Design a continuous function mapping [0,1] to [0,1] that remains constant on disjoint closed intervals while maintaining a non-decreasing, surjective property | Step-by-Step Solution

What We're Solving

We need to construct a continuous function f: [0,1] → [0,1] that:

• Is non-decreasing (f(x₁) ≤ f(x₂) whenever x₁ ≤ x₂)
• Is surjective (hits every point in [0,1])
• Remains constant on some collection of disjoint open intervals

This is a beautiful problem that combines several important concepts in real analysis!

The Approach

The key insight is to think of the Cantor function (also called the "Devil's Staircase")! Here's why this works:

• We'll remove middle thirds repeatedly (like constructing the Cantor set)
• On each removed interval, our function will be constant
• Between these intervals, the function will "jump" to cover all values in [0,1]
• The clever construction ensures continuity despite these constant pieces

Step-by-Step Solution

Step 1: Recall the Cantor Set Construction Start with [0,1] and repeatedly remove middle thirds:

• Remove (1/3, 2/3) from [0,1]
• Remove (1/9, 2/9) and (7/9, 8/9) from what remains
• Continue this process infinitely

Step 2: Understanding What We Have After all removals, we get:

• The Cantor set C (what remains - a "dust" of points)
• Countably many disjoint open intervals that were removed

Step 3: Construct the Function Define f(x) as follows:

• On each removed open interval, f is constant
• The constant values are chosen so that f increases from 0 to 1 as we move left to right
• On the Cantor set itself, f is defined to make the function continuous

Step 4: Why This Works

• Non-decreasing: By construction, f never decreases as x increases
• Continuous: The constant values on removed intervals are chosen to match the limits from endpoints
• Surjective: The constant values on the removed intervals are dense in [0,1], and combined with the Cantor set points, we hit every value in [0,1]

Step 5: Specific Construction The standard Cantor function works perfectly:

• On (1/3, 2/3): f(x) = 1/2
• On (1/9, 2/9): f(x) = 1/4
• On (7/9, 8/9): f(x) = 3/4
• And so on...

The Answer

The Cantor function (Devil's Staircase) is our solution! It's defined by:

• 1. Constructing the Cantor set by removing middle thirds
• 2. Setting f to be constant on each removed interval
• 3. Choosing these constant values to create a continuous, non-decreasing surjection from [0,1] to [0,1]

The function rises from 0 to 1 in a "staircase" pattern, staying constant on infinitely many disjoint open intervals.

Memory Tip

Think of it as a "magical staircase" - it climbs from bottom to top (surjective) without ever going down (non-decreasing), but spends "time" resting on infinitely many steps (constant on disjoint intervals). The "magic" is that it's still continuous despite all these flat steps!

This is one of the most elegant examples in real analysis showing how counterintuitive continuous functions can be! 🎯