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Denote: $[A]$ as the span of $A$.

Theorem: Every linearly independent subset of a vector space is a subset of a basis of a space.

Proof: Let $A$ be a linearly independent subset of a vector space $V$ and let $\,B\,$ be a basis of $\,V$.

Case 1. $B\subseteq [A]\,$. Then $\,[B]\subseteq [A]\,$. But $\,[B]=V\,$ , hence, $\,A=B\,$.

Case 2. $B\not\subseteq [A]$. Then there exists $\alpha \in B$ such that $\alpha \not\in [A]$. Thus $A\cup \{\alpha\}$ is linearly independent. Repeat the argument until an enlarged set is produced that spans $V$.

My question is: In case 2, how is it that $A$ is contained in a basis $B$? I just can't get the idea in the repeating the argument part. Thanks for your help.

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The argument as shown is a bit suspicious. Firstly, you need the existence of a basis $B$ in the first place. Some authors use the theorem to be shown here, i.e. that every linearly independent family can be extended to a basis, to prove just that (namely by extending the empty family). Secondly, "repeat until" may not work as easily as it sounds if the vector space is infinite-dimensional.

However, what happens here is that a linear independent faimily is extended (possibly repeatedly) to a bigger linear independent family, thus spanning larger and larger portions of the vector space and especiallycontaining more and more members of the basis assumed given. At least with finite dimensional vector spaces this process must end after finitely many steps, aand that means we have found an extended linear independent family that also spans the whole vetor space, i.e. a basis.

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That explains it very well. Thanks Sir Hagen von Eitzen. –  Philip Benj Marcoby Eragon Jan 6 '13 at 11:01
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The first case is incorrect, take any two different bases $A,B$.

Since $A$ is a basis $B$ is contained in $[A]$, but $A\neq B$.

In case $B$ you are adding one linearly independent vector to $A$, when $|A|=dim(V)$ you will be done.

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