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**Theorem 2.4.**Every vector space $V$ over a division ring $D$ has a basis and is therefore a free $D$-module. More generally every linearly independent subset of $V$ is contained in a basis of $V.$

I do not understand why the second statement of the theorem is a generalization of the first statement, could anyone clarify this for me please?

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    $\begingroup$ Because every Linearly independent set can be extended to a basis. $\endgroup$ – Sumit Mittal Dec 1 '17 at 14:09
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The first statement establishes that bases exist, and the second statement says more:

Not only do bases exist, but you have the freedom to extend any linearly independent set to a basis.

Obviously if you can extend any linearly independent set to a basis (you coudld just start with a single nonzero element, for example) then bases exist, as given in the first statement.

Another example of this would be

In a ring with identity, maximal ideals exist

and

In a ring with identity $R$, given any ideal $I\neq R$, there is a maximal ideal of $R$ containing $I$.

The second statement is clearly stronger.

There are situations where the first sort of thing holds, but not the second. For example, projective modules always have maximal submodules, but I believe I saw a post here once that showed you could not always find a maximal submodule containing a given submodule.

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