in B_{0}^{+}$. For every $\alpha<\kappa$ let $b_{\alpha}=\inf \{ d\in B_{\alpha}\mid d\geq b \} $, and let $\bar{b}=\sup\{b_{\alpha}\mid\alpha<\kappa\}$. Note that $\{b_{\alpha}\mid\alpha<\kappa\}$ is an ascending sequence, so in fact for every $\beta$, $\bar{b}=\sup\{b_{\alpha}\mid\beta\leq\alpha<\kappa\}$, and this is an element of $B_{\beta}$, so all-in-all $\bar{b}\in B_{\kappa}$. Now if $b'\sim b$ then for all large enough $\alpha$, $b_{\alpha}=b_{\alpha}'$ so $\bar{b}=\bar{b'}$. So the function $h([b])=\bar{b}$ is well defined, and since in particular $\bar{b}\geq b>0$, it is into $B_{\kappa}^{+}$. It is order preserving since if $[b]\leq[c]$ then for all large enough $\alpha$, we have $b_\alpha \leq c_\alpha $ so also $\bar{b}\leq\bar{c}$.

Let $[b],[c]\in B_{0}/{\sim}$. We want to show they are compatible iff $\bar{b}$ and $\bar{c}$ are compatible.

  • If $[b],[c]$ are compatible, $[d]\leq[b],[c]$, then for all large enough $\alpha$ $d_{\alpha}\leq b_{\alpha},c_{\alpha}$ so $\bar{d}\leq\bar{b},\bar{c}$.
  • If $[b],[c]$ are incompatible, it means that there is an unbounded $I\subseteq\kappa$ such that for $\alpha\in I$, $b_{\alpha}$ and $c_{\alpha}$ are incompatible. But this also implies that for every $\alpha,\beta\in I$ $b_{\alpha}$ and $c_{\beta}$ are incompatible (if $\alpha<\beta$ and there is e.g. $d\leq b_{\alpha},c_{\beta}$ then since $b_{\alpha}\leq c_{\beta}$ we get $d\leq b_{\beta},c_{\beta}$), so, as we are in a complete Boolean algebra, $$ \overline{b}\cdot\bar{c}=\sum_{\alpha\in I}b_{\alpha}\cdot\sum_{\beta\in I}c_{\beta}=\sum_{\alpha,\beta}b_{\alpha}\cdot c_{\beta}=0 $$ i.e. $\bar{b},\overline{c}$ are incompatible.