If I understand the whole Hamel basis idea correctly, there exists one such basis $B = \{v_\alpha\}_{\alpha \in I}$ for ${\mathbb R}$ (herein construed as a vector space of ${\mathbb Q}$), such that $1 = v_0 \in B$. (Here I'm using $I$ to denote some suitable index set; the $0$ subscript in $v_0$ just stands for one element of $I$.)

The span of $v_0 = 1$ in ${\mathbb R}$ is thus ${\mathbb Q}$.

This means that every $x \in {\mathbb R}$ can be expressed uniquely as a linear combination of the form:

$$x = q_0 + \sum_{\alpha \in I \backslash \{0\}} q_{\alpha} \cdot v_{\alpha}$$

Let $P_{\mathbb Q}:{\mathbb R}\rightarrow {\mathbb Q}$ be the projection of ${\mathbb R}$ onto ${\mathbb Q}$. Namely, using the same notation as in the expression for the decomposition of any real $x$ above,

$$P_{\mathbb Q}(x) = q_0$$

My first question is:

can it be proved or disproved whether the projection $P_{\mathbb Q}$, or at least its nullspace, ${\mathcal N}(P_{\mathbb Q})$, is independent of the choice of $B\backslash \{1\}$?

My second question is:

assuming that ${\mathcal N}(P_{\mathbb Q})$ is in fact independent of the choice of $B\backslash \{1\}$, is there a special name given to this subspace of ${\mathbb R}$?

  • $\begingroup$ This is an interesting question, but it's not really a question about set theory or about the axiom of choice. $\endgroup$
    – Asaf Karagila
    Commented Feb 16, 2013 at 14:46
  • $\begingroup$ Simpler question: take $V = \mathbb{Q}[\sqrt{2}]$ and a basis $\{1,\alpha\}$ of $V$. Does the null-space of the corresponding projection to $\mathbb{Q}$ depend on the irrational basis vector? $\endgroup$
    – Martin
    Commented Feb 16, 2013 at 14:48
  • $\begingroup$ @AsafKaragila: it is for me almost impossible to tell whether a question is about the AC, since it is often the case that the AC lurks where I least expect it... $\endgroup$
    – kjo
    Commented Feb 16, 2013 at 14:49
  • $\begingroup$ @kjo: Yes, but the tag is mainly for questions about explicit uses of AC, or independence results from it; not about unrelated issues where AC is used. If that was the case a lot more questions needed to be tagged [axiom-of-choice]! :-) $\endgroup$
    – Asaf Karagila
    Commented Feb 16, 2013 at 14:50
  • 1
    $\begingroup$ Note that Vitali sets are related to, but not exactly the same as, the null spaces you ask about in the second question. $\endgroup$ Commented Feb 16, 2013 at 16:27

1 Answer 1


The projection and its null space both depend on the choice of basis.

The set $\{1,\pi\}$ is linearly independent over $\mathbb Q$, so we can extend it to a basis. That basis projects $\pi$ to 0 and $1+\pi$ to 1.

On the other hand $\{1,1+\pi\}$ is also linearly independent and can also be extended to a basis. That one projects $\pi$ to $-1$ and $1+\pi$ to 0.

The second part of the question is then moot.


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