TITLE: When is $\dim(V^*)\gt \dim(V)$ $?$
QUESTION [5 upvotes]: Let $V$  be  an  infinite dimensional vector space  over  $\mathbb Z_2$  with  countably  infinite  basis  $\{e_1,e_2,e_3,.....,e_n,....\}$ and  write  $$V=S+T$$  s.t. $$S=\langle e_1 \rangle$$  and  $$T=\langle e_2,e_3,...,e_n,...\rangle.$$
Then $$T^*=S^0$$  is  easy  to  see but  how  can  I  prove  that $$T^*\cong V^*$$ and  $$\dim(V^*)\gt \dim(V)\ ?$$  
Clearly $V^*$  has  a  function  that  takes  $e_1$  to  $1$  and  the  rest  to  $0$  which  is  not   in  $T^*$ .  Although  that  is  not  going  to  stop  isomorphism  as  the  sets  are  infinite.
But  how  to  prove  $?$

REPLY [3 votes]: To address the question in your title: note that for any sequence $(x_n) \subset \{0,1\}^{\Bbb N}$, we can define an element of $V^*$ by
$$
e_n \mapsto x_n
$$
And all such elements are linearly independent.
As such, $V^*$ has dimension $2^{|\Bbb N|}$, which is uncountable and therefore greater than $\dim(V) = |\Bbb N|$.
Point to consider: why is $\sum x_n e_n$ not generally an element of $V$?