Proof of Theorem 9.2 of Rudin's Principles of Mathematical Analysis book I am trying to understand a part of the proof of Theorem 9.2 of Rudin's book "Principles of Mathematical Analysis". 
The theorem says: Let $r$ be a positive integer. If a vector space $X$ is spanned by a set of $r$ vectors, them $dim X \leq r$. 
The proof is by contradiction. He supposes that there is a space X that is spanned by a set $S_0$ with $r$ vectors and, at same time, there is an independent set $Q=\{y_1,...,y_{r+1}\}$ with $r+1$ vectors. 
Next, he defines a set $S_i$ obtained from $S_0$ by replacing $i$ of its elements by members of $Q$, without altering the the span. 
My question is: what guarantees, in this context, that it is possible to substitute elements of $S_0$ by elements of $Q$ without altering the span? 
Thanks in advance!
 A: I think the solution is the following: Let $n=dim X $ and consider the set $S_1=\{y_1,x_1,...,x_r\}$. This set is dependent (because dim X = r is the biggest number of independent vectors in a set). Thus,
$y_1= \sum_{i=1}^{n} \alpha_i x_i$ with some $\alpha_k\neq 0$ for some $k$. Then $x_k=\frac{1}{\alpha_k} y_1 + \sum_{i\neq k}^n \frac{\alpha_i}{\alpha_k} x_i$.
Given $z\in X$, we have $z=\sum_{i=1}^n \beta_i x_i$. Then 
$z= \frac{\beta_k}{\alpha_k} y_1 + \sum_{i\neq k}^{n} \gamma_k x_k$
That is, the set $S_1$ also span $X$.
A: If you read the proof carefully, the supposition is that this procedure works for some $i$ satisfying $0\leq i<r$. Since $S_0$ is assumed to span $X$, this supposition is guaranteed to hold.
Essentially he is inductively constructing these sets $S_k$ and the first sentence is the inductive hypothesis, where the the base case holds trivially.
A: There has been an explanation inductively in the Rudin's proof of the theorem.
First $S_0$ spans $X$ by assumption. Next, suppose that for some $0\le i<r$, 
$S_i$ spans $X$. Then it remains to show that $S_{i+1}$ also span $X$, which
has shown in the proof. So he concludes that $S_{r}$ spans $X$, which gives
a contradiction.
