Prove that a continuous and injective function on an interval is strictly monotone through a detailed case analysis | Step-by-Step Solution

🎯 What We're Solving:

We need to prove that if a function f is both continuous and injective (one-to-one) on an interval I, then f must be strictly monotone (either strictly increasing or strictly decreasing throughout the entire interval).

🧭 The Approach:

This is a beautiful theorem that connects three fundamental concepts in real analysis! Here's why this makes intuitive sense: imagine trying to draw a continuous curve that never repeats any y-value (injective) - you'll find you can't "wiggle up and down" because that would force you to repeat y-values. The curve must consistently go in one direction.

Our strategy will be a proof by contradiction combined with careful analysis of what happens when we assume the function isn't strictly monotone.

📝 Step-by-Step Solution:

Step 1: Set up the contradiction

• Assume f is continuous and injective on interval I, but f is NOT strictly monotone
• This means f is neither strictly increasing nor strictly decreasing on I
• Therefore, we can find points where f "changes direction"

Step 2: Find the troublesome points Since f isn't strictly monotone, there must exist points a < b < c in I such that:

• Either f(a) < f(b) > f(c) (f goes up then down), OR
• f(a) > f(b) < f(c) (f goes down then up)

Let's work with the first case: f(a) < f(b) > f(c)

Step 3: Apply the Intermediate Value Theorem Since f is continuous:

• On [a,b]: f increases from f(a) to f(b)
• On [b,c]: f decreases from f(b) to f(c)

Now, consider any value y between min{f(a), f(c)} and f(b).

Step 4: Show the contradiction By the Intermediate Value Theorem:

• There exists x₁ ∈ [a,b] such that f(x₁) = y
• There exists x₂ ∈ [b,c] such that f(x₂) = y

Since a < b < c, we have x₁ ≠ x₂, but f(x₁) = f(x₂) = y.

This violates the injectivity of f! 🎯

Step 5: Complete the proof The same argument works for the case f(a) > f(b) < f(c). Since assuming f is not strictly monotone leads to a contradiction with injectivity, we conclude that f must be strictly monotone.

✅ The Answer:

The theorem is proven by contradiction. A continuous, injective function on an interval cannot "change direction" because doing so would force it to repeat output values (by the Intermediate Value Theorem), violating injectivity. Therefore, it must be strictly monotone in one direction throughout the entire interval.

💡 Memory Tip:

Think of it as the "No U-Turns Theorem"! A continuous, one-to-one function can't make U-turns because the Intermediate Value Theorem would force it to hit the same height twice, breaking the one-to-one property. Once it picks a direction (up or down), it's committed! 🚗➡️

Encouragement: This proof beautifully showcases how different theorems work together - continuity gives us the Intermediate Value Theorem, and injectivity provides the constraint that creates our contradiction. You're connecting major concepts in real analysis! 🌟