Skip to main content
MathOverflow

Return to Question

added tag as this could be of interest to analysts
Link
Abdelmalek Abdesselam
Abdelmalek Abdesselam
  • 23k
  • 1
  • 67
  • 127
deleted 7 characters in body
Source Link
David E Speyer
David E Speyer
  • 156.2k
  • 14
  • 419
  • 763

Let $k$ be a field and $V$ a $k$-vector space. Then there is an injectiona map $V \to V^{\ast \ast}$, where $V^{\ast}$ is the dual vector space. If we are in ZFC and $\dim V$ is infinite, then this map is not surjective. As we learned in this question, there are models of $ZF$ where $V \to V^{\ast \ast}$ is an isomorphism when $V$ has a countable basis. I think the same argument shows that it is consistent with ZF that this is an isomorphism whenever $V$ has a basis.

Is it consistent with $ZF$ that $V \to V^{\ast \ast}$ is an isomorphism for all vector spaces $V$?

I ask because I'm teaching a rigorous undergrad analysis class. My students keep asking me whether they have to believe that $V \to V^{\ast \ast}$ can fail to be an isomorphism. Of course, I'm trying to change their intuition to point out why most mathematicians find the failure of isomorphism plausible and point out that there are more subtle ways to salvage the claim, such as Hilbert spaces, but I'd also love to be able to give them a choice free proof that there is some vector space where this issue comes up.

Let $k$ be a field and $V$ a $k$-vector space. Then there is an injection $V \to V^{\ast \ast}$, where $V^{\ast}$ is the dual vector space. If we are in ZFC and $\dim V$ is infinite, then this map is not surjective. As we learned in this question, there are models of $ZF$ where $V \to V^{\ast \ast}$ is an isomorphism when $V$ has a countable basis. I think the same argument shows that it is consistent with ZF that this is an isomorphism whenever $V$ has a basis.

Is it consistent with $ZF$ that $V \to V^{\ast \ast}$ is an isomorphism for all vector spaces $V$?

I ask because I'm teaching a rigorous undergrad analysis class. My students keep asking me whether they have to believe that $V \to V^{\ast \ast}$ can fail to be an isomorphism. Of course, I'm trying to change their intuition to point out why most mathematicians find the failure of isomorphism plausible and point out that there are more subtle ways to salvage the claim, such as Hilbert spaces, but I'd also love to be able to give them a choice free proof that there is some vector space where this issue comes up.

Let $k$ be a field and $V$ a $k$-vector space. Then there is a map $V \to V^{\ast \ast}$, where $V^{\ast}$ is the dual vector space. If we are in ZFC and $\dim V$ is infinite, then this map is not surjective. As we learned in this question, there are models of $ZF$ where $V \to V^{\ast \ast}$ is an isomorphism when $V$ has a countable basis. I think the same argument shows that it is consistent with ZF that this is an isomorphism whenever $V$ has a basis.

Is it consistent with $ZF$ that $V \to V^{\ast \ast}$ is an isomorphism for all vector spaces $V$?

I ask because I'm teaching a rigorous undergrad analysis class. My students keep asking me whether they have to believe that $V \to V^{\ast \ast}$ can fail to be an isomorphism. Of course, I'm trying to change their intuition to point out why most mathematicians find the failure of isomorphism plausible and point out that there are more subtle ways to salvage the claim, such as Hilbert spaces, but I'd also love to be able to give them a choice free proof that there is some vector space where this issue comes up.

Source Link
David E Speyer
David E Speyer
  • 156.2k
  • 14
  • 419
  • 763

Is it consistent with ZF that $V \to V^{\ast \ast}$ is always an isomorphism?

Let $k$ be a field and $V$ a $k$-vector space. Then there is an injection $V \to V^{\ast \ast}$, where $V^{\ast}$ is the dual vector space. If we are in ZFC and $\dim V$ is infinite, then this map is not surjective. As we learned in this question, there are models of $ZF$ where $V \to V^{\ast \ast}$ is an isomorphism when $V$ has a countable basis. I think the same argument shows that it is consistent with ZF that this is an isomorphism whenever $V$ has a basis.

Is it consistent with $ZF$ that $V \to V^{\ast \ast}$ is an isomorphism for all vector spaces $V$?

I ask because I'm teaching a rigorous undergrad analysis class. My students keep asking me whether they have to believe that $V \to V^{\ast \ast}$ can fail to be an isomorphism. Of course, I'm trying to change their intuition to point out why most mathematicians find the failure of isomorphism plausible and point out that there are more subtle ways to salvage the claim, such as Hilbert spaces, but I'd also love to be able to give them a choice free proof that there is some vector space where this issue comes up.