Demonstrate the compactness, Hausdorff property, and continuity of a functional mapping over a product space of continuous functions | Step-by-Step Solution

What We're Solving:

We need to analyze a product space Y formed by continuous functions from a topological space X to [0,1], then prove three key properties: Y is compact, Y is Hausdorff, and a certain mapping ψ is continuous. This is a beautiful example of how topology connects function spaces with product spaces!

The Approach:

This problem leverages Tychonoff's Theorem and properties of product topologies. We're essentially showing that when we collect all possible continuous functions as coordinates in a product space, the resulting space inherits nice topological properties. The strategy is to:

• 1. Recognize Y as a product of compact Hausdorff spaces
• 2. Apply fundamental theorems about product spaces
• 3. Use the product topology to establish continuity

Step-by-Step Solution:

Step 1: Understanding the Setup

• X is our given topological space
• C(X) is the set of all continuous functions f: X → [0,1]
• Y is the product space ∏_{f∈C(X)} [0,1]
• Each point in Y is a "tuple" indexed by functions in C(X)

Y assigns a value in [0,1] to each continuous function - it's like having infinitely many coordinate axes, one for each continuous function!

Step 2: Proving Y is Compact Since [0,1] is compact (it's closed and bounded in ℝ), and we're taking the product over all f ∈ C(X):

• Each factor [0,1] in the product is compact
• By Tychonoff's Theorem, any product of compact spaces is compact
• Therefore, Y = ∏_{f∈C(X)} [0,1] is compact

Step 3: Proving Y is Hausdorff Since [0,1] is Hausdorff (any metric space is Hausdorff):

• Each factor [0,1] in the product is Hausdorff
• The product of Hausdorff spaces is always Hausdorff
• Therefore, Y is Hausdorff

If y₁, y₂ ∈ Y are distinct, they differ in some coordinate f₀. We can separate them using open sets in that coordinate, then extend to the full product using the product topology.

Step 4: Proving ψ is Continuous The mapping ψ appears to be the evaluation map. For ψ: X → Y defined by ψ(x) = (f(x))_{f∈C(X)}:

• ψ(x) assigns to each point x ∈ X the "evaluation tuple" where coordinate f equals f(x)
• For continuity, we need inverse images of open sets to be open
• Basic open sets in Y have the form: finitely many coordinates restricted to open intervals
• If U is open in Y, then ψ⁻¹(U) involves only finitely many functions f₁,...,fₙ
• Since each fᵢ is continuous, ψ⁻¹(U) = ∩ᵢ fᵢ⁻¹(Vᵢ) where Vᵢ are open in [0,1]
• Finite intersections of open sets are open, so ψ is continuous

The Answer:

Y is compact by Tychonoff's Theorem (product of compact spaces [0,1]), Hausdorff because products of Hausdorff spaces are Hausdorff, and ψ is continuous because it's defined coordinatewise by continuous functions, making inverse images of basic open sets equal to finite intersections of open sets.

Memory Tip:

Remember "TPC": Tychonoff for compactness, Product preserves Hausdorff, Coordinatewise continuity! When you see product spaces of nice intervals like [0,1], these three properties almost always follow from the corresponding properties of the factors.

The beauty here is that we're embedding our original space X into this huge function space Y, and the topology works out perfectly to preserve all the properties we care about!