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This is an addition to Yemon Choi answer.

In fact, a real-valued function $f$ defined on a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$, for instance, extends uniquely to ad additive function, not to a linear function (it only needs to be linear over $\mathbb{Q}$). This can be seen readily defining $$F(x)=\sum_{i=1}^n q_i f(b_i),$$ where $b_i\in H$ for every $i$ and $q_i$ are $n$ rationals such that $\mathbb{R}\ni x=\sum_{i=1}^n b_iq_i$.

This is a classical tool to provide additive, discontinuous real functions, of course exploiting the axiom of choice.

Finally, a Hamel basis is not only "not write-downable": it can be almost as nasty as you want. For instance, you can prove under AC that there is a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$ and a function $f:H\to\mathbb{R}$ such that, for every real $x$, $f^{-1}(x)$ has nonempty intersection with every closed, uncountable set without containing any of them (= is a Bernstein set). Or you can prove that there exists a Hames basis which is not measurable and does not have the property of Baire, and much more...

All of this is very nicely covered in Chapter 7 of:

Ciesielski, K. (1997). Set theory for the working mathematician (No. 39). Cambridge University Press. ISO 690

This is an addition to Yemon Choi's answer.

In fact, a real-valued function $f$ defined on a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$, for instance, extends uniquely to ad additive function, not to a linear function (it only needs to be linear over $\mathbb{Q}$). This can be seen readily defining $$F(x)=\sum_{i=1}^n q_i f(b_i),$$ where $b_i\in H$ for every $i$ and $q_i$ are $n$ rationals such that $\mathbb{R}\ni x=\sum_{i=1}^n b_iq_i$.

This is a classical tool to provide additive, discontinuous real functions, of course exploiting the axiom of choice.

Finally, a Hamel basis is not only "not write-downable": it can be almost as nasty as you want. For instance, you can prove under AC that there is a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$ and a function $f:H\to\mathbb{R}$ such that, for every real $x$, $f^{-1}(x)$ has nonempty intersection with every closed, uncountable set without containing any of them (= is a Bernstein set). Or you can prove that there exists a Hamel basis which is not measurable and does not have the property of Baire, and much more...

All of this is very nicely covered in Chapter 7 of:

• Ciesielski, K. (1997). Set theory for the working mathematician (London Mathematical Society Student Texts No. 39). Cambridge University Press. https://doi.org/10.1017/CBO9781139173131

This is an addition to Yemon Choi answer.

In fact, a real-valued function $f$ defined on a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$, for instance, extends uniquely to ad additive function, not to a linear function (it only needs to be linear over $\mathbb{Q}$). This can be seen readily defining $$F(x)=\sum_{i=1}^n q_i f(b_i),$$ where $b_i\in H$ for every $i$ and $q_i$ are $n$ rationals such that $\mathbb{R}\ni x=\sum_{i=1}^n b_iq_i$.

This is a classical tool to provide additive, discontinuous real functions, of course exploiting the axiom of choice.

Finally, a Hamel basis is not only "not write-downable": it can be almost as nasty as you want. For instance, you can prove under AC that there is a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$ and a function $f:H\to\mathbb{R}$ such that $f^{-1}(x)$ has nonempty intersection with every closed, uncountable set without containing any of them (= is a Bernstein set). Or you can prove that there exists a Hames basis which is not measurable and does not have the property of Baire, and much more...

All of this is very nicely covered in Chapter 7 of:

Ciesielski, K. (1997). Set theory for the working mathematician (No. 39). Cambridge University Press. ISO 690

This is an addition to Yemon Choi answer.

In fact, a real-valued function $f$ defined on a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$, for instance, extends uniquely to ad additive function, not to a linear function (it only needs to be linear over $\mathbb{Q}$). This can be seen readily defining $$F(x)=\sum_{i=1}^n q_i f(b_i),$$ where $b_i\in H$ for every $i$ and $q_i$ are $n$ rationals such that $\mathbb{R}\ni x=\sum_{i=1}^n b_iq_i$.

This is a classical tool to provide additive, discontinuous real functions, of course exploiting the axiom of choice.

Finally, a Hamel basis is not only "not write-downable": it can be almost as nasty as you want. For instance, you can prove under AC that there is a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$ and a function $f:H\to\mathbb{R}$ such that, for every real $x$, $f^{-1}(x)$ has nonempty intersection with every closed, uncountable set without containing any of them (= is a Bernstein set). Or you can prove that there exists a Hames basis which is not measurable and does not have the property of Baire, and much more...

All of this is very nicely covered in Chapter 7 of:

Ciesielski, K. (1997). Set theory for the working mathematician (No. 39). Cambridge University Press. ISO 690

This is an addition to Yemon Choi answer.

In fact, a real-valued function $f$ defined on a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$, for instance, extends uniquely to ad additive function, not to a linear function (it only needs to be linear over $\mathbb{Q}$). This can be seen readily defining $$F(x)=\sum_{i=1}^n q_i f(b_i),$$ where $b_i\in H$ for every $i$ and $q_i$ are $n$ rationals such that $\mathbb{R}\ni x=\sum_{i=1}^n b_iq_i$.

This is a classical tool to provide additive, discontinuous real functions, of course exploiting the axiom of choice.

Finally, a Hamel basis is not only "not write-downable": it can be almost as nasty as you want. For instance, you can prove under AC that there is a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$ and a function $f:H\to\mathbb{R}$ such that $f^{-1}(x)$ has nonempty intersection with every closed, uncountable set (= is a Bernstein set). Or you can prove that there exists a Hames basis which is not measurable and does not have the property of Baire, and much more...

All of this is very nicely covered in Chapter 7 of:

Ciesielski, K. (1997). Set theory for the working mathematician (No. 39). Cambridge University Press. ISO 690

This is an addition to Yemon Choi answer.

In fact, a real-valued function $f$ defined on a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$, for instance, extends uniquely to ad additive function, not to a linear function (it only needs to be linear over $\mathbb{Q}$). This can be seen readily defining $$F(x)=\sum_{i=1}^n q_i f(b_i),$$ where $b_i\in H$ for every $i$ and $q_i$ are $n$ rationals such that $\mathbb{R}\ni x=\sum_{i=1}^n b_iq_i$.

This is a classical tool to provide additive, discontinuous real functions, of course exploiting the axiom of choice.

Finally, a Hamel basis is not only "not write-downable": it can be almost as nasty as you want. For instance, you can prove under AC that there is a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$ and a function $f:H\to\mathbb{R}$ such that $f^{-1}(x)$ has nonempty intersection with every closed, uncountable set without containing any of them (= is a Bernstein set). Or you can prove that there exists a Hames basis which is not measurable and does not have the property of Baire, and much more...

All of this is very nicely covered in Chapter 7 of:

Ciesielski, K. (1997). Set theory for the working mathematician (No. 39). Cambridge University Press. ISO 690