Question regarding an example of disjoint convex sets that cannot be separated If we consider two sets over $\ell^2$
$$
\Gamma = \{x\in \ell^2 |x_1 \geq n |x_n - n^{-\frac{2}{3}}|\}
$$
for $n\geq2$and $\Sigma$ being the $x_1$-axis. I need to show that $\Gamma$ and $\Sigma$ are disjoint convex sets for which no functional $g \in (\ell^2)'$ exists such that if $\gamma \in \Gamma$ and $\sigma \in \Sigma$, 
$$
g(\gamma)<g(\sigma)
$$
I believe that I need to show that $\Gamma - \Sigma$ is dense in $\ell^2$, but I'm not sure how to do this and how I can proceed from this. 
I've been stuck on this exercise for quite some time now, would appreciate any help.
 A: Disjointness of $\Gamma$ and $\Sigma$ is clear (also for the second part I suppose $\Sigma$ does not allow vectors with negative $x_1$ component). Convexity of $\Sigma$ is clear. Convexity of $\Gamma$: if $x, y \in \Gamma$, $\alpha, \beta \in [0,1]$ with $\alpha+\beta=1$
$$\alpha x_1 + \beta y_1 ≥ n(\alpha |x_n-1/n^{2/3}| + \beta |y_n -1/n^{2/3}|)≥ n\cdot\left| \alpha x_n + \beta y_n -(\alpha + \beta)n^{2/3} \right|$$
and $\alpha x + \beta y \in \Gamma$ follows from $\alpha + \beta = 1$.
Now for the second part:
Suppose there exists such a $g$. Note that any $\sigma \in \Sigma$ can be written as $\sigma_1 e_1$ with $\sigma_1≥0$. So $g(\sigma)=\sigma_1 g(e_1)$. It follows that $g(e_1)$ cannot be positive, as otherwise we could chose a $\sigma_1$ to make $g(\sigma)$ as big as we like, certainly bigger than some value of $g(\gamma)$.
Let $\gamma := \sum_{n\neq1} 1/n^{2/3} e_n \in \Gamma$ and denote $g(\gamma)=c$. This must be positive, as otherwise $g(0)=0>c$. From this positivity it follows that $g(e_n)\neq 0$ for at least one $n$ (linear functionals are continuous).
If $x \in \Gamma$ then $n |x_n -1/n^{2/3}|$ must be a bounded function, so write $x=a e_1 + \gamma + \sum_{n\neq 1} \frac{f(n)}n e_n $, in which $|f(n)|$ is bounded and $a≥\sup_n|f(n)|$. Then
$$g(x)≤c+\sum_{n\neq1}\frac{f(n)}ng(e_n)$$
Since $g(e_1)≤0$. But remember that $g(e_\tilde n)\neq0$ for some $\tilde n$ must hold. Take then $$f(n)=\begin{cases}-\frac{\tilde n\ c}{g(e_n)} & n=\tilde n\\0 & n\neq \tilde n \end{cases}$$
You then get $g(x)≤0=g(0)$.
