1. hw 4
The set of accumulation points of S ⊂ Rn is denoted by S0. (1) Let S = {(x, y) ∈ R2: x ≥ 1 and y ≥ 1}.
(a) Find int(S).
The set int(S) = {(x, y) : x > 1 and y > 1}. The proof is sketched as follows. Let U = {(x, y) : x > 1 and y > 1}. Then U is an open subset of S. Since int(S) is the larges open set containing in S, U ⊂ int(S). Since int(S) and U are both subsets of S, to show that int(S) is a subset of U, we only need to show that S \ int(S) contains S \ U = {(1, y) : y ≥ 1} ∪ {(x, 1) : x ≥ 1}. We leave it to the reader to verify that {(1, y) : y ≥ 1} ∪ {(x, 1) : x ≥ 1} is contained in ∂S and thus points of S \ U can not be interior points of S. Thus S \ U ⊂ S \ int(S).
(b) Find ∂S.
The boundary set
∂S = {(x, 1) : x ≥ 1} ∪ {(1, y) : y ≥ 1}.
(c) Show that S is closed by showing S contains ∂S.
Since ∂S = {(x, 1) : x ≥ 1} ∪ {(1, y) : y ≥ 1} is a subset of S = {(x, y) ∈ R2 : x ≥ 1 and y ≥ 1}, S is closed.
(d) Show that S is closed by showing Sc is open.
Since Sc = {(x, y) : x < 1 or y < 1} is a union of open sets, Scis open. Thus S is closed.
(2) Let S = {(x, y) ∈ R2: x = 0, 0 < y < 1}.
(a) Does S have interior points?
The set S does not have interior points, i.e. int(S) = ∅.
(b) Does S have isolated points?
No. Since every point of S is an accumulation point of S, S has no isolated point. (An isolated point of a set must belong to the set.)
(c) Dose S have accumulation points? Find S0. S0= {(0, y) : 0 ≤ y ≤ 1}.
(d) Find S.
S = S0. (e) Find int(S).
int(S) = ∅.
(f) Is S closed?
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No. It is because S 6= S.
(g) Is S open?
No. It is because int(S) = ∅ 6= S.
(3) Let Z2= {(m, n) ∈ R2: m, n ∈ Z}.
(a) Does Z2 have accumulation points?
No.
(b) Does Z2 have interior points?
No.
(c) Show that all points of Z2 are isolated points of Z2. The proof is similar to that of quiz 2, problem (2).
(d) Is Z2closed? If yes, prove it.
Yes. A0= ∅ ⊂ A. A subset of Rn is closed it and only if it contains its derived set.
(4) Let A and B be subset of Rn with A ⊆ B. Is it true that if x is an accumulation point of A, then x is also an accumulation point of B?
Yes.
(5) Find S0 the set of all accumulation points of S. Here
(a) S = {(p, q) ∈ R2 : p, q ∈ Q}. Hint: every real number can be approximated by a se- quence of rational numbers.
S0= R2.
Proof. The inclusion S0 ⊂ R2 follows from definition. Let us prove R2 ⊂ S0. Let P (x, y) be any point of R2. For > 0, we choose rational numbers p and q such that 0 < |x − p| < /√
2 and 0 < |y − q| < /√
2. Then (p, q) ∈ S and 0 < (p− x)2+ (q− y)2<2
2 +2 2 = 2.
Hence (p, q) ∈ B0(P, ) ∩ S. We find P is an accumulation point of S. Thus P ∈ S0. This shows that R2⊂ S0.
(b) S = {(m/n, 1/n) : m, n are integers with n 6= 0}.
S0 is the x-axis.
(6) Find the closure of A = {(x, y) ∈ R2: x > y2}.
The closure of A is
A = {(x, y) : x ≥ y2}.
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(7) Let A and B be subsets of Rn. Is it true that
cl(A ∩ B) = cl(A) ∩ cl(B)?
No. Let A = (0, 1) and B = (1, 2). Then A ∩ B = ∅ and cl(A ∩ B) = ∅. On the other hand, cl(A) = [0, 1] and cl(B) = [1, 2]. Hence cl(A) ∩ cl(B) = {1}.
If you want an example in R2, let
A = {(x, y) : x2+ y2< 1}, B = {(x, y) : x2+ y2> 1}.
Then A ∩ B = ∅ and cl(A ∩ B) = ∅. On the other hand, cl(A) = {(x, y) : x2+ y2≤ 1} and cl(B) = {(x, y) : x2+ y2≥ 1}. Thus cl(A) ∩ cl(B) = {(x, y) : x2+ y2= 1}.
