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Worksheet 7

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Question 1

Let be a topological space, and let be a finite sequence of compact subsets of . Is a compact subset of ?

Solution

Answer: Yes.

Claim: A finite union of compact sets is compact.

Proof:

Let be an open cover of . That is, is a collection of open sets in such that:

Step 1: Cover each compact set

For each , consider the restriction of to .

Since , we have that restricted to is an open cover of in .

Step 2: Extract finite subcovers

Since each is compact, there exists a finite subcover such that:

Let .

Step 3: Combine the finite subcovers

Consider the collection:

Since each is finite, is a finite collection (at most elements).

Moreover, , so it is a subcollection of the original cover.

Step 4: Verify the union is covered

By construction:

Conclusion

Therefore, is a finite subcover of . Since was an arbitrary open cover, is compact.

Question 2

Suppose is an infinite sequence of compact subsets of . What can you say about ?

Solution

Answer: The intersection may be empty, or non-empty, depending on the specific sequence.

Key Observation: Unlike finite intersections of compact sets, infinite intersections do not automatically preserve compactness.

Case 1: The intersection can be empty

Counterexample:

Let with the standard topology, and define:

Each is closed in and contains the point at infinity in the sense that it extends to infinity.

Actually, let us use a better example:

Wait, we need . Let me redefine:

Each is compact (closed and bounded interval in ).

However:

To see this, suppose . Then for all , which means for all .

But this is impossible: for any real number , we can choose large enough so that , contradicting .

Therefore, .

Case 2: The intersection can be non-empty

Example:

Let for each (considering only ).

Each is compact.

However:

which is compact.

But if we consider with both positive and negative indices, a similar construction gives a non-empty intersection.

Finite Intersection Property

What we can say:

If the sequence has the finite intersection property (meaning every finite sub-collection has non-empty intersection), then we can derive constraints on the intersection.

Specifically, in a compact Hausdorff space, a collection of compact sets with the finite intersection property has non-empty intersection.

However, for a general sequence without additional structure, can be empty or non-empty.

Conclusion

Summary:

  • A finite intersection of compact sets is compact
  • An infinite intersection of compact sets may be empty
  • An infinite intersection of compact sets may be non-empty but not necessarily compact
  • If the compact sets are nested (i.e., ), then is compact and non-empty (in a compact space)

Question 3

Prove: is a compact subset of for all .

Solution

Claim: The -sphere is a compact subset of for all .

Proof:

Step 1: Express as the inverse image of a closed set

Consider the Euclidean norm function:

This function is continuous (composition of continuous functions: the coordinate projections and the square root function).

The -sphere can be written as:

Step 2: Recognize as a closed set

The singleton is a closed set in .

Since is continuous, the preimage is closed in .

Therefore, is closed.

Step 3: Show is bounded

For any point , by definition:

Thus every point in lies at distance exactly 1 from the origin. In particular:

So , where is the closed ball of radius 2 centered at the origin.

Therefore, is bounded.

Step 4: Apply the Heine-Borel Theorem

By the Heine-Borel Theorem, a subset of is compact if and only if it is closed and bounded.

Since is both closed and bounded, is compact.

Question 4

Prove: (the set of orthogonal matrices in ) is a compact subset of .

Solution

Definition: The orthogonal group is defined as:

where is the identity matrix and is the transpose of .

Claim: is compact in .

Proof:

We identify with by writing a matrix as a vector of its entries.

Step 1: Show is closed

Define the map:

This map is continuous because:

  • Matrix transpose is continuous
  • Matrix multiplication is continuous (it is a polynomial in the entries)

The identity matrix is a single point (constant element) in .

Therefore:

is the preimage of a closed set under a continuous map, hence closed in .

Step 2: Show is bounded

For any matrix , we have .

Taking the Frobenius norm (Euclidean norm on ):

The Frobenius norm satisfies (submultiplicativity).

Since (transpose preserves the Frobenius norm):

The Frobenius norm of is:

Therefore:

Equivalently, for each matrix :

This means every entry satisfies .

Thus , a closed ball in , so is bounded.

Step 3: Apply the Heine-Borel Theorem

Since is closed and bounded in , by the Heine-Borel Theorem, is compact.

Question 5

Prove: is a compact subset of .

Solution

Definition: The special orthogonal group is defined as:

where is the determinant of .

Claim: is compact in .

Proof:

Step 1: Express as a closed subset of

Define the determinant map:

This map is continuous because the determinant is a polynomial function of the matrix entries.

We can write:

Since is closed in , the set is closed in .

Therefore, is the intersection of two closed sets, hence closed in .

Step 2: Show is bounded

Since , every matrix is orthogonal.

From the previous problem, we showed that for all :

Since , this bound applies to all as well.

Therefore, is bounded in .

Step 3: Apply the Heine-Borel Theorem

Since is closed and bounded in , by the Heine-Borel Theorem, is compact.

Question 6

Prove: .

Solution

Claim: Every special orthogonal matrix is an orthogonal matrix.

Proof:

By definition:

This definition explicitly states that consists of matrices in with an additional property (determinant equals 1).

Therefore, by definition, .

More explicitly: if , then:

  1. (by definition of ), and
  2. (additional constraint)

The first property alone is sufficient to conclude .

Question 7

Prove: (the set of unitary matrices in ) is a compact subset of .

Solution

Definition: The unitary group is defined as:

where is the conjugate transpose (Hermitian transpose) and is the identity matrix.

Claim: is compact in .

Proof:

We identify with by writing each complex entry as an ordered pair of real numbers (real and imaginary parts).

Step 1: Show is closed

Define the map:

This map is continuous because:

  • The conjugate transpose operation is continuous (coordinates are real linear combinations)
  • Matrix multiplication is continuous (polynomial in the entries)

The identity matrix is a single point in .

Therefore:

is the preimage of a closed set under a continuous map, hence closed in .

Step 2: Show is bounded

For any unitary matrix , we have .

Using the Frobenius norm (Euclidean norm on ):

For complex matrices, the Frobenius norm satisfies:

Since (conjugate transpose preserves the Frobenius norm):

Actually, more directly: if , then taking Frobenius norms:

where denotes the trace.

This gives:

But (the square of the Frobenius norm).

Therefore:

Equivalently, writing with (real and imaginary parts):

This means every entry satisfies .

Thus in , so is bounded.

Step 3: Apply the Heine-Borel Theorem

Since is closed and bounded in , by the Heine-Borel Theorem, is compact.

Additional Note: Relationship between , , and

These three groups form a natural hierarchy:

  1. consists of all orthogonal matrices (both orientation-preserving and orientation-reversing)
  2. consists of the orientation-preserving orthogonal matrices (those with determinant +1)
  3. is the complex analog of , consisting of unitary matrices

All three are compact, which makes them important in topology and differential geometry as examples of compact Lie groups.

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