Let A ⊂ ℝ be a non-empty subset.
If sup(A) ∈ A and inf(A) ∈ A, then A is closed.
False.
Take, for example, A = [0, 1[ ∪ ]1, 2].
The statement would be true for an interval.
Let A ⊂ ℝ be a non-empty subset.
If sup(A) ∉ A and inf(A) ∉ A, then A is open.
False.
Take A =]0, 1] ∪ [2, 3[.
Let A ⊂ ℝ be a non-empty subset.
If A is open, then its boundary ∂A is empty.
False.
Take A =]0, 1[, so that ∂A = {0, 1}.
Let A ⊂ ℝ be a non-empty subset.
If A = {x: 0 ≤ x2 < 4, x ∈ ℚ}, then A has no supremum in ℚ.
False.
√4 = 2 ∈ ℚ is the supremum of A.
Let A ⊂ ℝ be a non-empty subset. We denote by Å its interior, ¬A its closure and ∂A its boundary.
The boundary ∂A of A is closed.
True.
The boundary is ∂A = ¬A ∩ ¬(ℝ\A).
The closure of a set is closed and the intersection of two closed sets is closed.
Let A ⊂ ℝ be a non-empty subset. We denote by Å its interior, ¬A its closure and ∂A its boundary.
If a ∈ A, then a ∈ Å.
False.
Take A = {1}. We have 1 ∉ Å = ∅.
Let A ⊂ ℝ be a non-empty subset. We denote by Å its interior, ¬A its closure and ∂A its boundary.
If a ∈ ∂A, then a is a limit point of A.
False.
Take A={1}.
We have ∂A = {1} but 1 is not a limit point of A.
Consider the set E.
E ∪ {1} is closed.
True.
E ∪ {1} is the closure of E and the closure of a set is closed.
Consider the set E.
E has an infinite number of isolated points.
True.
E has infinitely many elements. They are all isolated since for each of them, we can
find a closed interval in R which contains only that point of E.
Consider the set E.
E° = E.
False.
Since E contains only isolated points, E° = ∅.
