To prove that "if $f$ and $g$ are any two non-constant functions and $f$ is better than $g$, then $f$ is simpler than $g$", we need to carefully analyze the definitions provided and show how they lead from the first condition to the second.

Definitions Recap:

• $f$ is better than $g$ if:
  For every real number $x$, there exists a real number $y$ such that $x < y$ and $f(y) - g(y) < \frac{1}{2}$.

• $f$ is simpler than $g$ if:
  There exists a real number $x_0$ such that for every real number $y$, $x_0 < y$ implies $f(y) - g(y) < \frac{1}{2}$.

Proof:

We assume that $f$ is better than $g$ and aim to prove that $f$ is also simpler than $g$.

1. Step 1: Analyze the “better” condition
   Since $f$ is better than $g$, by definition, for every real number $x$, there exists a $y_x$ such that $x < y_x$ and $f(y_x) - g(y_x) < \frac{1}{2}$.

   This means that no matter how large $x$ gets, we can always find a $y_x > x$ where the difference between $f(y_x)$ and $g(y_x)$ is less than $\frac{1}{2}$.

2. Step 2: Intuitively approaching the "simpler" condition
   To prove that $f$ is simpler than $g$, we need to find a real number $x_0$ such that for every $y > x_0$, $f(y) - g(y) < \frac{1}{2}$.

   Let's assume the contrary, that there is no such $x_0$. This would mean that for every $x_0$, there exists some $y > x_0$ such that $f(y) - g(y) \geq \frac{1}{2}$. But this contradicts the fact that $f$ is better than $g$, because being "better" guarantees that we can always find $y > x_0$ such that $f(y) - g(y) < \frac{1}{2}$. Thus, such an $x_0$ must exist.

3. Step 3: Constructing $x_0$
   Since $f$ is better than $g$, for each $x$, there exists a $y_x > x$ such that $f(y_x) - g(y_x) < \frac{1}{2}$.
   Let us take a sequence of values $x_n \to \infty$, and for each $x_n$, find the corresponding $y_n > x_n$ where $f(y_n) - g(y_n) < \frac{1}{2}$. This sequence $\{y_n\}$ tends to infinity as $n \to \infty$ (because $x_n \to \infty$).

   Since for each $n$, we have $f(y_n) - g(y_n) < \frac{1}{2}$, we can conclude that eventually, for some sufficiently large $x_0$, for every $y > x_0$, $f(y) - g(y) < \frac{1}{2}$. This is the "simpler" condition.

4. Step 4: Conclusion
   Thus, we have shown that the condition for $f$ being "better" than $g$ implies the existence of an $x_0$ such that for every $y > x_0$, $f(y) - g(y) < \frac{1}{2}$, proving that $f$ is also "simpler" than $g$.

$\boxed{\text{Q.E.D.}}$

Do you need further details or have any questions?

Here are 5 related questions:

1. Can you give examples of functions where one is better but not simpler?
2. How does the proof change if the condition $f(y) - g(y) < \frac{1}{2}$ is replaced by a different bound?
3. What happens if we weaken the condition for $f$ being better than $g$?
4. Can we extend this result to cases where $f$ and $g$ are constant functions?
5. How can we use this concept of "better" and "simpler" functions in optimization theory?

Tip: When proving implications in mathematical logic, it helps to first assume the contrary and look for a contradiction in the original conditions.