这是一份Sydney悉尼大学MATH1002/MATH1014/ MATH1902 的成功案例
Let $V$ be the vector space over $\mathbb{Z}{2}$ with basis $$ e{k}=(0, \ldots, 0,1,0, \ldots)
$$
where the 1 is in the $k$ th position. Thus, $V$ is the set of all infinite binary sequences with a finite number of 1 ‘s. Define the order $o(v)$ of any $v \in V$ to be the largest coordinate of $v$ with value 1 . Then $o(v)<\infty$ for all $v \in V$. Consider the dual vectors $e_{k}^{}$, defined (as usual) by $$ e_{k}^{}\left(e_{j}\right)=\delta_{k, j} $$ For any $v \in V$, the evaluation functional $\bar{v}$ has the property that $$ \bar{v}\left(e_{k}^{}\right)=e_{k}^{}(v)=0 \text { if } k>o(v)
$$
However, since the dual vectors $e_{k}^{}$ are linearly independent, there is a linear functional $f \in V^{ }$ for which $$ f\left(e_{k}^{}\right)=1
$$
for all $k \geq 1$. Hence, $f$ does not have the form $v$ for any $v \in V$. This shows that the canonical map is not surjective and so $V$ is not algebraically reflexive.
MATH1002/MATH1014/ MATH1902 COURSE NOTES ：
I) For $\tau, \sigma \in \mathcal{L}(V, W)$ and $a, b \in F$.
$$
(a \tau+b \sigma)^{x}=a \tau^{x}+b \sigma^{\times}
$$
2) For $\sigma \in \mathcal{L}(V, W)$ and $\tau \in \mathcal{L}(W, U)$.
$$
(\tau \sigma)^{x}=\sigma^{x} \tau^{x}
$$
3) For any imertible $\tau \in \mathcal{L}(V)$.
$$
\left(\tau^{-1}\right)^{x}=\left(\tau^{x}\right)^{-1}
$$
Proof of part 1) is left for the reader. For part 2), we have for all $f \in U^{*}$,
$$
(\tau \sigma)^{x}(f)=f(\tau \sigma)=\sigma^{\times}(f \tau)=\sigma^{\times}\left(\tau^{\times}(f)\right)=\left(\sigma^{\times} \tau^{\times}\right)(f)
$$$$
\tau^{x}\left(\tau^{-1}\right)^{x}=\left(\tau^{-1} \tau\right)^{\times}=\iota^{x}=\iota
$$
and in the same way, $\left(\tau^{-1}\right)^{x} \tau^{x}=\iota$. Hence $\left(\tau^{-1}\right)^{x}=\left(\tau^{x}\right)^{-1} . \square$