How are we allowed to "choose" some real number which obeys certain inequalities?

In the proof of Baby Rudin Theorem 1.21 about the existence of $$n$$th roots of positive numbers, we say the following:

Choose some $$h$$ which obeys $$0 and $$0.

Translating this into a naive set-theoretic context, we are saying that the set $$H := \left\{ h \mid 0 is nonempty. How do we argue this? Do we have to use something as "strong" as that the rationals are dense in the reals, or is there an easier way to see this?

• Well, you need to argue that $\frac {x-y^n}{ny^{n-1}}$ is positive. Clearly that depends on information about $x,y,n$.
– lulu
Mar 18 at 20:19
• Sorry, let me add that. Yes, that is proved (or rather, assumed) earlier. @lulu
– EE18
Mar 18 at 20:20
• Well, if it is positive, then what's the problem? Call that number $\alpha$ and let $h=\min \left(\frac {\alpha}2,\frac 12\right)$.
– lulu
Mar 18 at 20:21
• @lulu OOF, fields are closed. Of course. Thank you!
– EE18
Mar 18 at 20:21