1. Quiz 4 True or false. Prove or disprove the following statements.
(1) ( ) Let U be a subset of Rn. U is open if and only if ∂U = U \ U.
Proof. The statement is true. The key idea is that we need to use the decomposition of Rn into the following disjoint union of sets:
Rn= int(U ) ∪ ∂U ∪ ext(U ).
When U = ∅, ∂U = ∅ and U = U = ∅ and hence U \ U = ∅. The equation ∂U = U \ U holds when U = ∅.
Now assume that U 6= ∅.
Suppose ∂U = ∅ and U is open. Since U ⊃ ∅ = ∂U, U is also closed. Hence U = U. Thus U is both open and closed. We see that U \ U = ∅ = ∂U holds.
Suppose ∂U = ∅ and ∂U = U \ U holds. Then U = U. Thus U is closed. Let p ∈ U, p is never a boundary point of U (since ∂U = ∅). We can choose > 0 so that B(p, ) ∩ Uc = ∅.
This implies that B(p, ) ⊂ U. Since p is an arbitrary chosen point of U and p is an interior point of U, U is open. Thus we see that when ∂U = ∅, U is both open and closed. If U is open and ∂U = ∅, then U is open and closed.
Now assume that ∂U 6= ∅.
Suppose U is open. Then U = int(U ) Let p ∈ ∂U. Then B(p, ) ∩ Uc6= ∅ for any > 0. We see that p is never an interior point of U. Hence p is never a point of U. We find p ∈ U \ U.
This shows that ∂U ⊂ U \ U. If p ∈ U \ U, then p is not in U and p ∈ U and thus p is never an interior point of U. We see that B(p, ) ∩ Uc 6= ∅ for any > 0. We find p ∈ ∂U. We find U \ U ⊂ ∂U. We conclude that ∂U = U \ U when U is open.
Suppose ∂U = U \ U holds. Let p ∈ U. Then p is never a boundary point of U. There exists > 0 so that B(p, ) ∩ Uc = ∅. Thus B(p, ) ⊂ U. Hence p is an interior point of U.
Since p is arbitrarily chosen, U is open.
(2) ( ) Let A be a subset of Rn. Then ∂A = ∅ if and only if A is both open and closed.
Proof. This is already proved in the above problem.
(3) ( ) If U is an open subset of Rn, then U = int(U ).
Proof. The statement is false.
Consider
U = {(x, y) : 0 < x < 1, 1 < x < 2, −1 < y < 1}.
Then U = {(x, y) : 0 ≤ x ≤ 2, −1 ≤ y ≤ 1} and int(U ) = {(x, y) : 0 < x < 2, −1 < y < 1}.
We see
int(U ) 6= U.
(4) ( ) There exists a subset S of Rn so that ∂S = Rn.
Proof. The statement is true. Take S = Qn = {(r1, · · · , rn) : ri ∈ Q}. Then ∂S = Rn. (Prove it).
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(5) ( ) If A is a finite subset of Rn and B is a closed subset of Rn, A + B is closed.
Proof. The statement is true. At first show that x + B is closed for any x ∈ Rn. Then show that
A + B = [
x∈A
(x + B).
Use the closed set property: any finite union of closed sets is closed. (6) ( ) If A is an infinite subset of Rn and B is a closed subset of Rn, A + B is closed.
Proof. The statement is false. Please give a counterexample by yourself.
