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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 display=block> \|e_n\|\ = 1, \forall n \in \mathbb{N}\</math>. Then, we define
<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'' by defining ''T'' at the other vectors in the basis to be zero. ''T'' so defined will extend uniquely to a linear map on ''X'', and since it is clearly not bounded
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.
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