|
5
|
|
edited Nov 17 2010 at 3:36 |
It is consistent with the axioms of ZFC that there is a
Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ with
complexity $\Delta^1_2$ in the descriptive set theoretic
hierarchy. This is true, for example, in the constructible
universe $L$, where there is a $\Delta^1_2$ well-ordering
of the reals, as I explain in this MO
answer,
which is closely related to this question. Complexity $\Delta^1_2$ is a surprisingly low complexity, since such a set (and its complement) can both be obtained by starting with a certain closed set in $\mathbb{R}^3$, projecting it to $\mathbb{R}^2$, taking the complement, and projecting down to $\mathbb{R}$, and so it would seem to count as fairly explicit.
Meanwhile, there can never be a Hamel basis of $\mathbb{R}$
over $\mathbb{Q}$ that is Borel, that is, with complexity
$\Delta^1_1$, since from any Borel Hamel basis one can produce a
non-Lebesgue measurable set of the same complexity by the Vitali argument (remove
an element, take the span of the other elements, and
consider its cosets). But of course every Borel set is
Lebesgue measurable.
At the same time, it is a consequence of the existence of large
cardinals that every projective set of reals is Lebesgue
measurable, and in this case, there can be no projective
Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, again by the Vitali argument. The
projective hierarchy of sets arises by closing the Borel
sets under continuous images, as well as complements,
countable unions and intersections. Thus, in such a
situation, there can be no easily-described Hamel basis for
$\mathbb{R}$ over $\mathbb{Q}$.
|
|
|
|
|
4
|
|
edited Nov 17 2010 at 3:34 |
It is consistent with the axioms of ZFC that there is a
Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ with
complexity $\Delta^1_2$ in the descriptive set theoretic
hierarchy. This is true, for example, in the constructible
universe $L$, where there is a $\Delta^1_2$ well-ordering
of the reals, as I explain in this MO
answer,
which is closely related to this question. Complexity $\Delta^1_2$ is a surprisingly low complexity, since such a set (and its complement) can both be obtained by starting with a certain closed set in $\mathbb{R}^3$, projecting it to $\mathbb{R}^2$, taking the complement, and projecting down to $\mathbb{R}$, and so it would seem to count as fairly explicit.
Meanwhile, there can never be a Hamel basis of $\mathbb{R}$
over $\mathbb{Q}$ that is Borel, that is, with complexity
$\Delta^1_1$, since from any Hamel basis one can produce a
non-Lebesgue measurable set of the same complexity by the Vitali argument (remove
an element, take the span of the other elements, and
consider its cosets), and if the Hamel basis is Borel, then this set will be analytic $\Sigma^1_1$. . But of course every analytic Borel set is
Lebesgue measurable.
At the same time, it is a consequence of the existence of large
cardinals that every projective set of reals is Lebesgue
measurable, and in this case, there can be no projective
Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, again by the Vitali argument. The
projective hierarchy of sets arises by closing the Borel
sets under continuous images, as well as complements,
countable unions and intersections. Thus, in such a
situation, there can be no easily-described Hamel basis for
$\mathbb{R}$ over $\mathbb{Q}$.
|
|
|
|
|
3
|
|
edited Nov 17 2010 at 2:58 |
It is consistent with the axioms of ZFC that there is a
Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ with
complexity $\Delta^1_2$ in the descriptive set theoretic
hierarchy. This is true, for example, in the constructible
universe $L$, where there is a $\Delta^1_2$ well-ordering
of the reals, as I explain in this MO
answer,
which is closely related to this question. Complexity $\Delta^1_2$ is a surprisingly low complexity, since such a set (and its complement) can both be obtained by starting with a certain closed set in $\mathbb{R}^3$, projecting it to $\mathbb{R}^2$, taking the complement, and projecting down to $\mathbb{R}$, and so it would seem to count as fairly explicit.
Meanwhile, there can never be a Hamel basis of $\mathbb{R}$
over $\mathbb{Q}$ that is Borel, that is, with complexity
$\Delta^1_1$, since from any Hamel basis one can produce a
non-Lebesgue measurable set of the same complexity by the Vitali argument (remove
an element, take the span of the other elements, and
consider its cosets). , and if the Hamel basis is Borel, then this set will be analytic $\Sigma^1_1$. But of course every Borel analytic set is Lebesgue measurable.
At the same time, it is a consequence of the existence of large
cardinals that every projective set of reals is Lebesgue
measurable, and in this case, there can be no projective
Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, again by the Vitali argument. The
projective hierarchy of sets arises by closing the Borel
sets under continuous images, as well as complements,
countable unions and intersections. Thus, in such a
situation, there can be no easily-described Hamel basis for
$\mathbb{R}$ over $\mathbb{Q}$.
|
|
|
|
2
|
|
edited Nov 14 2010 at 19:57 |
It is consistent with the axioms of ZFC that there is a
Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ with
complexity $\Delta^1_2$ in the descriptive set theoretic
hierarchy. This is true, for example, in the constructible
universe $L$, where there is a $\Delta^1_2$ well-ordering
of the reals, as I explain in this MO
answer,
which is closely related to this question. Complexity $\Delta^1_2$ is a surprisingly low complexity, since such a set (and its complement) can both be obtained by starting with a certain closed set in $\mathbb{R}^3$, projecting it to $\mathbb{R}^2$, taking the complement, and projecting down to $\mathbb{R}$, and so it would seem to count as fairly explicit.
Meanwhile, there can never be a Hamel basis of $\mathbb{R}$
over $\mathbb{Q}$ that is Borel, that is, with complexity
$\Delta^1_1$, since from any Hamel basis one can produce a
non-Lebesgue measurable set of the same complexity by the Vitali argument (remove
an element, take the span of the other elements, and
consider its cosets). But of course every Borel set is
Lebesgue measurable.
Furthermore
At the same time, it is a consequence of the existence of large
cardinals that every projective set of reals is Lebesgue
measurable, and in this case, there can be no projective
Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$. \mathbb{Q}$, again by the Vitali argument. The
projective hierarchy of sets arises by closing the Borel
sets under continuous images, as well as complements,
countable unions and intersections. Thus, in such a
situation, there can be no easily-described Hamel basis for
$\mathbb{R}$ over $\mathbb{Q}$.
|
|
|
|
|
1
|
|
answered Nov 14 2010 at 19:47 |
It is consistent with the axioms of ZFC that there is a
Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ with
complexity $\Delta^1_2$ in the descriptive set theoretic
hierarchy. This is true, for example, in the constructible
universe $L$, where there is a $\Delta^1_2$ well-ordering
of the reals, as I explain in this MO
answer,
which is closely related to this question.
Meanwhile, there can never be a Hamel basis of $\mathbb{R}$
over $\mathbb{Q}$ that is Borel, that is, with complexity
$\Delta^1_1$, since from any Hamel basis one can produce a
non-Lebesgue measurable set of the same complexity (remove
an element, take the span of the other elements, and
consider its cosets). But of course every Borel set is
Lebesgue measurable.
Furthermore, it is a consequence of the existence of large
cardinals that every projective set of reals is Lebesgue
measurable, and in this case, there can be no projective
Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$. The
projective hierarchy of sets arises by closing the Borel
sets under continuous images, as well as complements,
countable unions and intersections. Thus, in such a
situation, there can be no easily-described Hamel basis for
$\mathbb{R}$ over $\mathbb{Q}$.
|
|
|
