Mistake in a proof of termination phase of Simplex algorithm in CLRS? There is a pseudo-code for Simplex algorithm in CLRS:

The proof consists from three-part loop invariant:

Proof We use the following three-part loop invariant:
At the start of each iteration of the while loop of lines 3–12,

the slack form is equivalent to the slack form returned by the call of INITIALIZE-SIMPLEX,
foreach $i \in B$, we have $b_i$ &gt;= 0, and
the basic solution associated with the slack form is feasible.


The initialisation and maintenance phases are clear for me.
But, the termination case seems(not sure) wrong to me:

So, my question is:
It is unclear for me that $\bar{x_i} = \infty $ if $i = e$. But, if we look at the pseudo-code, it is obvious that such thing happens only if $x_e \leq 0$. What am I missing ?
Because of line 9, $\Delta_l$ is going to be $\infty$ if and only if all the $\Delta_i$ with $i \in B$ are $\infty$. This can happen only when all the $a_{ie}$ with $i \in B$ are $ \le 0$. Hence, by construction the solution $\bar{x}$ satisfies $\bar{x}_i \ge 0$ for all $i$. Therefore it is both feasible and it has unbounded objective value. 

If you are confused by the $\infty$, as Marcus Ritt mentioned in a comment you can replace the $\infty$ in the definition of $\bar{x}$ with a very large value $M$ and consider what happens as $M$ tends to $\infty$.