Suppose that $k$ is a field. Consider the vector space $k^{\times \mathbb{N}}:= \{(x_0, x_1, x_2,…) ∣ x_i \in k\}$ and its subspace $k^{\oplus N}:= \{(x_0, x_1, x_2, . . .) \in k^\mathbb{N}∣ x_i ≠ 0$ for finitely many $i\}$. Prove that $(k^{\oplus \mathbb{N}})^*\simeq k^{\times \mathbb{N}}$. ($(k^{\oplus \mathbb{N}})^*$ is dual space of $k^{\oplus\mathbb{N}}$)
Attempt:
We can take bases of $k^{\oplus\mathbb{N}}$ as infinite dimensional vectors such that every basis has a coordinate $1$ on the place where $x_i\neq 0$ and all other coordinates are $0$. Since vectors in $k^{\oplus\mathbb{N}}$ have finitely many coordinates that are not equal to $0$, then the dual bases are $f_1=x_{i_1}, f_2=x_{i_2},…, f_n=x_{i_n}$, which is finite. My question is that why this finite dimensional vector space can be isomorphic to infinite dimensional vector space $k^{\times\mathbb{N}}$.
Best Answer
You have the wrong idea about dual bases. For any $i\in \mathbb{N}$, there is a linear form $f_i$ which takes $x=(x_0,x_1,\dots)$ to $x_i\in k$. So there are an infinite number of those $f_i$. You seem to be confused by the fact that for a given vector $x$, only a finite number of the $f_i(x)$ will be non-zero. This is true: if you fix $x$, only a finite number of $f_i$ matter; but for any $i$, $f_i$ is non-zero on some vector, so it is not a trivial linear form.
This being said, the subtlety of the exercise lies elsewhere: the $f_i$ are linearly independent, but they are not a basis of the dual (there is no such thing as a "dual basis" in infinite dimension, unless topology enters the mix). In infinite dimension, the dual is bigger than the original space.
To see this, take any linear form $f$. It is characterized by its values $f(e_i)$ where $e_i=(0,\dots,0,1,0,\dots)$. And for any sequence $(a_i)$, you can define $f$ by the fact that $f(e_i)=a_i$. Even when an infinite number of $a_i$ are non-zero. This is why the dual is isomorphic to $k^{\times \mathbb{N}}$ and not $k^{\oplus \mathbb{N}}$ (which would be the case if we only allowed the sequences $a_i$ with a finite number of non-zero terms).
The key point to understand is that even if all $a_i$ are non-zero, for a fixed $x$, we have a finite decomposition $x=\sum x_ie_i$, so $f(x)=\sum x_ia_i$ is a finite sum. So even though $f$ requires an infinite number of parameters, when applying $f$ to any vector, only a finite number of these parameters intervene.