Showing $\mathbb{R}^2$ has a Hamel basis using Zorn’s lemma

axiom-of-choicehamel-basisset-theory

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 $B$?

Best Answer

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.

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