Exercise on algebra filtrations In Radford's book on Hopf-Algebras he states the following exercise (5.6.1).

Show that if $(A,\mathcal{F})$ is a filtered algebra over $k$, then $1\in V_0$.

The definition of a filtered algebra is given as follows.

Let $A$ be a (unital) algebra over $k$. A filtration $\mathcal{F}$ is a family of subspaces $\left\{V_n\right\}_{n=0}^{\infty}$ such that $A=\cup_{i\geq 0}V_i$, $V_0\subset V_1\subset V_2\subset\dots $ and $V_n\cdot V_m\subset V_{n+m}$ for all $n,m\geq 0$.

Now consider $A=\mathbb{R}^2$ as a $2$-dimensional algebra over $\mathbb{R}$ with pointwise multiplication. Then $1_A=(1,1)\in A$. Now consider $V_0=\mathbb{R}\times \left\{0\right\}$ and $V_1=\mathbb{R}\times \mathbb{R}$. Then $V_0\subset V_1$ is a filtration of $A$ but $1_A\notin V_0$. So clearly the exercise is wrong. 
Now I know what filtrations are, and most of the times one just assumes that $k\subset V_0$. (Here we identify $k$ by the subalgebra generated by $1_A$). But why is this exercise stated as such in Radford's otherwise brilliant book? Or am I missing something?
 A: I think that it is a matter of definition: I have looked at both M. Sweedler's book on Hopf algebras (p.230, 1969 edition) and E.Abe's book on Hopf algebras (p.20, 1977 edition) and they both seem to include the assumption: $1\in V_0$,  as part of the definition of a filtration on an algebra $A$ to make it a filtered algebra. 
If you do not include it as part of the definition, which is the case in your definition (I do not have a copy of Radford's book available at the time of writing this), then I doubt that $1\in V_0$ can be extracted as a consequence. Your counterexample shows that. 
P.S.: However, note that, not including the assumption $1\in V_0$ as a part of the definition of a filtration, leads on other problematic situations as well: The associated graded algebra $gr A$, of a filtered algebra $A$ whose filtration $\mathcal{F}$ is given by $V_i\subset V_{i+1}\subset A$, $i\geq 0$, with $V_i\cdot V_j\subset V_{i+j}$, is defined through:
$$
gr A=\bigoplus_{i\geq 0}A_i, \ \ \ \ A_i=V_i\Big/V_{i-1}, \textrm{ for } i>0 \textrm{ and } A_0=V_0 \ \ \ \ 
$$
If $1\in V_0$, then $gr A$ constructed above, is a $\mathbb{Z}$-graded algebra (with zero negative components). It is called the associated graded algebra of the filtered algebra. 
But the above, does not constitute a consistent $\mathbb{Z}$-grading in situations where $1\notin V_0$ (your counterexample for instance), since, in a $\mathbb{Z}$-graded algebra we must necessarily have $1\in A_0$. (More generally, for any group $G$ with neutral $\theta$ and any $G$-graded algebra $A$ we necessarily have $1_A\in A_\theta$. This is a direct consequence of the definition of grading of an algebra $A$ by a group $G$).   
A: You are correct. Given an algebra, you can always define a filtration on $A$ by $V_0 = \{0 \}$ and $V_i = A$ for $i > 0$. Then this will be a filtration according to the definition given but $1 \notin V_0$ (unless $A = \{ 0 \}$).
