0
$\begingroup$

Show that any non-null vector space has a basis.

What I am trying to do -Taking a spanning set which spans the vector space then if the spanning set is linearly independent then it form basis and we are done but if it is linearly dependent then we can find a vector of that set which can be written as a linear combination of other vectors we remove that vector now if the remaining set of vectors ar linearly independent then we are done otherwise repeat the same process till we reach a stage when only linearly independent set of vector remains left and thus it form's a basis. Don't know right or wrong.

$\endgroup$
2
$\begingroup$

In your argument you are making a choice to remove elements and repeat the process. This requires axiom of choice for an infinite dimensional vector space.
Hint: Take a non trivial vector space $V$. Since it is non trivial there is a non zero vector in it. Now make a set of linearly independent subsets of $V$. Show that this is partially ordered(poset) by inclusion and each poset has an upper bound(hint: think of nion). Apply Zorns lemma to show there is a maximal element.

$\endgroup$
  • $\begingroup$ Is there are any other than what u have shown.As it is given where the use of lemme what u have stated is not allowed. $\endgroup$ – Daniel Feb 1 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.