/sci/ - Science & Math

>>11636692
>just banter and Scholze drama
The authentic /mg/ experience.
Also, a dude asked about Quillen functors yesterday on /sqt/. Someone go help him.

>>11486054
Proof. Assume that [math]v = \Sigma_{i=1}^n k_i b_i[/math] and [math]v = \Sigma_{j=1}^l k_j b_j[/math]. Then, [math]0=v-v = \Sigma_{i=1}^n k_i b_i - \Sigma_{j=1}^l k_j b_j[/math], and linear independence guarantees all [math]k[/math] zero.
Theorem three: Any non-trivial vector space admits a basis.
Proof: We order the set of linearly independent subsets by inclusion. Because the vector space is non-trivial, it contains some nonzero v, and then [math]\{ v \}[/math] is a linearly independent subset, which shows that the set of linearly independent subsets is non-empty.
Assume we have a chain [math]B_{ \lambda}[/math], [math]\lambda \in \Lambda [/math]. We claim that [math]B = \cup _{\lambda \in \Lambda} B_{\lambda}[/math] is linearly independent. Assume it isn’t. Then, for some [math]\{ b_1 , \cdots , b_n \} \subseteq B[/math] and [math]\{ k_1 , \cdots , k_n \} \subseteq K[/math], we have [math]\Sigma_{i=1}^n k_ib_i = 0[/math]. Because [math]\{ b_1 , \cdots , b_n \} [/math] is a finite set, there is some [math]\eta \in \Lambda[/math] such that [math]\{ b_1 , \cdots , b_n \} \subseteq B_{ \eta}[/math]. Then [math]\Sigma_{i=1}^n k_ib_i = 0[/math] implies all [math]k_i[/math] zero because [math]B_{ \eta}[/math] is linearly independent, and [math]B[/math] is linearly independent.
By Zorn’s lemma, there’s a maximal linearly independent subset of [math]V[/math], a basis.