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I recognize the fact that the natural map from an infinite dimension vector space $V$ to it's double dual space $V^{**}$ need not necessarily to be surjective because we don't have that the $\dim V$ = $\dim V^{*}$.

However, what is a simple example to show that it is not surjective in this case?

My first instinct was to pick the vector space $V = (\mathbb{Z}/\mathbb{2Z})^{\mathbb{N}}$ just a simple vector space.

However, when picking up a functional in the dual space which wouldn't be isomorphic to the natural map for any $v \in V$ I wasn't able to write one down.

Can someone help me come up with an example / some intuition as to why this happens?

Thanks.

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You can pick any infinite dimensional vector space. Say, take $V=k[x]$ to be just polynomials in one variable. It is countable-dimensional. If you take its dual, you get $V^*=k[[x]]$, which is already uncountable-dimensional. When taking double dual $V^{**}$ it will be even bigger. So there is no way $V$ will be isomorphic to $V^{**}$, you don't even have to worry about how exactly the map $V\to V^{**}$ looks like.

And this is true in general for infinite-dimensional vector spaces. Their linear duals are getting too big.

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  • $\begingroup$ Is there any example which directly relates to what it looks like with the linear map as well? - And thanks so much, makes more sense now. $\endgroup$ – Nick R Mar 11 '14 at 16:14
  • $\begingroup$ In my example with polynomials, the canonical map $V\to V^{**}$ is a map, which sends each basis vector $x^n$ to the linear function $f_n$ on $V^*=k[[x]]$, defined by $f_n(\sum\limits_{i=0}^\infty c_ix^i)=c_n$. $\endgroup$ – Sasha Patotski Mar 11 '14 at 16:21

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