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Vladimir
Vladimir
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Another useful example is the statement that $M$ and $M^2$ have the same power for any infinite set $M$. This is proved with the use of the axiom of choice. This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the $l_1$-basis in $G$ has the same power as $M$. (See paper of mine in: Siegel W., Duplij S. and Bagger J., eds. (2004), Concise Encyclopedia of Supersymmetry And Noncommutative Structures in Mathematics and Physics, Berlin, New York: Springer-Verlag; math-ph/0009006.) To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.

Another useful example is the statement that $M$ and $M^2$ have the same power for any infinite set $M$. This is proved with the use of the axiom of choice. This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the $l_1$-basis in $G$ has the same power as $M$. To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.

Another useful example is the statement that $M$ and $M^2$ have the same power for any infinite set $M$. This is proved with the use of the axiom of choice. This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the $l_1$-basis in $G$ has the same power as $M$. (See paper of mine in: Siegel W., Duplij S. and Bagger J., eds. (2004), Concise Encyclopedia of Supersymmetry And Noncommutative Structures in Mathematics and Physics, Berlin, New York: Springer-Verlag; math-ph/0009006.) To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.

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Vladimir
Vladimir
  • 371
  • 2
  • 8

Another useful example is the statement that $M$ and $M^2$ have the same power for any infinite set $M$. This is proved with the use of the axiom of choice. This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the (Schauder$l_1$-like) basisbasis in $G$ has the same power as $M$. To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.

Another useful example is the statement that $M$ and $M^2$ have the same power for any infinite set $M$. This is proved with the use of the axiom of choice. This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the (Schauder-like) basis in $G$ has the same power as $M$. To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.

Another useful example is the statement that $M$ and $M^2$ have the same power for any infinite set $M$. This is proved with the use of the axiom of choice. This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the $l_1$-basis in $G$ has the same power as $M$. To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.

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Vladimir
Vladimir
  • 371
  • 2
  • 8

Another useful example is the statement that $M$ and $M^2$ have the same power for any infinite set $M$. This is proved with the use of the axiom of choice. This fact have some applications in supermathematics. If infinite set $M$ is the index set for generators of a Grassmann-Banach algebra $G$ with an $l_1$-norm then the (Schauder-like) basis in $G$ has the same power as $M$. To circumvent the axiom of choice in this example one needs to consider $M$ to be a well-ordered set. But in general the construction of $G$ does not need a linear ordering in $M$. Of course, when physicists calculate Berezin's integral the power of $M$ and the axiom of choice are under the carpet of the calculations but the foundations (the flooring under the ``carpet'') need the axiom of choice.