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m →General existence theorem: Sequence of Hamel basis vectors needs to be normalized, otherwise we cannot guarantee that the expression n\|e_n\|\ will go to infinity. Also, clearer wording. |
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Discontinuous linear maps can be proven to exist more generally, even if the space is complete. Let ''X'' and ''Y'' be [[normed space]]s over the field ''K'' where <math>K = \R</math> or <math>K = \Complex.</math> Assume that ''X'' is infinite-dimensional and ''Y'' is not the zero space. We will find a discontinuous linear map ''f'' from ''X'' to ''K'', which will imply the existence of a discontinuous linear map ''g'' from ''X'' to ''Y'' given by the formula <math>g(x) = f(x) y_0</math> where <math>y_0</math> is an arbitrary nonzero vector in ''Y''.
If ''X'' is infinite-dimensional, to show the existence of a linear functional which is not continuous then amounts to constructing ''f'' which is not bounded. For that, consider a [[sequence]] (''e''<sub>''n''</sub>)<sub>''n''</sub> (<math>n \geq 1</math>) of [[linearly independent]] vectors in ''X'', which we normalize, i.e. <math>\|e_n\|\ = 1, \forall n\!\in\!\mathbb{N}\</math>. Then, we
<math display=block>T(e_n) = n\|e_n\|\,</math>
for each <math>n = 1, 2, \ldots</math> Complete this sequence of linearly independent vectors to a [[basis (vector space)|vector space basis]] of ''X''
Notice that by using the fact that any set of linearly independent vectors can be completed to a basis, we implicitly used the axiom of choice, which was not needed for the concrete example in the previous section
== Role of the axiom of choice ==
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