Mistake in proof about vector spaces Hello this is my first post on here and I have a question about an error in a proof. My Linear Algebra professor was covering Linear Independence, Basis, and Dimension for Vector Spaces. We got to a proof from the textbook and he said the proof is wrong. He said to try and figure out what is wrong with it. I have been trying to figure out what is wrong and after thinking about it, talking with peers, and with my professor I have narrowed it down to a single sentence within the proof. I think that it is a wrong word choice. He said it is very subtle. I was hoping some of you can provide some insight.  
THM 6.11: Let $W$ be a subspace of a finite-dimensional vector space $V$. Then: (a) $W$ is finite-dimensional and $\dim W \le \dim V$. (b) $\dim W = \dim V$ if and only if $W = V$.
He has a problem with the proof of part (a) which is as follows. 
Proof (a) Let $\dim V = n$. If $W = \{\mathbf{0}\}$, then $\dim(W) = 0 \le n = \dim V$. If $W$ is nonzero, then any basis $\mathcal{B}$ for $V$ (containing $n$ vectors) certainly spans $W$, since $W$ is contained in $V$. But $\mathcal{B}$  can be reduced to a basis $\mathcal{B'}$ for $W$ (containing at most $n$ vectors), by Theorem 6.10(f). Hence, $W$ is finite-dimensional and $\dim W \le n = \dim V$. 

Theorem 6.10(f) states: Let $V$ be a vector space with $\dim V = n$. Then: (f) Any spanning set for $V$ can be reduced to a basis for $V$. 

My professor said that me and my peers were right about the error being in the sentence, "But $\mathcal{B}$  can be reduced to a basis $\mathcal{B'}$ for $W$ (containing at most $n$ vectors), by Theorem 6.10(f)." I believe that the error has something to do with with the word reduced. I asked him and he said that I was on the right track but he didn't give me a clear answer. What is wrong with this proof? More specifically the sentence above? Any help would be much appreciated. (Textbook: Linear Algebra: A Modern Introduction by David Poole 4th edition. The proof is found in section 6.2: page 456)
 A: Your professor is correct. 
The problem is with the word " reduced" 
Let us look at an example.
Let W be the subspace of $\mathbb {R}^2$  spanned by $\{(1,1)\}.$
Let $ B=\{(1,0),(0,1)\}$ be a basis for  $\mathbb {R}^2$
Then $B$ spans $W$, but we can not reduce $B$ to a basis for $W$ 
A: Maybe there is a difference in use of terminology here but I would say that the problem is with the statement "$B$ spans $W$".
The way I have always seen the term used is as follows.


*

*We define ${\rm span}(B)=\{\hbox{linear combinations of vectors in $B$}\}$.

*Then "$B$ spans $W$" means that ${\rm span}(B)$ equals $W$.


In the proof you have quoted, they appear to be saying that "$B$ spans $W$" means that ${\rm span}(B)$ contains $W$.
So it is a question of using the precise meaning of terminology.  BTW, I would not agree with your professor that it is a "very subtle" error, IMHO it is a blatant error :)
A: The error is the following. Theorem 6.10(f) says that if you have a set that spans $V$, you can remove elements ("reduce") until you get a basis. In the proof of 6.11, the elements in the basis of $V$ may not belong to $W$, so the argument does not apply. 
For example, maybe $V=\mathbb R^3$, and you take the canonical basis $\{(1,0,0),(0,1,0),(0,0,1)\}$. Then if $W=\{(x,y,z): x=y=z\}$, none of the three elements in the basis belongs to $W$. 
The usual argument is that a basis of $W$ will be linearly independent, and so it will also be linearly independent in $V$, and it follows that $\dim W\leq \dim V$. 
