Claim: with the standard topology is not locally compact.
Proof:
Recall that a topological space is locally compact if every point has a compact neighborhood; that is, there exists a neighborhood of such that the closure is compact.
For with the standard topology, we must show that there exists at least one point that does not have a compact neighborhood.
Step 1: Analyze compact sets in
By the Heine-Borel Theorem, a subset is compact if and only if is closed and bounded.
Step 2: Check neighborhoods of an arbitrary point
Consider any point and any neighborhood of .
Since is a neighborhood of in the standard topology, there exists an open interval for some .
The closure of any neighborhood of must contain an open interval around , and hence cannot be bounded.
For instance, if is a neighborhood of , then either:
The upper end of is unbounded, or
The lower end of is unbounded, or
Both ends are unbounded.
In any case, is unbounded.
Step 3: Conclude non-compactness
Since is unbounded, by the Heine-Borel Theorem, is not compact.
This holds for every neighborhood of every point .
Conclusion
Since no point in has a compact neighborhood, is not locally compact.
Question 2
Is a closed subset of a locally compact space locally compact? Justify.
Solution
Answer: Yes.
Claim: Let be a locally compact topological space and let be a closed subset. Then (with the subspace topology) is locally compact.
Proof:
We must show that for every point , there exists a neighborhood of in whose closure in is compact.
Step 1: Fix a point in
Let (arbitrary).
Step 2: Use local compactness of
Since is locally compact, there exists a neighborhood of in such that (the closure of in ) is compact.
Step 3: Construct a neighborhood in
Consider . Since is open in and has the subspace topology, is open in .
Moreover, and , so . Thus is a neighborhood of in .
Step 4: Analyze the closure in
The closure of in (denoted ) is related to the closure in by:
Since is closed in :
Step 5: Verify compactness
Since , we have:
Now, is compact by assumption.
A closed subset of a compact space is compact, so we need to show that is closed in .
Since is closed in (as the closure of a set in ) and is closed in , the set is closed in , and hence closed in the subspace .
Therefore, is compact.
Conclusion
For every , we have constructed a neighborhood of in with compact closure in . Thus is locally compact.
Question 3
Let be a locally compact topological space. Let be continuous and open. If is locally compact, where , then . Is locally compact? Justify.
Solution
Answer: Yes, is locally compact.
Claim: Let be locally compact and be continuous and open. Then the image (with the subspace topology from ) is locally compact.
Proof:
We must show that every point has a locally compact neighborhood in .
Step 1: Fix a point in
Let be arbitrary. Then there exists some such that .
Step 2: Use local compactness of
Since is locally compact and , there exists a compact neighborhood of in . That is:
is a neighborhood of
is compact
Step 3: Apply the open map property
Since is an open map, is open in (with the subspace topology).
Actually, we use a stronger property: The image of a compact space under a continuous map is compact.
Therefore, is a compact subset of .
Step 4: Construct a neighborhood in
Since and , we have .
The set is compact (from Step 3) and contains .
Moreover, there exists an open set in containing such that .
Since is open, is open in , and hence open in (with the subspace topology).
Step 5: Verify local compactness at
The set is:
An open neighborhood of in
Its closure in is (using continuity of )
Since is compact and is a closed subset of :
Conclusion
For every , there exists an open neighborhood of in with compact closure. Thus is locally compact.
Question 4
Let and be locally compact Hausdorff spaces. Let be a homeomorphism. Let and denote the one-point compactifications of and respectively. Is (the natural extension of ) a homeomorphism? Justify.
Solution
Answer: Yes, is a homeomorphism.
Claim: If is a homeomorphism between locally compact Hausdorff spaces, then the extension to their one-point compactifications is also a homeomorphism.
Proof:
Step 1: Recall the one-point compactification
For a locally compact Hausdorff space , the one-point compactification is defined as:
where is a point not in , and the topology on consists of:
All open sets of , together with
Sets of the form where is a compact subset of
Step 2: Define
The natural extension is defined as:
Step 3: Prove is continuous
We must verify that is continuous.
Case 1: For any open set in , the set is open in (since is a homeomorphism, hence continuous). This is also open in .
Case 2: For a basic open set in where is compact in , we compute:
Since is a homeomorphism, is compact in (the continuous image of compact is compact under the inverse map).
Therefore:
This is open in by definition of the one-point compactification topology.
Thus is continuous.
Step 4: Prove is bijective
Injective: Since is injective and for all , and , the map is injective.
Surjective: For any , if , then (since is surjective). If , then .
Thus is bijective.
Step 5: Prove is open (or prove is continuous)
By a symmetric argument, the inverse map defined as:
is continuous (by the same reasoning as Step 3).
Conclusion
Since is a continuous bijection with continuous inverse, is a homeomorphism.
Question 5
Is (the positive integers with the standard subspace topology from ) locally compact? Determine a one-point compactification of .
Solution
Answer: Yes, is locally compact. Its one-point compactification is with an appropriate topology.
Part A: Local Compactness of
Claim: with the subspace topology from is locally compact.
Proof:
In the subspace topology from , a set is open in if and only if it is the intersection of an open set in with .
For any point , the singleton is open in the subspace topology. To see this, note that:
Since is open in , the set is open in .
Therefore, is a neighborhood of in .
Moreover, is compact (any finite set is compact).
Thus, every point has a compact neighborhood, so is locally compact.
Part B: One-Point Compactification of
Construction:
Let where is a point not in .
The topology on consists of:
All open sets of (in the subspace topology from )
Sets of the form where is a compact subset of
Since compact subsets of are precisely the finite subsets (as is discrete in its subspace topology), the topology on consists of:
Arbitrary subsets of
Sets of the form where is a finite subset of
Equivalently, the sets containing that are open are exactly those of the form where is a cofinite subset of (a subset whose complement is finite).
Verification:
Let be an open cover of .
Since , there exists some containing .
By the definition of the topology, for some finite set .
Thus, .
Since is finite, we can choose finitely many sets from to cover the points in (possibly with some redundancy).
Then is a finite subcover of covering all of .
Therefore, is compact.
Topological Structure:
The one-point compactification can be visualized as:
where:
Points in are isolated (every singleton is open)
The point has neighborhoods of the form for any
This space is homeomorphic to the convergent sequence space: the set of natural numbers together with a limit point at infinity.