**Proof**: *(related to "Zorn's Lemma")*

We will prove the Zorn’s lemma by assuming the opposite, leading to a contradiction. While doing so, we will use the axiom of choice.

- Let $(V,\preceq )$ be a poset.
- Assume, the Zorn’s lemma is false. This means that although every chain $S\subseteq X$ has an upper bound, $V$ has no maximal element.
- Take $S_0:=\{a_0\}$ as an example of a chain in $V$ with $a_0\in V.$
- By assumption, $V$ contains at least one upper bound $u$ of $S_0$, i.e. $u\in V$ and $a_0\preceq u.$
- By assumption, none of the existing upper bounds $u$ can be maximal, i.e. there is at least one $x\in V$ with $x\succ u.$
- By the axiom of choice, we can choose $x$ from the set of all existing elements $x\in V$ with $x\succ u$ and set $a_1:=x.$
- By construction, $a_1\succ a_0$ and we can construct a new chain $S_1:=\{a_0,a_1\}.$

- By analogy, we can construct a new chain $S_3:=\{a_0,a_1,a_2\}$ with $a_0\prec a_1\prec a_2.$
- This process can be repeated “endlessly”, and holds even for an infinite chain $S$, and the axiom of choice ensures that we can extend $S$ by those elements of $X$ which are greater than any upper bound we have found for $S$ in $X$.
- This contradicts our the assumption that
*every*chain $S$ has an upper bound in $X$ but still $X$ has no maximal elements. - Therefore, the assumption is false and $X$ has to have at least one maximal element.

q.e.d

| | | | created: 2019-02-02 15:58:18 | modified: 2019-02-02 16:07:39 | by: *bookofproofs* | references: [979], [1038], [8055]

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[8055] **Hoffmann, D.**: “Forcing, Eine EinfÃ¼hrung in die Mathematik der UnabhÃ¤ngigkeitsbeweise”, Hoffmann, D., 2018

[979] **Reinhardt F., Soeder H.**: “dtv-Atlas zur Mathematik”, Deutsche Taschenbuch Verlag, 1994, 10

[1038] **Wille, D; Holz., M **: “Repetitorium der Linearen Algebra”, Binomi Verlag, 1994

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