# base for finite dimensional vector space is not infinite dimensional vector space?

Let $V$ be a infinite dimensional vector space over a field $K$, and $\{v_i\}_{i\in I}$ be a basis of $V$. For each $i\in I$, let $f_i: V\to K$ be defined by $f_i(v_j)=\delta_{ij}$. Prove that $\{f_i\}_{i\in I}$ is linearly independent but does not span the dual space $V^*$.

I can prove that $\{f_i\}_{i\in I}$ is linearly independent. But I cannot find a counterexample in $V^*$ that is not spanned by $\{f_i\}_{i\in I}$. Does anyone has any idea on how to create this counterexample? This is the difference between finite and infinite dimensional vector space, right?

Thank you very much!

• Remember that linear combinations only have finitely many non-zero coefficients, even in infinite-dimensional vector spaces. Mar 21 '15 at 21:30
Let $V$ the real vector space generated by an infinite set $A$. By definition, $$V:=\{f:A\rightarrow\mathbb{R}\,\mbox{ s.t. } f(a)\neq 0\mbox{ for finitely many }a\in A\}$$ A basis of $V$ is $\{e_a\}_{a\in A}$, where $e_{a_i}(a_j)=\delta_{i,j}$. Let's denote by $e_a'$ the corresponding elements of $V^*$, i.e. $e_{a_i}'(e_{a_j})=\delta_{i,j}$. Now consider $\psi\in V^*$, defined by $$\psi(f):=\sum_{a\in A}f(a)$$ Observe that $\psi$ is well defined cause the sum involves only a finite number of terms. Suppose now that $\psi$ can be written as a (finite) linear combination of $e_a'$. Hence $$\psi=\sum_{a\in B}\lambda_ae_a'$$ where $B$ is a finite subset of $A$, and $\lambda_a$ are scalars. Taking $\bar{a}\not\in B$ (it exists cause $A$ is infinite while $B$ is finite), and applying the last identity to $e_\bar{a}$, we get $1=0$, contradiction. In general, $V^*\equiv \mathbb{R}^A$ (the set of all real-valued functions on $A$), while the space spanned by $\{e_a'\}_{a\in A}$ is isomorphic to $V$ (the set of real-valued functions on $A$ which are $0$ for all but finitely many $a\in A$).