5
$\begingroup$

This question already has an answer here:

Is it possible to construct a function $f \colon \mathbb{R} \to \mathbb{R}$ such that $$f(x + y) = f(x) + f(y)$$ and $f$ is not continuous?

$\endgroup$

marked as duplicate by azimut, M Turgeon, Amitesh Datta, user61527, Alex Wertheim Oct 11 '13 at 2:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Haha, you saw the continuous case and you wondered, right? Good question. $\endgroup$ – Patrick Da Silva Oct 11 '13 at 0:10
  • 1
    $\begingroup$ By "construct" you mean to actually write it down? $\endgroup$ – GEdgar Oct 11 '13 at 0:15
  • $\begingroup$ To construct? No. You'd need to invoke the axiom of choice. $\endgroup$ – TBrendle Oct 11 '13 at 0:16
  • $\begingroup$ patrick, could you please get put of my mind? thanks :D $\endgroup$ – user67133 Oct 11 '13 at 0:25
3
$\begingroup$

"Fix a basis for $\mathbb R$ as a $\mathbb Q$-vector space" (which exists under the axiom of choice, but under weaker axioms I have no idea what happens). The condition $f(x+y) = f(x) + f(y)$ is equivalent to $f$ being $\mathbb Q$-linear, so you're asking if there exists a non-trivial discontinuous map of $\mathbb Q$-vector spaces between $\mathbb R$ and itself. If you map the basis elements to other basis elements in a discontinuous way, you will obtain such a map.

Added : A quick way to see that "a discontinuous way of doing it" exists, is that the set of $\mathbb Q$-linear maps that are also $\mathbb R$-linear has cardinality of the reals, where as the set of all $\mathbb Q$-linear maps (or in other words, the number of maps between the basis elements) has cardinality $|\mathbb R|^{|\mathbb R|}$. To understand why, well of course the basis for $\mathbb R$ as a $\mathbb Q$-vector space has cardinality $\le |\mathbb R|$, but if it was countable, then $\mathbb R$ would be countable because all of its elements would be a linear combination of elements in a countable set with countably many possible coefficients. Since the basis for $\mathbb R$ as a $\mathbb Q$-vector space is uncountable, the set of all maps from a basis to another also is. Therefore there must be at least one map between the two bases which generates a discontinuous linear map.

Hope that helps,

$\endgroup$
0
$\begingroup$

Yes. The real numbers $\mathbb{R}$ is a vector space over the field $\mathbb{Q}$. Therefore there is a basis $\mathcal{B}$. Suppose you have any function $f:\mathcal{B}\to \mathbb{R}$. This can be extended $\mathbb{Q}$-linearly to all of $\mathbb{R}$. This is a place to start.

$\endgroup$