That seems quite opaque: It's a way of computing a quantity rather than telling what exactly it is or even motivating it. It also leaves completely open the question of why such a function exists and is well-defined. The properties you give are sufficient if you're trying to put a matrix in upper-triangular form, but what about other computations? It also gives no justification for one of the most important properties of the determinant, that $\det(ab) = \det a \det b$.
I think the best way to define the determinant is to introduce the wedge product $\Lambda^* V$ of a finite-dimensional space $V$. Given that, any map $f:V \to V$ induces a map $\bar{f}:\Lambda^n V \to \Lambda^n V$, where $n = \dim V$. But $\Lambda^n V$ is a $1$-dimensional space, so $\bar{f}$ is just multiplication by a scalar (independent of a choice of basis); that scalar is by definition exactly $\det f$. Then, for example, we get the condition that $\det f\not = 0$ iff $f$ is an isomorphism for free: For a basis $v_1, \dots, v_n$ of $V$, we have $\det f\not = 0$ iff $f(v_1\wedge \cdots \wedge v_n) = f(v_1) \wedge \cdots \wedge f(v_n) \not = 0$; that is, iff the $f(v_i)$ are linearly independent. Furthermore, since $h = fg$ has $\bar{h} = \bar{f}\bar{g}$, we have $\det(fg) = \det f \det g$. The other properties follow similarly. It requires a bit more sophistication than is usually assumed in a linear algebra class, but it's the first construction of $\det$ I've seen that's motivated and transparently explains what's otherwise a list of arbitrary properties.
The authors state that the empty set spans the zero subspsace $\{ 0 \}$ by convention.
However, this really depends on your definition of subspace spanned by a set. The definition I use is the following:
the subspace spanned by a set $S \subset V$ is defined to be the intersection of all subspaces of $V$ that contain $S$. That is, if $\langle S \rangle$ denotes the subspace spanned by $S$, then
$$
\langle S \rangle := \bigcap_{S \subset W \leq V} W,
$$
where $W \leq V$ indicates that $W$ is a subspace of $V$. So, if $S$ is the empty set, then the zero subspace $\{ 0 \}$ contains the empty set, and every vector space contains the zero subspace, so $\langle \emptyset \rangle = \{ 0 \}$.
For the second question, there appears to be a typo. The sentence should read:
By theorem $(5.1)$ any set of $n+1$ vectors in $V$ is linearly dependent, and since set consisting of a single nonzero vector in linearly independent, it follows that, for some integer, $m \geq 1$, $V$ contains linearly independent vectors $b_1,\dots,b_m$ such that any set of $m+1$ vectors in $V$ is linearly dependent.
Perhaps that should clear the confusion. To elaborate on why this corrected statement is true, proceed by contradiction:
suppose it is false that
for some integer, $m \geq 1$, $V$ contains linearly independent vectors $b_1,\dots,b_m$ such that any set of $m+1$ vectors in $V$ is linearly dependent.
What would this mean? This means that for each $m \geq 1$, if $b_1,\dots,b_m$ is any set of $m$ linearly independent vectors, then there is a vector $b_{m+1}$ such that $b_1,\dots,b_{m+1}$ is also linearly independent. However, $(5.1)$ says that this is not possible for $m = n$, where $n$ is the size of the given generating set of $V$.
Edit: based on the comments requesting clarification.
I am not sure that the statement under consideration is of the form "(not P) or Q". I always prefer to reason out the negation in a step-by-step fashion rather than work with formal statements and the rules for their negation. It leads to less confusion, at least in my mind.
Now, the negation of
There exists $m \geq 0$ such that ~blah~.
is
For every $m \geq 0$ we have ~not blah~.
Here ~blah~ is
There exists a set of linearly independent vectors $b_1,\dots,b_m$ such that ~foo~.
So, ~not blah~ is
For any set of linearly independent vectors $b_1,\dots,b_m$, we have ~not foo~.
Here, ~foo~ is
Any set of $m+1$ vectors in $V$ is linearly dependent.
So, ~not foo~ is
Some set of $m+1$ vectors in $V$ is linearly independent.
So, the negation of the statement in consideration is:
For every $m \geq 0$, we have that for any set of linearly independent vectors $b_1,\dots,b_m$, we have some set of $m+1$ vectors in $V$ is linearly independent.
There is no loss of generality in taking the set of $m+1$ linearly independent vectors in $V$ to be of the form $b_1,\dots,b_{m+1}$ because any subset of a linearly independent set is linearly independent. So, if we start with the set $b_1,\dots,b_m$ of $m$ linearly independent vectors and get $v_1,\dots,v_{m+1}$ a set of $m+1$ linearly independent vectors as per the claim, then in particular $v_1,\dots,v_m$ is a set of $m$ linearly independent vectors such that there exists $v_{m+1}$ such that $v_1,\dots,v_{m+1}$ is linearly independent. So, we might as well relabel the $v_i$'s as $b_i$'s and proceed inductively, since this does not change the proof.
Hope this helps. Feel free to reply in the comments for any clarifications.
Best Answer
It seems that (I) is used implicitly when using the words "selecting the maximal subset" there.
[[ We might have to see the whole argument to be certain , though I think this is where the author used (1) ]]
Certain Properties will not allow selecting "selecting the maximal subset" , like this Eg 1 :
Let Set $X=\{1,2,-3,4,-5,6,-7,-8\}$
We want Sub-Set $Y$ where the Sum is Positive.
We can check that $Y$ can be $\{1,2\}$ , $\{1,2,4\}$ , $\{1,-3,4\}$ , $\{-3,4,-5,6\}$ , where Sum is indeed Positive.
Let $Y=\{1,-3,4\}$ , with Positive Sum.
Yet , there are Sub-Sets of $Y$ whose Sum goes Negative too : $\{1,-3\}$ , $\{-3\}$
Hence , we can not talk about "selecting the maximal subset" here.
When we include new element $-5$ to $Y$ , then Sum goes Negative.
Yet , we can add two new elements $-5$ & $6$ to $Y$ , where Sum will remain Positive.
It reiterates that we can not talk about "selecting the maximal subset" here.
Consider this Eg 2 :
We have list of numbers $Z = [+5,-5,+5,-5,+5]$ where we want to pick out "contiguous" numbers with Positive Sum.
We have various ways to try that :
There is not much meaning to unique maximal subset here too.
[[ I am just making explanatory examples to show that maximal subset may or may not exist , the answer will stand without these examples too ]]
When the author talks about "selecting the maximal subset" with the Property "strongly linearly independent" , then (I) implicitly allows it.
When we have made the selection , it can not occur that we can add new element which makes it lose that Property yet we can add two new elements to retain that Property.
In other words , "selecting the maximal subset" will make sense here.