Aℚ (the rational numbers) with the subspace topology from ℝ
BAn infinite-dimensional Hilbert space with the norm topology
Cℝ with its standard topology
DNo non-compact space can be locally compact, by definition
ℝ is locally compact because every point x has the compact neighborhood [x−1, x+1] (compact by Heine-Borel). ℚ is not locally compact — compact subsets of ℚ have empty interior, so no compact set can contain an open neighborhood of any rational point. Infinite-dimensional Hilbert spaces are not locally compact because the closed unit ball is not compact. Option D is false — local compactness is strictly weaker than compactness; ℝ itself is the canonical example.
Question 2 Multiple Choice
Why is local compactness the exactly right hypothesis needed to construct the one-point (Alexandroff) compactification of a Hausdorff space?
AIt guarantees the space is metrizable, which the compactification requires
BIt ensures there are enough compact sets to define neighborhoods of the added point ∞ in a way that makes the resulting space Hausdorff
CIt is needed so that every continuous function on the original space extends continuously to the compactification
DIt guarantees the original space is already compact, making the construction trivial
In the one-point compactification, neighborhoods of ∞ are defined as sets whose complement in X is compact. For the resulting space X* to be Hausdorff, every point x in X must have a compact neighborhood disjoint from a neighborhood of ∞ — which is exactly the definition of local compactness. Without local compactness, the construction fails to produce a Hausdorff space because there aren't enough compact sets to separate x from ∞.
Question 3 True / False
Every compact topological space is also locally compact.
TTrue
FFalse
Answer: True
If X is compact, then X itself is a compact neighborhood of every point — the whole space is open in itself and is compact by assumption. So every point trivially has a compact neighborhood. Local compactness is thus a strictly weaker condition than compactness: every compact space is locally compact, but not vice versa (ℝ is locally compact but not compact).
Question 4 True / False
The real numbers ℝ fail to be locally compact because no bounded subset of ℝ is compact.
TTrue
FFalse
Answer: False
This gets it backwards: ℝ *is* locally compact. Moreover, closed and bounded subsets of ℝ *are* compact (by Heine-Borel), which is exactly why ℝ is locally compact — every point x has the compact neighborhood [x−1, x+1]. What ℝ lacks is *global* compactness: the whole space has no finite subcover for the cover {(−n, n)}, so ℝ is not compact. Local compactness asks only for compact neighborhoods at each point, not for the whole space to be compact.
Question 5 Short Answer
Why is ℚ (the rationals with the subspace topology from ℝ) not locally compact, even though ℝ is? What is the structural reason?
Think about your answer, then reveal below.
Model answer: A locally compact space requires compact neighborhoods around every point. In ℚ, any open set around a rational number q contains an interval (a, b) ∩ ℚ. But compact subsets of ℚ have empty interior — no compact subset of ℚ can contain an open interval of rationals. The reason is that any interval in ℚ contains Cauchy sequences converging to irrationals, so the interval cannot be compact (a compact metric space must be complete, but ℚ is not). There is therefore no compact set in ℚ that contains an open neighborhood of any point.
The key structural fact is that every compact subset of ℚ is nowhere dense in ℚ. This means no open set in ℚ is contained in a compact set — the condition for local compactness fails at every point. The difference from ℝ is that ℝ is complete: closed bounded sets are compact, so [x−1, x+1] works as a compact neighborhood. ℚ's incompleteness destroys this.