Application of Zorn's Lemma to linearly independent subsets of a vector space

Question

Let's say I have a vector space $V$, and a linearly independent subset $L$ of $V$ along with a set $S$ that spans $V$ such that $$L \subseteq S.$$ Let $\Gamma = \{ K \subseteq V \ | \ L \subseteq K \text{ and } K \text{ is linearly independent}\}$ and put a partial order on $\Gamma$ by set-theoretic inclusion. I'm trying to conclude that $\Gamma$ contains a maximal element, i.e. that there is a maximal linearly independent subset of $V$ containing $L$.

My idea was this, choose any chain $L \subseteq K_1 \subseteq K_2 \dots$ of elements in $\Gamma$. If this chain doesn't have an upper bound, then at some point we'd have $V$ to be a proper subset of $K_i$ for some $i$ a contradiction, so every chain has an upper bound that being $V$.

Is this correct? I don't think so, the reason being that if $V$ has cardinality of $\mathbb{R}$ and $L$ is finite, then it could be possible that we have a chain $L \subseteq K_1 \subseteq K_2 \dots$ with $K_i$ adding $i$ points to $L$ and $K_i$ linearly independent, and then this chain never seems to have an upper bound and it can continue without us ever reaching the contradiction above.

Answer 1

What you have to prove is that there exists an element of $\Gamma$ that is an upper bound for the chain $K_i$.

Rather than an argument by contradiction, it is easier to do a direct construction of the upper bound: the union $\bigcup_i K_i$ is an element of $\Gamma$; and it is an upper bound for the chain. Both of these are relatively easy to verify.

Let me also add that by writing the chain in the form $K_1 \subseteq K_2 \subseteq\cdots$, you are fooling yourself a bit: you do indeed want to consider chains which are linearly ordered by inclusion; however, you do not want to restrict yourself to countable chains. The correct form of the chain is as a subset $\cal K \subset \Gamma$ such that the $\subseteq$ relation on $\Gamma$ restricts to a total order on $\cal K$: for all $K,K' \in \cal K$, either $K \subseteq K'$ or $K' \subseteq K$. Then the upper bound for the chain is simply $\bigcup_{K \in \cal K} K$ (and you still need to verify that this is an element of $\Gamma$).

Answer 2

The main problem with your argument is that $V$ doesn't qualify as an upper bound of a chain in $\Gamma$, because no vector space is, as a set, linearly independent. The upper bound of the chain must be a member of $\Gamma$.

Less important, but not a detail to be neglected, is that a chain in $\Gamma$ need not be countable.

A chain $\mathcal{C}$ in $\Gamma$ is a subset of $\Gamma$ such that, if $K,L\in\mathcal{C}$, then either $K\subseteq L$ or $L\subseteq K$.

The bulk of the proof lies in the following lemma.

Lemma. A subset $K$ of $V$ is linearly independent if and only if every finite subset of $K$ is linearly independent.

The proof is essentially applying the definition of linearly independent set. Do it.

Now, if $\mathcal{C}$ is a chain in $\Gamma$, set $$ L=\bigcup\mathcal{C}=\{v\in V: v\in K\text{, for some }K\in\mathcal{C}\} $$ You want to prove that $L\in\Gamma$ and then it will clearly be an upper bound of $\mathcal{C}$, because by definition we have $K\subseteq L$, for every $K\in\mathcal{C}$. By the lemma, you just need to take $v_1,\dots,v_n\in L$; then $v_i\in K_i$, for some $K_i\in\mathcal{C}$, $i=1,2,\dots,n$. Since $\mathcal{C}$ is a chain, the set $\{K_1,\dots,K_n\}$ has a largest element, let it be $K$.

Now you should be able to finish.

Answer 3

You just have to show that any chain $\mathcal{K} \subseteq \Gamma$ has an upperbound (in $\Gamma$) and then Zorn applies.

Basic fact about chains $\mathcal{K}$ in partially ordered sets: if $K_1,\ldots,K_n$ are finitely many members of $\mathcal{K}$ then for some $j \in \{1,2,\ldots,n\}$ we have $K_i \le K_j$ for all $i \in \{1,2\ldots,n\}$. This can be shown by induction on $n$. (i.e. In a linearly ordered set every finite subset has a maximum (and a minimum too).

Now, given your chain in $\Gamma$, let $K=\bigcup \mathcal{K}$. It's clear that if this set is in $\Gamma$ (i.e. linearly independent), it will be an upperbound for $\mathcal{K}$ (even the smallest one, though we don't even need that), as $L \subseteq K = \bigcup \mathcal{K}$ is obvious when $L \in \mathcal{K}$.

Now suppose $K$ were not linearly independent, then there would be a finite subset $\{v_1,\ldots,v_n\}$ of $K$ and as many scalars $a_i, i=1,\ldots, n$ such that not all $a_i$ are $0$ and $\sum_{i=1}^n a_i v_i = 0$.

As all $v_i$ are in $K=\bigcup \mathcal{K}$, by the definition of the union we can find for each $1 \le i \le n$ some $K_i \in \mathcal{K}$ such that $v_i \in K_i$. By the basic fact quoted above, for some fixed $j \in \{1,\ldots,n\}$ we have that all $K_i \subseteq K_j$, but then the linear relation for the $v_i$ already held in $K_j$, which contradicts that $K_j \in \mathcal{K}\subseteq \Gamma$, because it's not linearly independent! This contradiction shows that $K$ is linearly independent, and so is the required upperbound in $\Gamma$.