Assume the Exchange Theorem and prove the following: Assume the vector space V is finitely generated. Then there is a natural number n such that the length of a linearly independent sequence is less than or equal to n.

Theorem: Assume V is spanned by a sequence $(v_1, . . . , v_n)$ of length n. Then any sequence $(w_1, . . . , w_{n+1})$ of length $n + 1$ is linearly dependent.



Let x be the cardinality of a linearly independent sequence

Assume a vector space V is finitely generated by a spanning sequence $B=(v_1,...,v_k)$. By the exchange theorem, if we append a $v_j \in V\B$, $B$ becomes linearly dependent, so there doesn't exist a linearly independent sequence of cardinality greater than $k$.

If $B$ is linearly independent $|B|=k=x$.

If $B$ is linearly dependent then,there exists maximal linearly independent spanning sub-sequence $(v_1,..,v_j)$ for $j<k$,so $j=x$

How does that look?

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    $\begingroup$ Doesn't the exchange theorem tell you exactly that the cardinality of a spanning set is an upper bound for the cardinality of a linearly independent set? What is the statement you call “Exchange theorem”? $\endgroup$ – egreg May 23 '16 at 20:17

I would not go such a way because what does it mean to have V-B. One is a space the other is a sequence. You can do V- span (B). Depending on how formal you want to be you can make many arguments.

Proof by contradiction assuming the $\{w_1,\dots,w_{n+1}\}$ is Linearly independent. Say dim $V= n$ because there are $n$ elements in the basis. Now $w_i \in V$ for all i. Now $span (\{w_1,\dots,w_{n+1}\})$ is a subspace of V. Well if you are a subspace of a space $dim V \geq dim W$ where W is any subspace of V. Now if let $W= span (\{w_1,\dots,w_{n+1}\})$. Well because this list is linearly independent they form a basis for W ie dim $W=n+1$.

This is a contradiction $dim V \geq dim W$, but you get $n \geq n+1$.


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