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This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.

See my answer to Universally measurable sets of $\mathbb{R}^2$Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.

This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.

See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.

This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.

See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.

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Martin
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This is more ofshould probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.

See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.

This is more of a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.

See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details.

This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.

See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.

Source Link
Martin
  • 1.3k
  • 13
  • 18

This is more of a comment to Michael Greinecker's answer, but I do not have the necessary privileges.

Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.

Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.

See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details.