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Iosif Pinelis
Iosif Pinelis
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$\newcommand\R{\Bbb R}$Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set?

Of course, such a set $S$, if it exists, must be of measure $0$.

"The axiom of choice implies the existence of sets of reals that do not have the perfect set property", so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?

On the other hand, it "is well-known that assuming the countable axiom of choice that every uncountable Borel set contains a [nonempty] perfect set." So, a set $S$ in question, if it exists, cannot be Borel.

$\newcommand\R{\Bbb R}$Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set?

Of course, such a set $S$, if it exists, must be of measure $0$.

"The axiom of choice implies the existence of sets of reals that do not have the perfect set property", so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?

$\newcommand\R{\Bbb R}$Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set?

Of course, such a set $S$, if it exists, must be of measure $0$.

"The axiom of choice implies the existence of sets of reals that do not have the perfect set property", so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?

On the other hand, it "is well-known that assuming the countable axiom of choice that every uncountable Borel set contains a [nonempty] perfect set." So, a set $S$ in question, if it exists, cannot be Borel.

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LSpice
LSpice
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$\newcommand\R{\Bbb R}$

Assuming$\newcommand\R{\Bbb R}$Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set?

Of course, such a set $S$, if it exists, must be of measure $0$.

"The axiom of choice implies the existence of sets of reals that do not have the perfect set property", so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?

$\newcommand\R{\Bbb R}$

Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set?

Of course, such a set $S$, if it exists, must be of measure $0$.

"The axiom of choice implies the existence of sets of reals that do not have the perfect set property", so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?

$\newcommand\R{\Bbb R}$Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set?

Of course, such a set $S$, if it exists, must be of measure $0$.

"The axiom of choice implies the existence of sets of reals that do not have the perfect set property", so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?

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Iosif Pinelis
Iosif Pinelis
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An uncountable measurable subset of $\Bbb R$ containing no nonempty perfect set

$\newcommand\R{\Bbb R}$

Assuming the axiom of choice, is there an uncountable Lebesgue-measurable subset $S$ of $\R$ that contains no nonempty perfect set?

Of course, such a set $S$, if it exists, must be of measure $0$.

"The axiom of choice implies the existence of sets of reals that do not have the perfect set property", so that there is an uncountable subset of $\R$ that contains no perfect set; but can such a set be Lebesgue-measurable?