Prove that compact sets in a metric space are necessarily closed by demonstrating that a set with a limit point not in the set cannot be compact | Step-by-Step Solution

🎯 What We're Solving:

We need to prove that every compact set in a metric space must be closed. We'll do this by showing the contrapositive: if a set is NOT closed, then it cannot be compact.

🧭 The Approach:

This is a proof by contradiction! Here's our game plan:

• We'll assume a set K is compact but NOT closed
• If K isn't closed, it must have a limit point that's not in K
• We'll construct an open cover of K that has NO finite subcover
• This contradicts compactness, so our assumption must be wrong!

The key insight is using the "troublesome" limit point to create an open cover that can't be finitely reduced.

📝 Step-by-Step Solution:

Step 1: Set up the contradiction

• Assume K is compact but NOT closed
• Since K isn't closed, there exists a limit point x of K such that x ∉ K
• A set is closed iff it contains all its limit points!

Step 2: Understand what this limit point means

• Since x is a limit point of K, every neighborhood of x contains points from K
• But x ∉ K, so we have points of K arbitrarily close to x, but x itself isn't in K

Step 3: Construct a problematic open cover For each point y ∈ K, we'll create an open ball around y that "avoids getting too close to x":

• Let d(x,y) be the distance from x to y
• Since x ∉ K and K is the set we're covering, d(x,y) > 0 for all y ∈ K
• Create the open ball B(y, d(x,y)/2) around each point y ∈ K

Step 4: Verify this is indeed an open cover

• Each B(y, d(x,y)/2) is open (open balls are open!)
• Every point y ∈ K is in its corresponding ball B(y, d(x,y)/2)
• Therefore, {B(y, d(x,y)/2) : y ∈ K} covers K

Step 5: Show no finite subcover exists Suppose we could cover K with finitely many of these balls: B(y₁, d(x,y₁)/2), B(y₂, d(x,y₂)/2), ..., B(yₙ, d(x,yₙ)/2)

• Let r = min{d(x,y₁)/2, d(x,y₂)/2, ..., d(x,yₙ)/2}
• Since we only have finitely many distances, r > 0
• The ball B(x,r) around x is disjoint from our finite collection of balls!
• But since x is a limit point of K, B(x,r) must contain some point z ∈ K
• This point z cannot be covered by our finite subcover - contradiction!

Step 6: Conclude Since assuming K is compact but not closed leads to a contradiction, every compact set must be closed.

✅ The Answer:

Theorem: Every compact set in a metric space is closed.

Proof Summary: We proved the contrapositive. If K is not closed, then K has a limit point x ∉ K. We constructed an open cover using balls of radius d(x,y)/2 around each y ∈ K. This cover cannot have a finite subcover because any finite collection leaves a neighborhood of x uncovered, yet this neighborhood must contain points of K. Therefore, non-closed sets cannot be compact.

💡 Memory Tip:

Think of compactness as "everything stays nicely contained" - if you have points escaping to a limit point outside your set, you can't maintain that "finite control" that compactness requires. The limit point becomes a "hole" that prevents finite subcovering! 🌟