# Proof

### "$\Leftarrow$"

• Assume Zorn's Lemma is true.
• Let $X$ be a set of disjoint and non-empty sets $X=\{V_i\mid i\in I,i\neq j\Longleftrightarrow V_i\cap V_j=\emptyset\}$ for some index set $i\in I$. We want to construct a set $Y$ containing exactly one element from each of these sets, i.e. we want to derive the axiom of choice.
• We define $\mathcal F$ as the set of all non-empty subsets of the respective disjoint sets $V_i$: $$\mathcal F:=\{T_i\subseteq V_i\mid i\in I,\; T_i\neq\emptyset\}.$$
• Obviously, $(\mathcal F,\preceq)$ is a poset if and only if we set $$T\preceq T^\prime\Longleftrightarrow T^\prime\subseteq T,$$ i.e. we consider $T^\prime$ greater or equal $T$ if $T^\prime$ is a (non-empty) subset of $T.$
• Note that $T,T^\prime\in\mathcal F$ are only comparable by "$\preceq$" if they are subsets of the same original set $V_i$ for some $i\in I$ and one is contained in the other. This is because all $V_i$ are disjoint.
• By assumption, the Zorn's lemma is correct, i.e. if every chain $C\subseteq \mathcal F$ that has an upper bound, then $\mathcal F$ contains at least one maximal element.
• We have therefore first to check if every chain in $\mathcal F$ has an upper bound. * Let $C\subseteq\mathcal F$ be a chain, i.e. for all $T,T^\prime\in C$ we have $T^\prime\preceq T$ or $T\preceq T^\prime.$ * By construction, since all pairs $T$ and $T^\prime$ in the chain $C$ are only comparable, if they are subsets of the same original set $V_i$ for some $i\in I$ and one is contained in the other, we can identify exactly one $V_i\in\mathcal F$ for which $C$ is a chain. Therefore, can we write $C_i$ for each chain $C\in\mathcal F.$ * In other words, in $F$, there exist only chains $C_i$ that can be built from the subsets of the respective $V_i,$ which are contained in each other. * Since all subsets are non-empty, by construction, for each chain $C_i$ we can build the set intersection $B_i:=\bigcap_{T\in C_i} T.$ * $B_i$ is both, non-empty and, by construction, an element of each $C_i$ for all $i\in I.$ * Thus, for all $i\in I,$ $B_i$ is an upper bound of $C_i$ since $B_i\subseteq T$ for all $T\in C_i.$ * Thus, every chain in $\mathcal F$ has an upper bound.
• It follows now from Zorn's lemma that $\mathcal F$ has at least one maximal element.
• In particular, it has at least one maximal element for each $i\in I,$ namely a $B_i,$ since only the subsets inside a single $V_i$ are comparable.
• Now, for all $i\in I$, the found example of a maximal element $B_i$ must be a singleton, i.e. containing exactly one element of $V_i,$ otherwise it would either be empty (which it is not by construction) or not maximal with respect to the partial order $\preceq.$
• Set $Y:=\bigcup_{i\in I} B_i.$ We have constructed a set $Y$ that contains exactly one element from the original disjoint sets $V_i,$ $i\in I.$
• It follows the axiom of choice.

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### References

#### Bibliography

1. Piotrowski, Andreas: Own Research, 2014