# Application of Zorn's lemma.

• mahler1
In summary, in this exercise, the task is to prove that a linear transformation f:S→R can be extended to all the vector space V by using Zorn's lemma. This is done by defining a partially ordered set and showing that it has a maximal element. The proof involves verifying that the relation defined is an order relation, and then showing that every totally ordered set has an upper bound. There may be some issues with the proof, but the overall idea is correct.

#### mahler1

1. Homework Statement .
Let V be a vector space over the field R and S be a subspace of V. Now let f:S→R be a linear transformation. Prove that T can be extended to all the vector space V, i.e., there exists f':V→R such that f' restricted to S equals f.

3. The Attempt at a Solution .
In this exercise I am supposed to use Zorn's lemma to prove the existence of f'. I define a partially ordered set as follows: P={(U,T) where U is a subspace of V which contains the subspace S and T is a linear transformation T:U→R that satisfies T restricted to S equals f}. Then, I define the relation (U,T)≤(U',T') if and only if UcU' and T' restricted to U equals T. I need to verify that the relation I defined is indeed an order relation on P, but I didn't have problems with that; it's very easy to check for reflexivity, antisymmetry and transitivity.

I want to show that every totally ordered set of P has an upper bound, so let P' be a totally ordered set of P. If I choose the set X to be the union of all elements (U',T') of P', then (U',T')≤X for all (U',T') in P'. Now I can apply Zorn's lemma: P has a maximal element which I will denote M=(W,T'') where W is a subspace of V and T'' is a linear transformation, T'':W→R. By definition of maximal element, W is a subspace which includes all subspaces which contain the subspace S and T'' is a linear transformation that satisfies T'' restricted to S equals f. To complete the proof, I need to show that the domain of T'' is all the vector space V. Suppose V is not the domain of T''. So the domain W has to be a proper subspace of V. Then there is an element v in V which is not in W and Wu<v> is another subspace of V which includes W. I define h to be the linear transformation on Wu{v} as: h restricted to W equals T'' and h(v)=0. Then (W,T'')≤(Wu<v>,h). This is absurd since (W,T'') was the maximal element of P. Then V must be the domain of T'' and T'' restricted to S equals f so T'' is the linear transformation I was looking for.

Well, I know my proof is verbose, but is it correct?

You'd need to show that $W\cup <v>$ is a subspace (it's not in general) and that $h$ is linear. I reckon you're on the right track, though.