I am reading Approximate linear discrimination via logistic modeling in the Section 8.6.1 of B & V's Convex Optimization book. On Page 428,
$$ \operatorname{minimize} \ -l(a, b) \tag{8.27} $$
with variables $a$, $b$, where $l$ is the log-likelihood function
$$ \begin{aligned} &l(a, b)=\sum_{i=1}^{N}\left(a^{T} x_{i}-b\right)-\sum_{i=1}^{N} \log \left(1+\exp \left(a^{T} x_{i}-b\right)\right)-\sum_{i=1}^{M} \log \left(1+\exp \left(a^{T} y_{i}-b\right)\right) \end{aligned} $$
It says that if two sets can be linearly seperated, i.e., if there exist $a$, $b$ with $a^T x_i > b$ and $a^T y_i < b$, then the optimization problem (8.27) is unbounded below. Why is it unbounded below for this case?
We can rewrite $l(a,b)$ as \begin{align} l(a,b) = \overbrace{\sum_{i=1}^N \log \bigg(\frac{e^{a^\top x_i - b}}{1+e^{a^\top x_i - b}}\bigg)}^A + \overbrace{\sum_{j=1}^M \log\bigg( \frac{1}{1+e^{a^\top y_j - b}} \bigg)}^B. \end{align}
Now consider dropping indices and considering a single $x$ and $y$. In this case, if we plot the shape of $A$, we see that $A\to-\infty$ as $a^\top x - b \to -\infty$ but that $A\to0$ as $a^\top x-b\to+\infty$. On the other hand, $B\to-\infty$ as $a^\top y - b\to+\infty$. Thus, if we can find $a,b$ such that $a^\top x < b$ and $a^\top y>b$, then we have that $t(a^\top x-b) <0 $ and $t(a^\top y-b) >0$ for any $t>0$. Sending $t\to\infty$, we have that $A\to-\infty$ and $B\to-\infty$. Hence $A+B$ is unbounded below.