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I don't understand why this statement is wrong:

$V$ is a vector space, and $W$ is a subspace of $V$. $K$ is a basis of $V$. We can manage to find a subset of $K$ that will be a basis of $W$.

Sorry if my English is bad... and if you can show me an example of something that contradicts it, it'd be great.

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    $\begingroup$ a basis has finitely many subsets, but a vector space usually has infinitely many subspaces $\endgroup$ – mercio Jun 30 '13 at 9:38
  • $\begingroup$ this argument is wonderful! $\endgroup$ – Ant Jun 30 '13 at 12:36
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Take for example $V = \mathbb{R}^2$, $W = \{(x,y)\in\mathbb{R}^2 : x = y\}$, and $K = \{(1,0),(0,1)\}$.

$W$ is a subspace of $V$, but there is no subset of $K$ that gives a basis for $W$.

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A simple counter-example is $V=\mathbb R^2$, $W$ = the $y$-axis, and the basis of $V = \{ (1,0),(1,1)\}$.

Clearly neither of the two linearly independent vectors in this basis lie along the $y$-axis.

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