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Question

Give an example of an infinite dimensional vector space $V$ and show that it is NOT isomorphic to $V^\ast$.

Attempt

Consider the vector space of all polynomials with real coefficients, $\mathbb{R}[x]$. This is an infinite-dimensional vector space since there is no finite set of polynomials that can form a basis for $\mathbb{R}[x]$. We claim that $\mathbb{R}[x]$ is not isomorphic to its dual, $\mathbb{R}[x]^\ast$.

The dual space $\mathbb{R}[x]^\ast$ consists of all linear functionals from $\mathbb{R}[x]$ to $\mathbb{R}$. Notice that $\mathbb{R}[x]$ has countable basis (e.g. the set of monomials $\{1,x, x^2,…\}$). However, if we define linear functionals that evaluate the polynomials at each real number $r\in \mathbb{R}$ by $L_r(p)=p(r)$ for $p\in \mathbb{R}[x]$, we see that there are uncountably many such functionals (since there are uncountably many choices for $r$).

Mr Prof
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    Countable basis doesn't imply that the space itself it countable. $\mathbb{R}[x]$ is clearly uncountable, as it contains $\mathbb{R}$. So I don't get the contradiction here. – Mark Apr 01 '24 at 00:37
  • @Mark The intended contradiction is that the functionals are linearly independent, and so the minimal cardinality of a basis for the dual must be uncountable. – legionwhale Apr 01 '24 at 01:15
  • legionwhale: Whatever you may think was the "intended" contradiction was not stated in the question. – Dan Asimov Apr 01 '24 at 03:07

2 Answers2

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I think that your argument should work, though you should also prove that the functionals $L_r$ are bounded, so that they lie in the dual space. I'm not sure if this would be easy or not.

Here is an indirect proof which works by essentially the same observation. As you say, $\mathbb{R}[x]$ has a countable (Hamel) basis. However, the dual $\mathbb{R}[x]^*$ is a Banach space, so any Hamel basis for $\mathbb{R}[x]^*$ must be uncountable by the Baire category theorem. Therefore, the spaces cannot be isomorphic.

legionwhale
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Your reasoning seems correct. Let's formalize your statements with infinite cardinals. Since you consider $V = \Bbb{R}[x]$ with the monomial basis, one has $\dim V = |\Bbb{N}| = \aleph_0$ and thus $|V| = |\Bbb{R}|^{\dim V} = \mathfrak{c}^{\aleph_0} = \mathfrak{c}$. Next, given that $V^*$ contains the linear functionals over $V$, as you recalled, one finds that $|V^*| = |\Bbb{R}|^{|V|} = \mathfrak{c}^\mathfrak{c} = 2^\mathfrak{c}$, hence $|V| < |V^*|$ and no possible surjection and a fortiori isomorphism. This result could be stated equivalently by saying $\dim V < \dim V^* = |\Bbb{R}|^{\dim V} = \mathfrak{c}$ (cf. "Erdős–Kaplansky theorem").

Abezhiko
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