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As the thread was bumped anyway, might as well http->https the link.
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Asaf Karagila ♦
Asaf Karagila ♦
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Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.

The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.

Note that you don't need to go as far as Mycielski and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See http://karagila.org/2020/countable-sets-of-reals/https://karagila.org/2020/countable-sets-of-reals/ for details.)

Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.

The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.

Note that you don't need to go as far as Mycielski and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See http://karagila.org/2020/countable-sets-of-reals/ for details.)

Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.

The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.

Note that you don't need to go as far as Mycielski and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See https://karagila.org/2020/countable-sets-of-reals/ for details.)

corrected a typo
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bof
bof
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Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.

The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.

Note that you don't need to go as far as MycielskyMycielski and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See http://karagila.org/2020/countable-sets-of-reals/ for details.)

Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.

The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.

Note that you don't need to go as far as Mycielsky and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See http://karagila.org/2020/countable-sets-of-reals/ for details.)

Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.

The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.

Note that you don't need to go as far as Mycielski and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See http://karagila.org/2020/countable-sets-of-reals/ for details.)

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Asaf Karagila ♦
Asaf Karagila ♦
  • 39.7k
  • 8
  • 134
  • 282

Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.

The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.

Note that you don't need to go as far as Mycielsky and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See http://karagila.org/2020/countable-sets-of-reals/ for details.)