Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I really need some help with this proof..

Let V be a vector-space, $(A_i)_{i\in N}$ a sequence of linearly-independent subsets of V with properties: $ A_i \subseteq A_{i+1}$ (for all $in \in N$ )

Show: $\bigcup_{i \in N} A_i $ is linearly-independent.


My ideas:

Assuming that $\bigcup_{i \in N} A_i $ is linearly-dependent, I should come to a contradiction.

If $\bigcup_{i \in N} A_i $ is l.p. then there's at least one $k_i \not = 0$ so that $\sum_{i=1}^n (k_i * v_i) = 0$ (with $v_i$ being a vector in V)

The set $\bigcup_{i \in N} A_i $ is infinitely. So a finite set of $\bigcup_{i \in N} A_i $ should also be linearly-dependent.

I now wanted to show the contradiction, but I'm still not sure whether I used correct steps so far..

Thank you very much!

share|cite|improve this question
Notice that a finite subset of $\bigcup_i A_i$ is a subset of $A_{i_0}$ for some $i_0$ large enough. – Siméon Nov 25 '13 at 11:38
Assume $\{v_1,\dots,v_k\}\subset\bigcup_{n\in\boldsymbol N}A_{n}$ are lin. dep. Now let $m:=\min_{l\in\boldsymbol N}(\{v_1,\dots,v_k\}\subset A_l\}$. – Michael Hoppe Nov 25 '13 at 11:42
According to Siméon: Noticing this I come to a point where a subset of $A_{i_0}$ should be linearly dependent. However, every $A_i$ is lin.ind. - ?! – Vazrael Nov 25 '13 at 18:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.