You use only one "less than" sign.
Consider this as an analogy: Just as we would not say: $a = 5$, $b = 5$, so therefore $a + b$ "equals equals" $10$, but rather, we'd conclude that $a + b = 10$. In your case, we would not say $ab$ "is less than less than" $50$, but rather, $ab < 50$.
For any two real numbers, $x, y$, one and only one of the following is true:
- $x < y\quad$ OR (x less than y)
- $x = y\quad$ OR (x equals y)
- $x > y\quad$ (x is greater than y)
EDIT following question edit:
To answer the question in the comment below: There is no largest number preceding 50: For every $k < 50$, there is a $j$ such that $k < j < 50$, and there is an $m$ such that $k < j < m < 50...$ and on and on and on...for every proposed "largest number $n$ preceding 50", there is a larger number than $n$ preceding 50.
In this case, for example, take any possible choice of $ab < 50$. Then there exists $x = \large \frac {50 + ab}{2}$ $> ab$ but nonetheless, $x < 50$. If we then let $ab = x$, there exists $y = \large\frac {50 + x}{2}$ $> x,$ with $y< 50$...and so on.
The least upper bound for $ab$, in this case, is not in the set of possible values for $ab$. Think of it like this: $ 0 < a < 5,\;\;0 < b < 10, \;\; \implies \;\; ab \in (0, 50),$ where $(0, 50)$ is the open interval of reals greater than $0$ but less than $50$ (the interval of all real numbers from $0$ to $50$, excluding $0$ and $50$:
$ab \in \{x \in \mathbb{R}: 0 < x < 50\}$