Chain that doesn't contain maximal element?

Question

I cannot find a proof or example that every chain contains a maximal element (from Friedberg et al). A collection of sets is called a chain if for each pair of sets A and B in the chain, A ⊆ B or B ⊆ A; and maximal element being a member M of C that is contained in no member of C other than M itself.
Wouldn't the maximal element in any chain just be the set X such that all other members of C are a subset of X?

Answer 1

Let me give an example regarding the existence of basis of (possibly not finite dimensional) vector space, since your resource is Linear Algebra, Friedbert et al.

Let $V$ be a space of real-coefficient polynomials. (This is a vector space over $\mathbb{R}$.)

Let $\mathcal{F}$ be the collection of linearly independent subsets of $V$. For example $\{1, x, x^3 + 3x\} \in \mathcal{F}$ and $\{1, x^2, 3^2 + 2\} \not\in\mathcal{F}$.

Consider a chain $\mathcal{C}$ of linearly independent sets, $1 \subset \{1, x\} \subset \{1, x, x^2\} \subset \cdots$. This chian does not have a maximal element in itself. (Assume some element (say the nth element $\{1, x, \cdots, x^{n-1}\}$) is the maximal. This contradicts to $ \{1, x, \cdots, x^{n}\} \in \mathcal{C}$.)

But the union of all of them, $\{1, x, x^2, \cdots, x^n, \cdots \}$ is in $\mathcal{F}$. (You should prove this, i.e. you should prove that this is a linearly independent subset of $V$.)