0
$\begingroup$

This question already has an answer here:

The hint given is that we can assume the set of real numbers has a basis. I've seen several other questions of this form, but all of them use terminology which I am unfamiliar with ($Q$-linear, extend linearly to $\mathbb{R}$ etc).

Edit : I know what a Hamel basis is, and I don't have to use the axiom of choice to show that it exists or anything. I'm essentially told that I can assume it exists.

$\endgroup$

marked as duplicate by Arthur, Clayton, Arnaud Mortier, Mostafa Ayaz, Arnaud D. Feb 15 '18 at 22:33

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.

  • $\begingroup$ WHat do you mean by basis in this context? $\endgroup$ – José Carlos Santos Feb 15 '18 at 7:05
  • $\begingroup$ A set of linearly independent vectors which span the reals where each scalar is rational $\endgroup$ – Saad Feb 15 '18 at 7:07
  • $\begingroup$ That's what a basis of a $\mathbb Q$-vector space is. Do you think that the irrational numbers are such a space? $\endgroup$ – José Carlos Santos Feb 15 '18 at 7:10
  • $\begingroup$ You must read about Hammel basis and Axiom of Choice.You may refer to the book Functional Equations and how to solve them Springer $\endgroup$ – Shubhrajit Bhattachrya Feb 15 '18 at 7:15
  • 1
    $\begingroup$ Then, you'll probably have to learn a bit more to understand the reasoning. It isn't a trivial thing you could explain to a toddler. $\endgroup$ – Professor Vector Feb 15 '18 at 7:17
1
$\begingroup$

Well, not disregarding the other answers, I suggest that it is better understood in the following order. Firstly, there is a

Lemma Let $f:\mathbb{R}\to \mathbb{R}$ additive ($f(x_1+x_2)=f(x_1)+f(x_2)$ identically) and continuous at zero, then $f$ is linear i.e. $$ (\exists a\in \mathbb{R})(f(x)=a.x) $$

Then assuming the existence of a Hamel $\mathbb{Q}$-basis $(b_i)_{i\in I}$ of $\mathbb{R}$, remark that the indexing set $I$ cannot be finite (otherwise $\mathbb{R}$ would be countable, or if $I=\emptyset$, $\mathbb{R}$ would be a singleton !) and pick $i_0\not= i_1$ in $I$. Then the mapping defined by $f(b_i)=b_i$ if $i\notin \{i_0,i_1\}$ and $f(b_j)=b_{1-j}$ (exchange $i_0,i_1$) is discontinuous. The final argument goes as follows.

If $f$, constructed as above, were continuous, we would have $f(x)=ax$ for some $a\in \mathbb{R}$. From the fact the basis is not finite, one can choose $i_2\notin\{i_0,i_1\}$ and by definition $f(b_{i_2})=b_{i_2}$ which entails $a=1$ ($b_{i_2}\not=0$ as element of a basis). Now (still supposing $f$ continuous) $f$ must be the identity which contradicts $f(b_{i_0})=b_{i_1}$. Hence this $f$ is not continuous.

Final remark The lemma shows that among additive functions $\mathbb{R}\to \mathbb{R}$ the class of ($\mathbb{R}$-) linear functions is exactly the same as the class of continuous ones. That's why answers of this question are the same even if their wording is different.

$\endgroup$
  • $\begingroup$ Sorry, I don't see how it's discontinuous.. $\endgroup$ – Saad Feb 15 '18 at 21:27
  • $\begingroup$ @Saad I have detailed the final argument. $\endgroup$ – Duchamp Gérard H. E. Feb 15 '18 at 22:59
  • $\begingroup$ Ah! That makes sense, especially since we were asked to show the lemma in a previous result. $\endgroup$ – Saad Feb 15 '18 at 23:01
  • $\begingroup$ @Saad Do not hesitate. $\endgroup$ – Duchamp Gérard H. E. Feb 16 '18 at 6:29
1
$\begingroup$

Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$ and let $H$ be a Hamel basis. Then for any real $x$ there exist a unique finite collection (the cardinality depends on $x$) $h_1,\dots,h_n\in H$ and $q_1,\dots,q_n\in \mathbb{Q}$ such that $$x=\sum_{k=1}^nq_k h_k.$$ Now for $h\in H$, define $f(h)\in\mathbb{R}$ in some way, but be sure that $f(g_1)/f(g_2)\not=g_1/g_2$ for at least two distinct $g_1,g_2\in H$ (this will be used to prove discontinuity). Then for all $x\in \mathbb{R}$, let $$f(x):=\sum_{k=1}^nq_k f(h_k).$$ Now verify that $f$ is additive and show that $f$ is not continuous at the origin.

P.S. Since $f(g_1)/f(g_2)\not=g_1/g_2$, there exist $\epsilon>0$ and rational sequences $(q_n)_n$, $(p_n)_n$ such that $|f(q_ng_1)-f(p_ng_2)|=|q_nf(g_1)-p_nf(g_2)|>\epsilon$ and $\lim_{n\to\infty}|q_ng_1-p_ng_2|=0$.

$\endgroup$
  • $\begingroup$ Do you mean $f(h)$ must be defined in an intelligent way, or will any $f(h)$ work? Also, I've no idea how to show this is discontinuous.. $\endgroup$ – Saad Feb 15 '18 at 7:25
  • $\begingroup$ @Saad See my revised answer. $\endgroup$ – Robert Z Feb 15 '18 at 7:27
  • $\begingroup$ How do we know what $h$ is? $\endgroup$ – Saad Feb 15 '18 at 7:30
  • $\begingroup$ We just need that a Hamel basis EXISTS (this is implied by the axiom of choice). $\endgroup$ – Robert Z Feb 15 '18 at 7:42
  • 1
    $\begingroup$ Here is an argument which makes Robert's construction discontinuous and does not use the fact that continuous additive functions are multiples of the identity: instead of the condition imposed on $f(g_1)/f(g_2)$ observe that H is uncountable. Otherwise $\mathbb R$ would become countable. Any uncountable set has a fi nite limit point. Let $\{h_n\}$ be a convergent sequence in H. When you define f on H take $f(h_n)=n$. Then f is not continuous because $\lim f(h_n)=\infty$ $\endgroup$ – Kavi Rama Murthy Feb 15 '18 at 7:53
0
$\begingroup$

Let $f:\mathbb R\to \mathbb R$ be additive. The form any integer $n$ you get $f(n)=f(1+\ldots+1)=nf(1)$ so that values on integers are determined by value at $1$. Furthermore, for any integer $q$ you also have:

$$ f(1)=f(\frac 1 q+\ldots+\frac 1 q)=f(\frac 1 q)+\ldots+f(\frac 1 q)=qf(\frac 1 q) $$

from which you get that $f(\frac 1 q)=\frac 1 q f(1)$ and putting the two together you end up with $$ f(\frac p q)=\frac p q f(1) $$ which tells you that this function is determined by its values at $1$. This means exactly that the function is $\mathbb Q$-linear. Now if this function is also continuous, once its value on $\mathbb Q$, which is dense in $\mathbb R$ is given all its values regimen and by continuity $f(a)=af(1)$ for every real $a$. But if the function is not continuous then if you want to extend it to $\mathbb R$ you can do it in a variety of possible ways, essentially depending on a choice of a basis of $\mathbb R$ as a $\mathbb Q$-vector space.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.