Discontinuous linear map: Difference between revisions

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== A nonconstructive example ==
 
An algebraic basis for the [[real number]]s as a vector space over the [[rationals]] is known as a [[Hamel basis]] (note that some authors use this term in a broader sense to mean an algebraic basis of ''any'' vector space). Note that any two [[commensurability (mathematics)|noncommensurable]] numbers, say 1 and <math>\pi</math>, are linearly independent. One may find a Hamel basis containing them, and define a map <math>f : \R \to \R</math> so that <math>f(\pi) = 0,</math> ''f'' acts as the identity on the rest of the Hamel basis, and extend to all of <math>\R</math> by linearity. Let {''r''<sub>''n''</sub>}<sub>''n''</sub> be any sequence of rationals which converges to <math>\pi</math>. Then lim<sub>''n''</sub> ''f''(''r''<sub>''n''</sub>) = π, but <math>f(\pi) = 0.</math> By construction, ''f'' is linear over <math>\Q</math> (not over <math>\R</math>), but not continuous. Note that ''f'' is also not [[measurable function|measurable]]; an [[Additive map|additive]] real function is linear if and only if it is measurable, so for every such function there is a [[Vitali set]]. The construction of ''f'' relies on the axiom of choice.
 
This example can be extended into a general theorem about the existence of discontinuous linear maps on any infinite-dimensional normed space (as long as the codomain is not trivial).