Is 0 a limit point of the set A = {1, 1/2, 1/3, 1/4, …} ⊂ ℝ (with the usual topology)?
ANo — 0 is not in A, so it cannot be a limit point of A
BNo — the sequence 1/n approaches 0 but never equals 0, so the definition fails
CYes — every open interval around 0 contains infinitely many points of A different from 0
DYes — 0 is the greatest lower bound of A, which automatically makes it a limit point
The definition of limit point requires every open set containing x to contain a point of A other than x itself. For x = 0: any interval (−ε, ε) contains 1/n for all sufficiently large n, and none of these equal 0. So 0 is a limit point. Option A is the classic error — membership in A is irrelevant to limit-point status. A limit point does not need to belong to the set.
Question 2 Multiple Choice
A set A is closed if and only if which condition holds?
AEvery point of A is a limit point of A
BA contains no isolated points
CEvery limit point of A belongs to A
DA equals its own interior
A set is closed iff it contains all its limit points — equivalently, A = Ā. Option A is wrong: closed sets can contain isolated points (points that are not limit points), such as a finite set like {1, 2, 3}. Option B is also wrong for the same reason. The key is whether limit points that A 'accumulates toward' are captured inside A.
Question 3 True / False
Most point in a set A is a limit point of A.
TTrue
FFalse
Answer: False
Isolated points are members of A that are NOT limit points: they have some open neighborhood containing no other point of A. For example, in A = {0} ∪ (1, 2), the point 0 belongs to A but every small open interval around it contains no other element of A, so 0 is isolated — not a limit point.
Question 4 True / False
A limit point of a set A may lie outside of A.
TTrue
FFalse
Answer: True
The definition requires every neighborhood of x to contain a point of A different from x — x itself need not be in A at all. For instance, 0 is a limit point of the open interval (0, 1) even though 0 ∉ (0, 1). This is precisely why the closure Ā = A ∪ A′ may be strictly larger than A: it must add the limit points that A 'approaches' but does not yet contain.
Question 5 Short Answer
Why does the definition of limit point require the nearby point of A to be 'different from x itself'? What would go wrong without this requirement?
Think about your answer, then reveal below.
Model answer: Without the 'different from x' clause, every point in A would trivially satisfy the definition: just take any neighborhood and find x ∈ A as the required element. The requirement forces x to be genuinely approached by other points of A — a real accumulation point. Without it, the distinction between isolated points (in A but surrounded by gaps) and true limit points (surrounded by other elements of A) would collapse.
The clause does real work: it rules out isolated points from being called limit points. An isolated point is in A but has a neighborhood containing no other element of A — with the 'different from x' requirement, it fails the limit-point test. This distinction matters for the closure operation and for characterizing closed sets.