we say that function f is BETTER than function g if:
for every real number x, there exists a real number y such that x < y AND f(y) - g(y) < 1/2

we say that function f is SIMPLER than function g if:
there exists a real number x such that for every real number y, x < y => f(y) - g(y) < 1/2

CLAIM 1
PROVE THAT "if f and g are any two NON-CONSTANT functions and f is better than g, then f is simpler than g.”

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.

we say that function f is BETTER than function g if: for every real number x, there exists a real number y such that x < y AND f(y) - g(y) < 1/2. We say that function f is SIMPLER than function g if: there exists a real number x such that for every real number y, x < y => f(y) - g(y) < 1/2. CLAIM 1 PROVE THAT "if f and g are any two NON-CONSTANT functions and f is better than g, then f is simpler than g."

Explore the mathematical proof that demonstrates if f is better than g for two non-constant functions, then f is also simpler than g. This problem delves into concepts of real analysis, function comparison, and inequality relations. Ideal for advanced undergraduate or graduate-level students.

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