Krdytkyu

To get some intuition for Zorn's lemma I want to use it explicitly in the proof of the following theorem in the case when $X = mathbb{R}^2$:

Every vector space $X neq { 0}$ has a Hamel basis.

So let $M$ be the set of all linearly independent subsets of $mathbb{R}^2$. So as far as I can see $M$ contains two types of elements:

1. All non-zero vectors in $mathbb{R}^2$, as each non-zero vector is linearly independent as a set consisting of just itself.


2. All groups of vectors such as ${(1,0),(0,1)},{(2,0),(0,1)},{(2,0),(0,2)},$ etc.. as each of these elements is a linearly independent set. Or we could write this as $M$ contains contains all groups of vectors ${(a,b),(c,d)}$ where $(a,b)$ and $(c,d)$ are linearly independent vectors.

Now supposedly the elements in $M$ are partially ordered by set inclusion. Then we can apply Zorn's lemma (Kreyszig, p. 211) which states that every if every chain $C subset M$ has an upper bound, then $M$ has at least one maximal element $B$. In particular, the upper bound in our case is the union of all elements of $X$ which are elements of $C$.

But I don't see how the elements of $M$ are ordered by set inclusion? Take the single vector elements from $1.$ above. None of these elements contain each other! Any single vector element of $M$ such as, say, ${(5,6)}$, only contains itself. It's the same for the elements of $M$ with two components. E.g. ${(2,0),(0,1)}$ only contains itself.

So what am I missing, how can Zorn's lemma be used to show that $mathbb{R}^2$ has a Hamel basis? Have I got the wrong elements in $M$? What would the chains $C$ look like and what is the maximal element $B$?

Since there are at most three elements in a chain, $varnothing,{x},{x,y}$, every chain is finite. Every finite linear order has a maximum. If a chain has a maximum, then it has an upper bound: that maximum.

Of course, this is confusing and one should never make use of Zorn's lemma or some other abstract choice principles when finite objects are in play. With finite objects we can just do things by hand, for the most part.

More to the point, most of the uses of Zorn's lemma rely, to some extent, on the understanding that we appeal to some property that is compact. If it holds for finite chains, it holds for all chains. For example in the general proof of existence of Hamel bases, if the union of an increasing chain of sets is not linearly independent, this is already revealed by finitely many of them in the chain, so we can take the largest of those finitely many.

So while it is perfectly valid to apply to Zorn's lemma to find a basis for $Bbb R^2$ over $Bbb R$, it is also somewhat odd, since in general the use of Zorn's lemma relies on the knowledge that finite objects behave fine with respect to the property we wish to find a maximal set for.

Well, the set ${ v,w}$ contains both ${v}$ and ${w}$. In particular your maximal chains look like:
$${ v,w} supset { v} quad textrm{ and } quad { v,w} supset { w}$$
And so your maximal elements and basis are:${ v,w}$ where by definition the 2 vectors are independent

Yes, Zorn's lemma can be used for that. Consider the set $M$. It is partially orderd by inclusion. Every subset of $mathcal{P}(S)$, where $S$ is some set, is partially orderd by inclusion. Then, take a chain $C$ in $M$. Now, there aren't lots of choices for chains here. You have three possibilites:

1. 
   $C=bigl{{v}bigr}$ where $vneq0$;


2. 
   $C=bigl{{v,w}bigr}$, where $v$ and $w$ are two vectors such that none of them is a multiple of the other one;


3. 
   $C=bigl{{v},{v,w}bigr}$, where $v$ and $w$ are two vectors such that none of them is a multiple of the other one.

And you're done. There are no other possibilites.

For Zorn's Lemma, you only need a partial order and not a total one. In particular, this means that we don't need any pair of elements of $M$ to be comparable to each other to apply Zorn's Lemma.

It is true that some elements are comparable under $subseteq$ though. For example $A = {(1,0)} in M$ and $B = {(1,0),(0,1)}$ and we have $A subseteq B$. In fact, ${A,B}$ is a chain in $M$ and $B$ is its maximal element.

Finally, it's worth noticing that you really aren't using Zorn's lemma in this case because chains in $M$ are always finite so you can prove this result without choice. However, essentially the exact same proof (removing specific references to $mathbb{R}^2$) works for general $X$ and there you will need Zorn's lemma because in general chains will be very large.

The claim is that if $X$ be a vector space over a field $F$ and if $L$ is a linearly independent subset of $X$, there exists a basis of $X$ containing $L.$ The proof begins as follows:

Let $mathscr A$ be the collection of all linearly independent $sets$ of vectors containing $L$ (so $mathscr A$ is not empty,) ordered by inclusion. So using your example, we would have, for $L={(1,0),(0,1)}$ we have $ Lsubset {(1,0),(0,1),(2,0)}subset {(1,0),(0,1),(2,0),(0,2)}.$

A chain $C$ in $mathscr A$, is a collection of sets of linearly independent vectors, so $C$ is a $subset$ of $mathscr A$. And $C$ has the property that if $C_1$ and $C_2$ are members of $C$, then either $C_1subseteq C_2$ or $C_2subseteq C_1$. To apply Zorn's Lemma, you need only to show that the evident upper bound for $C$, namely $bigcup C$ is a set of linearly independent vectors.