[proofplan] All five parts follow from the [Steinitz Exchange Lemma](/theorems/373). Part (i) applies it in both directions to get equality of basis sizes. Part (ii) shows an independent set of size $n$ must span (otherwise it extends to a larger independent set, contradicting Steinitz). Part (iii) shows a spanning set of size $n$ must be independent (otherwise it reduces, contradicting Steinitz). Parts (iv) and (v) use iterative removal or extension. [/proofplan] [step:Show all bases have the same number of elements (part (i))] Let $B = \{e_1, \ldots, e_n\}$ and $B' = \{f_1, \ldots, f_k\}$ be two bases for $V$. Since $B$ is linearly independent and $B'$ spans $V$, the [Steinitz Exchange Lemma](/theorems/373) gives $n \leq k$. Since $B'$ is linearly independent and $B$ spans $V$, the same result gives $k \leq n$. Hence $n = k$. [/step] [step:Show a linearly independent set of size $n$ is a basis (part (ii))] Let $S = \{e_1, \ldots, e_n\}$ be linearly independent. Suppose $S$ does not span $V$. Then there exists $v \in V \setminus \langle S \rangle$. The set $S' = \{e_1, \ldots, e_n, v\}$ is linearly independent: if $\sum_{i=1}^{n}\lambda_i e_i + \muv = \mathbf{0}$ with $\mu \neq 0$, then $v = -\mu^{-1}\sum \lambda_i e_i \in \langle S \rangle$, contradicting the choice of $v$; so $\mu = 0$, and independence of $S$ forces all $\lambda_i = 0$. Now $S'$ is an independent set of size $n + 1$. Let $B$ be any basis for $V$, which has $n$ elements by part (i). The [Steinitz Exchange Lemma](/theorems/373) applied with $S'$ independent and $B$ spanning gives $n + 1 \leq n$, a contradiction. Hence $S$ spans $V$ and is a basis. [/step] [step:Show a spanning set of size $n$ is a basis (part (iii))] Let $S = \{f_1, \ldots, f_n\}$ span $V$. If $S$ is linearly dependent, by the [Linear Dependence Characterisation](/theorems/371) some $f_j$ is a linear combination of the others. Removing $f_j$ gives a spanning set of size $n - 1$. But any basis has $n$ elements by part (i), and the [Steinitz Exchange Lemma](/theorems/373) applied to this basis (independent, size $n$) and the reduced spanning set (size $n - 1$) gives $n \leq n - 1$, a contradiction. Hence $S$ is linearly independent and therefore a basis. [/step] [step:Extract a basis from a finite spanning set (part (iv))] Let $S = \{f_1, \ldots, f_m\}$ span $V$. If $S$ is linearly independent, it is a basis. Otherwise, by the [Linear Dependence Characterisation](/theorems/371), some element is a linear combination of the others; remove it. The remaining set still spans $V$ and has $m - 1$ elements. Repeat until the set is linearly independent. The process terminates because the size decreases at each step and cannot drop below $\dim V = n$ (an independent set of size $n$ already spans by part (ii), and Steinitz prevents independent sets from exceeding $n$). The resulting independent spanning set is a basis. [/step] [step:Extend a linearly independent set to a basis (part (v))] Let $S = \{e_1, \ldots, e_k\}$ be linearly independent with $k \leq n$. Let $B = \{f_1, \ldots, f_n\}$ be a basis for $V$. By the [Steinitz Exchange Lemma](/theorems/373) applied with $S$ independent and $B$ spanning, after reordering: \begin{align*} \{e_1, \ldots, e_k, f_{k+1}, \ldots, f_n\} \end{align*} spans $V$. This set has $n = \dim V$ elements, so by part (iii) it is a basis extending $S$. [/step]