Separation theorems and polynomial sets I am not really sure about the solution of the following exercise:

Let $V = \mathcal P ([0,1])$, $V \ni v = p(x) = \sum_{k=0}^{\deg p} a_k x^k$, the vector space of polynomials defined on $[0,1]$, let $A, B \subset V$ defined by $A = \{ p \in \mathcal P : a_{deg\,p} > 0 \}$ and $B = \{ p \in \mathcal P : a_{deg \, p} < 0 \}$. Show that $A$ and $B$ are disjoint convex sets but does not exist a closed hyperlane $H$ that separates (in any sense) $A$ and $B$.

Now, $A$ and $B$ are obviously convex and disjoint. Clearly, the solution relies on applying the so called geometric forms of the Hahn-Banach theorem (in terminology of Brezis's famous book "Functional analysis, Sobolev spaces and partial differential equations"), often called also separation theorems (as in Rudin, "Functional analysis"). So we have to show that $A$ and $B$ are not open nor closed. In fact, if one was open, we could apply the first geometric form of the Hahn-Banach theorem and hence show that there exists a closed hyperplane $H$ that separates $A$ and $B$, and if one was compact and the other (at least) closed, we could apply the second form, obtaining that such an hyperplane exists and stricly separates $A$ and $B$. (since compact implies closed and bounded, not closed or not bounded implies not compact, so both of them must be closed in order to apply the theorem.)
Now, taking sequences such that $a_{deg\,p_n}$ tends to zero when $n$ tends to infinity, it's easy to see that $A$ and $B$ can not be closed (they are unbounded too). To show that they are not open, I think it's easier to prove that their (relative) complement is not closed in $V$, by finding a sequence in the complement that converges out of it. Suggestions? 
(I believe it's a very simple problem, but until now I have not found a suitable sequence! Thank you for any in reasoning mistake highlighted!) 
 A: If two convex sets $C,\,D$ in a topological vector space $E$ can be separated by a closed hyperplane $H$, that means there is a continuous real linear form $\lambda \colon E \to \mathbb{R}$ such that (without loss of generality) $H = \lambda^{-1}(1)$, $\lambda(c) < 1$ for all $c \in C$, and $\lambda(d) \geqslant 1$ for all $d\in D$, then there is a convex open set that contains $C$ and doesn't intersect $D$. The immediate example of such a set is $\lambda^{-1}((-\infty,\, 1))$.
Now, in our setting, we can show that for any TVS topology on $V$, the sets $A$ and $B$ are both dense, and that means they can't be separated by a closed hyperplane by the above reasoning.
To see that $A$ and $B$ are both dense, note that every neighbourhood of $0$ contains polynomials of arbitrarily high degree - since a neighbourhood of $0$ is absorbing - and since it contains a balanced neighbourhood of $0$, polynomials of arbitrarily high degree with both positive and negative leading coefficient.
Thus, for every polynomial $p$ and every neighbourhood $U$ of $0$, the neighbourhood $p + U$ of $p$ intersects both, $A$ and $B$, hence both are dense.
