Every vector space has a basis.

 

https://en.wikipedia.org/wiki/Basis_%28linear_algebra%29

 

 

Thm. Every vector space has a basis. (even in the case of infinite dimension)

 

 

proof) Let $V$ be a vector space other than $\{0\}$. Pick a vector $x\ne 0$ from $V$. Consider the set

 

$X=\{S\subseteq V|x\in S,S$ is linearly independent$\}.$

 

Then $(X,\subseteq )$ is an ordered set. Obviously $X$ is nonempty because $\{x\}\in X$. By Hausdorff maximal principle (see below), $X$ has a maximal chain $m$. Let $M:=\bigcup _{s\in m}s$, an upper bound of $m$.

 

Two things to check:

 

1) Is $M$ linearly independent?

 

Because the set $m$ is a chain of sets of linearly independent vectors, $M$ also is linearly independent. For a more accurate proof, pick arbitrary $n$ vectors $v_1,v_2,\dots ,v_n$ from $M$. Then there exists a set $M_1$ in $m$ which have all $v_i$'s. All sets in $m$ are linearly independent, so is $M_1$. All subsets of $M_1$ are also linearly independent, so is a set only with $v_i$'s.

 

2) Does $M$ span $V$?

 

If not, there exists a vector $v\in V\setminus \text{span }M$. Then $\{v\}\cup M$ also linearly independent. Since $x\in \{v\}\cup M$ and $M⊊\{v\}\cup M$, $M$ is not an upper bound of $m$. Contradiction.

 

---

 

 

$\text{[Z]}$ Zorn's lemma: If every chain of a nonempty ordered set $X$ has an upper bound, then $X$ has a maximal element.

 

 

Caution: There may be more than one maximal element. For example, in the first set above, both $b$ and $c$ are maximal.

 

$\text{[H]}$ Hausdorff maximal principle: Every ordered set has a maximal chain.

 

Caution: There may be more than one maximal chain. Again in the first set above, both $\{a,b\}$ and $\{a,c\}$ are maximal chains. The second set has 6 maximal chains. Find them!

 

---

 

 

Zorn's lemma and Hausdorff maximal principle are logically equivalent.

 

 

proof) $\text{[Z]}\to \text{[H]}$

 

Let $X$ be a nonempty ordered set, and $\mathcal{C}$ be a set of all chains of $X$. Pick any chain ${\mathcal{C}}_{\mathcal{1}}$ from $\mathcal{C}$, i.e., a chain of chains of $X$. The set ${\mathcal{C}}_{\mathcal{1}}$ is an ordered set with order relation $\subseteq$. Then the set

 

$$\bigcup _{c\in {\mathcal{C}}_{\mathcal{1}}}c$$

 

is the upper bound of the chain ${\mathcal{C}}_{\mathcal{1}}$, so, by Zorn's lemma, $\mathcal{C}$ has a maximal element which is a maximal chain of $X$.

 

Caution: Union of all $c$'s above may not be maximal. Consider $X=\{1,2,3\}$ and a chain of chains ${\mathcal{C}}_{\mathcal{1}}=\{\{1\},\{1,2\}\}$. Union of all elements of ${\mathcal{C}}_{\mathcal{1}}$ is not a maximal chain of $X$, but is an upper bound of ${\mathcal{C}}_{\mathcal{1}}$ of which the existence is premise of Zorn's lemma.

 

$\text{[H]}\to \text{[Z]}$

 

Let $X$ be a nonempty ordered set. Suppose that every chain of $X$ has an upper bound. By Hausdorff, $X$ has a maximal chain $C$, so $C$ also has an upper bound $m$. If there exists $x\in X$ such that $x>m$, then $\{x\}\cup C$ also a chain. But this contradicts with the maximality of $C$. Therefore $m$ is a maximal element of $X$.

 

References.

http://www.math.snu.ac.kr/~kye/lecture_V/V_set/set_10b.mp4

https://youtu.be/oGC_BU5Erkk?si=0JV0tJG9kb4DtkRK

 

Note.

If you find something wrong, please ignore them. XD

 

Special Thanks to 이보.