Questions: First Countable Spaces

Question 1

In a first-countable topological space, which statement about the closure of a set A is correct?

• A. The closure of A equals A itself, since first-countable spaces have a countable basis
• B. A point x belongs to cl(A) if and only if there exists a sequence of points in A that converges to x
• C. Closure must be computed using all neighborhoods, not just a countable basis
• D. Closure in a first-countable space equals the set of isolated points of A

Answer: (B)

First countability makes sequences sufficient to detect closure: if x ∈ cl(A), you can construct a sequence in A converging to x by picking a point from each basis element B₁ ∩ A, B₂ ∩ A, …. Conversely, if a sequence in A converges to x, then x is in the closure. This sequential characterization of closure fails in general topological spaces, where nets or filters are needed.

Question 2

A topologist asserts: 'f is continuous at x if and only if xₙ → x implies f(xₙ) → f(x) for every sequence xₙ → x.' In which spaces is this sequential criterion both necessary and sufficient?

• A. All topological spaces — sequential continuity always characterizes topological continuity
• B. Only in metric spaces, where the open-ball structure is essential
• C. In any first-countable space — metric spaces are a special case of this broader condition
• D. Only when both the domain and codomain are second-countable

Answer: (C)

The sequential criterion for continuity is equivalent to topological continuity precisely in first-countable spaces. Metric spaces are first countable (use balls of radius 1/n), which is why analysis students learn sequential continuity first — it works everywhere analysis is done. In non-first-countable spaces, a function can satisfy all sequential tests yet still be discontinuous at x.

Question 3

In any topological space, a point x belongs to the closure of a set A if and only if some sequence of points in A converges to x.

• T. True
• F. False

Answer: False

This sequential characterization of closure holds only in first-countable spaces. In general topological spaces — like ℝ^ℝ with the product topology, which is not first countable — a point can be in cl(A) without any sequence from A converging to it. To detect closure in full generality, one needs nets or filters, which generalize sequences by allowing uncountable indexing sets.

Question 4

Every metric space is first countable because the open balls {y : d(x, y) < 1/n} for n = 1, 2, 3, … form a countable neighborhood basis at each point x.

• T. True
• F. False

Answer: True

Any open set containing x must contain an open ball around x (by definition of the metric topology), and that ball must contain B(x, 1/n) for sufficiently large n. So the countable collection of rational-radius balls is a neighborhood basis at x. This is why all of real analysis — operating in metric spaces — can be done entirely with sequences.

Question 5

Why does first countability matter — what goes wrong with sequential reasoning in spaces that fail to be first countable?

Model answer: In a first-countable space, countable sequences of neighborhoods suffice to detect all topological properties: closure, limit points, and continuity. Without first countability, sequences can be 'too thin' — a point can lie in the closure of a set with no sequence from that set converging to it, and a discontinuous function can satisfy every sequential test. The fix requires nets or filters, which generalize sequences by allowing uncountable index sets. First countability marks the boundary between the sequential world of metric spaces and the more exotic behavior of general topology.

The failure of sequential intuition outside first-countable spaces reflects a genuine structural limitation: sequences sample countably many approximations, but non-first-countable spaces have neighborhoods that countable sequences cannot fully probe. Recognizing this boundary is what distinguishes fluency in metric-space analysis from fluency in general topology.