Does $\{a:a=bc\text{ for }b,c\in\mathbb{Q},\text{ and }b\le t,\text{ }c\le u\}$ contain all rationals $a\le tu?$ Note that $t,u\in\mathbb{R},$ and $t,u>0.$ What if, for some $a\le tu,$ all possible candidates $b,c$ reside in outer scope? If the above statement (in the title) is true, the proof must be the form of contradiction, showing that there is no such case. However, other than that, I have no clue where to start the proof from, so any hints would be greatly appreciated.

This was used in this text. On page 5 in the proof that $(r^t)^u=r^{tu}$ (for real numbers) the author writes:

So let $a \le tu$ be rational and assume further that $a > 0$. In this case we can write $a = bc$ for $b, c \in \mathbb Q$
  and $b \le t$, $c \le u$. 

 A: all. This appears indeed to be a problem. I think the simplest fix is to show that for all real numbers $a$, the sets $\{r^b : b \in \mathbb{Q}, b \leq a\}$ and $\{r^b : b \in \mathbb{Q}, b < a\}$ have the same supremum. This way, you can use the argument posted above to the second set. These suprema are equal essentially by a continuity argument. I can provide some details for this if it is helpful. (I emailed them to the two people who emailed me.)
I wrote these notes sometime around 2012 or so when I was teaching analysis at Princeton during my postdoc. They are part of a larger document which basically trudges through Rudin's Principles of Mathematical Analysis. These properties of real powers are part of an exercise that is in Chapter 2, I think.
A: HINT: What is the set in question when $t=u=-1$?
Added: Now that we have it, I’ll address the underlying question of why the following statement is true:

So let $a \le tu$ be rational and assume further that $a > 0$. In this case we can write $a = bc$ for $b, c \in \mathbb Q$
  and $b \le t$, $c \le u$. 

If $a=tu$, then of course we can take $b=t$ and $c=u$, so assume that $a<tu$. 

Added2: When I fixed an the statement below to reflect the fact that $t$ and $u$ are not required to be rational, I didn’t even think about the previous statement, which of course can fail. As others have pointed out, that problem can’t be fixed: the assertion is false as it stands, and we really need either strict inequalities or rational $t$ and $u$. The assertion is also flawed in that it doesn’t require $t$ and $u$ to be positive, but that was clearly the intent, so I just used the OP’s original version in which they are assumed to be positive.

Since $t>0$, we can divide through by $t$ to get $\frac{a}t<u$. Let $c$ be any rational between $\frac{a}t$ and $u$, and let $b=\frac{a}c$; it’s straightforward to verify that $b<t$.
A: The claim is not true if you allow $u$ and $v$ to be negative. (This was already pointed out in another answer.)

EDIT: Even after the reformulation the claim is not true. (Which I did not notice and posted an incorrect proof before.)
To see this, consider $t=u=\sqrt2$ and $a=2$.
If we want to get $a=bc$ for some positive rational numbers $b$, $c$, then necessarily $b<\sqrt2$, $c<\sqrt2$ and, consequently $bc<2=a$.

If we allow $b$ and $c$ to have negative values, then the claim is trivial, since we could take $b=-1$ and $c=-a$.

Let us try to prove this claim, which should be valid:

If $a<tu$ for $a\in\mathbb Q$, $t,u\in\mathbb R$ such that $t,u>0$, then there exist rational numbers $b,c\in\mathbb Q$ such that $a=bc$ and $0<b<t$, $0<c<u$.



*

*Let us choose first some $b\in\mathbb Q$ such that $a<bu<tu$. (In the other words, take any rational number $b$ between $a/u$ and $t$.)

*We can now simply put $c=a/b$. Then we have $a=bc$ and
$$c = \frac ab < u.$$

A: That proof is busted. The statement that you quote is simply false if $t$ and $u$ are irrational and $a=tu$ is rational (assuming that we require $b$ and $c$ to be non-negative, which is implied in the proof).
I am not saying that it would be difficult to fix the proof, but as it stands, it is invalid.
(I have notified the author, Professor Damron.)
