Question on proof of every finitely generated vector space has a basis The following is a proof of the theorem given in Curtis's Linear Algebra book:

First we consider the case in which $V$ consists of the zero vector
  alone. Then the zero vector spans $V$ but cannot be a basis. In this
  case we agree that the empty set is a basis for $V$, so that the
  dimension of $V$ is zero. Now, let $V \ne \{ 0 \}$ be a vector space
  with $n$ generators. By theorem $(5.1)$ (see below) 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, $m\ge 1$, $V$ contains linearly independent vectors
  $b_1, \ldots , b_m $ such that any set of $m+1$ vectors in $V$ is
  linearly independent. We prove that $\{ b_1, \ldots , b_m \}$ is a
  basis for $V$, and for this it is enough to show, for any vector $b \in V$, that $b \in S(b_1, \ldots , b_m )$. Because of the properties
  of the set $\{ b_1, \ldots , b_m \}$, $\{ b_1, \ldots , b_m , b \}$ is
  a linearly dependent set. Since $\{ b_1, \ldots , b_m \}$ is linearly
  independent, Lemma $(7.1)$ implies that $b \in S(b_1, \ldots , b_m )$
  and the theorem is proved.

Here are the theorems used in the proof:

$(5.1)$ Theorem. Let $S$ be a subspace of a vector space $V$ over a
  field $F$, such that $S$ is generated by $n$ vectors $\{ a_1 , \ldots
, a_n \}$. Suppose $\{ b_1 , \ldots , b_m \}$ are vectors in $S$, with
  $m > n$. Then the vectors $\{ b_1 , \ldots , b_m \}$ are linearly
  dependent.
$(7.1)$ Lemma. If $\{ a_1 , \ldots , a_m\}$ is a linearly dependent
  and if $\{ a_1 , \ldots , a_{m-1} \}$ is linearly independent, then
  $a_m$ is a linear combination of $a_1 , \ldots , a_{m-1}$.

My questions:


*

*How does the empty set span $\{ 0 \}$? The subspace generated by $\{\}$ seems to be an empty set. Am I wrong here?

*The author claims that "it follows that, for some integer, $m\ge 1$, $V$ contains linearly independent vectors $b_1, \ldots , b_m $ such that any set of $m+1$ vectors in $V$ is linearly independent."? But how do I find such linearly independent vectors?
 A: 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.
