Section 11*: Problem 8 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

A typical use of Zorn’s lemma in algebra is the proof that every vector space has a basis. Recall that if $A$ is a subset of the vector space $V$ , we say a vector belongs to the span of $A$ if it equals a finite linear combination of elements of $A$ . The set $A$ is independent if the only finite linear combination of elements of $A$ that equals the zero vector is the trivial one having all coefficients zero. If $A$ is independent and if every vector in $V$ belongs to the span of $A$ , then $A$ is a basis for $V$ .

(a) If $A$ is independent and $v\in V$ does not belong to the span of $A$ , show $A\cup\{v\}$ is independent.

(b) Show the collection of all independent sets in $V$ has a maximal element.

(a) If it were not independent, there would be a non-trivial linear combination equal to the zero vector. This combination must include $v$ with a non-zero coefficient, as $A$ is independent. But this leads to the contradiction as it implies that $v\in span(A)$ .

(b) Let $\mathcal{A}$ be the collection of all independent sets in $V$ partially ordered by the proper inclusion. Then $B\in\mathcal{A}$ iff any non-trivial linear combination of any finite number of vectors in $B$ is not the zero vector iff any finite subset of $B$ is in $\mathcal{A}$ . Therefore, $\mathcal{A}$ is of finite type and Tukey’s lemma applies.

(c) The maximal set found in (b) is a basis, as it is independent, and if there were a vector not in its span, then, according to (a), the set would not be maximal.

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Section 11*: Problem 8 Solution

Section 11*: Problem 8 Solution