Infimum of partitions-set Let $A\neq \varnothing$ and denote by $\operatorname{Part}(A)$ the set of partitions of $A$. For $S,S' \in\operatorname{Part}(A),$ we set
$$
 S\leq S' :\!\!\iff \forall C \in S: \exists C'\in S': C\subseteq C'.
$$
Let $T\subseteq \operatorname{Part}(A)$. Show that $T$ has an infimum with respect to $\leq$.

I would actually want to show that
$$
\inf T = \left\{\bigcap_{S\in T} C_S\neq \varnothing : \forall S \in T, C_S\in S\right\}.
$$
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?
 A: Let $X=\left\{\bigcap_{S\in T} C_S\neq \varnothing : C_S\in S\right\}.$ Observe that $X\in\operatorname{Part}(A)$.
First of all, we show that $X\le S$, for all $S\in T$. Let $L\in X$, then $L=\bigcap_{S\in T} C_S$, with $C_S\in S$. Obviously, $L\subseteq C_S$ for all $S\in T$, that means that $X\le S$ for all $S\in T$. So $X$ is a lower bound.
Now we show that $X$ is the greatest lower bound, i.e. for all $Y\in\operatorname{Part}(A)$ such that $Y\le S$ for all $S\in T$, $Y\le X$. Suppose $Y$ is such an element, and let $M\in Y$. Now, because $Y\le S$, for all $S\in T$, we have $M\subseteq C_S$, for some $C_S\in S$, for all $S\in T$. Obviously $M\subseteq\bigcap_{S\in T}C_S$, and $\bigcap_{S\in T}C_S\in X$. So $Y\le X$, and $X$ is the infimum of $T\subseteq\operatorname{Part}(A)$.
A: SKETCH: For each $S\in\operatorname{Part}(A)$ and $a\in A$ let $[a]_S$ be the part of $S$ that contains $a$. Observe that $S\le S'$ iff $[a]_S\subseteq[a]_{S'}$ for each $a\in A$. Given $T\subseteq\operatorname{Part}(A)$, for each $a\in A$ let $C_a=\bigcap_{S\in T}[a]_S$, and let $C=\{C_a:a\in A\}$.

*

*Show that $C\in\operatorname{Part}(A)$. It’s clear that $\bigcup C=A$, so you just have to show that if $a,b\in A$, and $C_a\ne C_b$, then $C_a\cap C_b=\varnothing$.

Clearly $C_a\subseteq[a]_S$ for each $S\in T$, so at this point you can conclude that $C\le S$ for each $S\in T$.

*

*Then show that if $L$ is a lower bound for $T$, then $L\le C$. You can do this by showing that if $L$ is a lower bound for $T$, then it must be true that $[a]_L\subseteq C_a$ for each $a\in A$.

None of this requires the axiom of choice.
