The set A = {1/n : n = 1, 2, 3, ...} ⊂ ℝ with the standard topology. Which point is a limit point of A?
A1, because it equals 1/1 and is in A
B0, because every open interval around 0 contains points of A other than 0
C1/2, because it is in A
DEvery element of A is its own limit point
0 is not in A, but every neighborhood of 0 contains 1/n for sufficiently large n — infinitely many points of A distinct from 0. That satisfies the definition. The elements of A like 1/2 are isolated: the interval (1/3, 2/3) contains 1/2 but no other point of A, so 1/2 does NOT meet the definition of limit point.
Question 2 Multiple Choice
Point x is a limit point of set A if and only if:
Ax ∈ A and some neighborhood of x contains another point of A
BEvery neighborhood of x contains at least one point of A (including possibly x itself)
CEvery neighborhood of x contains a point of A distinct from x
Dx is the limit of a convergent sequence of distinct points in A
The definition of limit point requires every neighborhood to contain a point of A DIFFERENT from x. Option B is too weak — it would make any point of A trivially a limit point of itself. Option D is too restrictive — sequences do not characterize limit points in all topological spaces, only in first-countable ones.
Question 3 True / False
If x is a limit point of A, then x should be an element of A.
TTrue
FFalse
Answer: False
Limit points need not belong to A. For example, 0 is a limit point of A = {1/n : n ∈ ℕ} but 0 ∉ A. The closure cl(A) = A ∪ A' explicitly includes limit points that lie outside A — that is precisely what closure adds.
Question 4 True / False
A set is closed if and only if it contains all of its limit points.
TTrue
FFalse
Answer: True
This is the characterization of closed sets via limit points: A is closed ⟺ A' ⊆ A ⟺ cl(A) = A. A closed set contains every point that is 'approached' by elements of the set. This connects the abstract definition of closed set to the intuition of a set that contains its boundary.
Question 5 Short Answer
Why does the definition of limit point require every neighborhood to contain a point of A DIFFERENT FROM x? What goes wrong if you drop that clause?
Think about your answer, then reveal below.
Model answer: Without the 'different from x' requirement, every point of A would trivially be a limit point of A, since every neighborhood of x ∈ A contains x itself. The notion would collapse: every set would contain all its 'limit points,' all sets would be closed by the new definition, and the distinction between isolated points and genuine accumulation points would disappear.
The clause is what distinguishes isolated points (in A but not approached by other elements of A) from genuine limit points. An isolated point has a neighborhood containing no other element of A; a limit point has OTHER elements of A in every neighborhood, no matter how small.