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Infimum of partitions-set

Mathematics Asked by Zuy on January 21, 2021

Let $Aneq varnothing$ and denote by $operatorname{Part}(A)$ the set of partitions of $A$. For $S,S’ inoperatorname{Part}(A),$ we set
$$
Sleq S’ :!!iff forall C in S: exists C’in S’: Csubseteq C’.
$$

Let $Tsubseteq operatorname{Part}(A)$. Show that $T$ has an infimum with respect to $leq$.


I would actually want to show that
$$
inf T = left{bigcap_{Sin T} C_Sneq varnothing : forall S in T, C_Sin Sright}.
$$

Am I allowed to write this without using the Axiom of Choice?

If not, what other way is there to prove the existence of $inf T$, if any at all?

2 Answers

SKETCH: For each $Sinoperatorname{Part}(A)$ and $ain A$ let $[a]_S$ be the part of $S$ that contains $a$. Observe that $Sle S'$ iff $[a]_Ssubseteq[a]_{S'}$ for each $ain A$. Given $Tsubseteqoperatorname{Part}(A)$, for each $ain A$ let $C_a=bigcap_{Sin T}[a]_S$, and let $C={C_a:ain A}$.

  • Show that $Cinoperatorname{Part}(A)$. It’s clear that $bigcup C=A$, so you just have to show that if $a,bin A$, and $C_ane C_b$, then $C_acap C_b=varnothing$.

Clearly $C_asubseteq[a]_S$ for each $Sin T$, so at this point you can conclude that $Cle S$ for each $Sin T$.

  • Then show that if $L$ is a lower bound for $T$, then $Lle C$. You can do this by showing that if $L$ is a lower bound for $T$, then it must be true that $[a]_Lsubseteq C_a$ for each $ain A$.

None of this requires the axiom of choice.

Correct answer by Brian M. Scott on January 21, 2021

Let $X=left{bigcap_{Sin T} C_Sneq varnothing : C_Sin Sright}.$ Observe that $Xinoperatorname{Part}(A)$.

First of all, we show that $Xle S$, for all $Sin T$. Let $Lin X$, then $L=bigcap_{Sin T} C_S$, with $C_Sin S$. Obviously, $Lsubseteq C_S$ for all $Sin T$, that means that $Xle S$ for all $Sin T$. So $X$ is a lower bound.

Now we show that $X$ is the greatest lower bound, i.e. for all $Yinoperatorname{Part}(A)$ such that $Yle S$ for all $Sin T$, $Yle X$. Suppose $Y$ is such an element, and let $Min Y$. Now, because $Yle S$, for all $Sin T$, we have $Msubseteq C_S$, for some $C_Sin S$, for all $Sin T$. Obviously $Msubseteqbigcap_{Sin T}C_S$, and $bigcap_{Sin T}C_Sin X$. So $Yle X$, and $X$ is the infimum of $Tsubseteqoperatorname{Part}(A)$.

Answered by Antonio Ficarra on January 21, 2021

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