My Math Forum A Full Proof of Zorn's Lemma
 User Name Remember Me? Password

 Applied Math Applied Math Forum

 December 22nd, 2011, 11:02 PM #1 Newbie   Joined: Jan 2011 Posts: 20 Thanks: 0 A Full Proof of Zorn's Lemma Statement A (Zorn's Lemma) -> Let S be a non-empty inductively ordered set. Then S has a maximal element -------------------- Let A be a set of non-empty totally ordered subsets of S. Then A is not empty since any subset of S with 1 element belongs to, if $X,Y\inA$, define ${X}\leq{Y}$ to be $X= Y$. Then A is partially ordered and is strictly inductively ordered. Let $T={ {A_{{i}}}$ be a totally ordered subset of A. Let $Z=\bigcup {X_{i}}$, then Z is totally ordered. Let ${x},{y}\in{Z}$, then ${x}\in{X_{i}}$ and ${y}\in{Y_{j}}$ for some $i,j\in{I}$. Since T is totally ordered, say ${X_{i}}\subset{X_{j}}$. Then $x,y\in{X_{j}}$ and since ${X_{j}}$ is totally ordered this implies $x\leq{y}$ or $y\leq{x}$, thus Z is totally ordered and is the supremum of T in A. To conclude the proof, I prove something else: ------------ Statement B -> "Let A be a non empty strictly inductively ordered set. Then A has a maximal element" Suppose A doesn't have a maximum element. For each $x\in{A}, \exists {y_{x}} \in A$ such that $x<{y_{x}}$ Let $f:A\rightarrow{A}$ such that $f(x)={y_{x}}$ , $\forall x\in A$ This contradicts a well known theorem, "Let A be a non empty partially ordered and strictly inductively ordered set. Let $f:A\rightarrow{A}$ be an increasing function. Then there exists an element $g\in{A}$ such that $f(g)=g$ And the proof for Statement B is complete, A has a maximum element g, this means g is a maximum totally ordered subset of S. Now, let k be an upper bound for g in S. Then k is the maximum element of ${S_{i}}$, for if $x\in S$ and $k\leq x$, then $g \cup {x}$ is totally ordered. Thus $x\in g$ and $x\leq{k}$, hence x = k (maximum element) --- Let me know your thoughts on whether this proof is correct.
 December 25th, 2011, 09:28 AM #2 Senior Member   Joined: Jun 2011 Posts: 298 Thanks: 0 Re: A Full Proof of Zorn's Lemma You introduced too many sets into your work. $S$ consists of a chain such that $X\preceq Y$ or $Y\preceq X$. Take you pick. You can create a set of ordinal numbers, say $\alpha$ , and $S$ is isomorphic to the set of ordinals. $\alpha$ has a maximal, and so does $S$.

 Tags full, lemma, proof, zorn

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post alfred_oh Computer Science 1 May 14th, 2013 12:15 PM xoofee Real Analysis 2 July 24th, 2012 01:30 AM krausebj0 Number Theory 56 February 22nd, 2012 11:50 PM Iran_rezayi Number Theory 1 December 6th, 2009 01:36 PM tmn50 Math Events 10 May 8th, 2009 03:59 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top