Is the closed interval [0, 2] a neighborhood of the point 1 in ℝ with the standard topology?
ANo — neighborhoods must be open sets, and [0, 2] is closed
BYes — [0, 2] contains an open set (e.g., (0.5, 1.5)) that contains 1
CNo — a neighborhood must be an open ball centered at the point
DOnly if we restrict to the subspace topology on [0, 2]
A neighborhood of x is any set containing an open set that contains x — the neighborhood itself need not be open. [0, 2] contains the open interval (0.5, 1.5) which contains 1, so it qualifies as a neighborhood of 1. Option A is the most common misconception: students assume 'neighborhood' means 'open set,' but the definition is deliberately more general. Open balls centered at x are the canonical examples of neighborhoods but are not required.
Question 2 Multiple Choice
A sequence (xₙ) in a topological space X converges to x under the neighborhood definition when:
AEvery open set in X eventually contains all terms of the sequence
BFor every neighborhood N of x, there exists M such that xₙ ∈ N for all n > M
CThe sequence enters every open ball around x and never leaves
DFor some neighborhood N of x, almost all terms of the sequence lie in N
Convergence to x requires that every neighborhood of x eventually contains all terms — for any N containing an open set around x, there exists M with xₙ ∈ N for all n > M. Option A is too strong: requiring every open set (not just those containing x) makes no sense. Option C adds a 'never leaves' condition not in the definition, and 'open ball' is metric-specific. Option D uses 'some neighborhood,' but convergence requires all neighborhoods of x.
Question 3 True / False
Every open set containing x is a neighborhood of x.
TTrue
FFalse
Answer: True
By definition, a neighborhood N of x is any set containing an open set U with x ∈ U ⊆ N. If N is itself open and contains x, take U = N: N contains the open set N which contains x. So every open set containing x satisfies the neighborhood definition. The class of neighborhoods is a superset of 'open sets containing x' — it includes non-open sets too, as long as they contain an open cushion around x.
Question 4 True / False
A neighborhood of a point is expected to itself be an open set.
TTrue
FFalse
Answer: False
This is the central misconception. The definition requires only that a neighborhood N of x contain some open set U with x ∈ U ⊆ N — N itself need not be open. For example, [0, 1] is a neighborhood of 0.5 in ℝ (it contains (0.25, 0.75) which contains 0.5), even though [0, 1] is closed. Insisting that neighborhoods be open would collapse the concept to 'open sets containing x,' losing the flexibility that makes neighborhoods useful.
Question 5 Short Answer
Why do topologists define neighborhoods as sets that *contain* an open set around x, rather than simply requiring neighborhoods to be open sets containing x? What flexibility does this gain?
Think about your answer, then reveal below.
Model answer: Requiring neighborhoods to be open would make them identical to 'open sets containing x,' adding no new concept. Allowing any set that contains an open cushion around x separates 'local' from 'open': a neighborhood captures what happens near x without requiring the global property of openness. This allows closed intervals, half-open intervals, and other non-open sets to serve as neighborhoods, making arguments more flexible. It also connects naturally to the basis: a neighborhood of x just needs to contain some basis element containing x, letting local arguments reduce to checking finitely describable sets.
The deeper payoff is in continuity and convergence: f is continuous at x iff preimages of neighborhoods of f(x) are neighborhoods of x — this formulation works identically whether sets are open or not. The generalization also opens the door to filter theory, where the collection of all neighborhoods of x forms the prototypical example of a filter.