1. Quiz 3

True or false. If the statement is true, prove it; if not, disprove it or give a counterexample.

(1) ( ) The set

{(x, y) ∈ R^{2}: (x, y) 6= (0, 0)}

is an open subset of R^{2}.

Proof. The statement is true. The proof is given as follows. Let U = {(x, y) ∈ R^{2}: (x, y) 6=

(0, 0)}. By definition,

U = R^{2}\ {(0, 0)}.

Denote (0, 0) by 0. We show that U is open by proving that every point of U is an interior point of U. Let p be a point of U. Then p 6= 0. Thus d(p, 0) > 0. Choose = d(p, 0)/2. Then

> 0. Claim that B(p, ) is contained in U. To show that B(p, ) is contained in U, we need to show that every element of B(p, ) is an element of U. To show that a point belong to U, we only need to show that that point is not the origin, i.e. q ∈ U if and only if q 6= 0. Let q ∈ B(p, ). Then d(p, q) < . By triangle inequality,

d(q, 0) > d(p, 0) − d(p, q) > d(p, 0) − = 2 − = > 0.

Thus q 6= 0. We find q ∈ U. Therefore B(p, ) ⊂ U. This shows that p is an interior point of U. We conclude that every point of U is an interior point of U. (2) ( ) The point p(1, 1) is an interior point of

{(x, y) ∈ R^{2}: x^{2}+ 4y^{2}≤ 16}.

Proof. The statement is true. Let S be the given set. Take = 1/2. Claim that B(p, ) ⊂ S.

In fact, we can prove that B(p, ) is contained in the interval I = {(x, y) : 1/2 < x, y < 3/2}

and prove that I is contained in S. The proof of the fact that the ball B(p, ) is contained in I is given in the solution of homework 1. Now, let us prove that I is contained in S. If (x, y) ∈ I, then 0 < x, y < 3/2. Thus

x^{2}+ 4y^{2}<9
4 + 4 ·9

4 = 45 4 < 16.

This shows that (x, y) ∈ S.

(3) ( ) The point p(1, 2) is a boundary point of

{(x, y) ∈ R^{2}: xy > 1}.

Proof. The statement is false. Let U be the given set. Then U is open by homework 1.

Since p ∈ U, p is an interior point of U. We can choose > 0 so that B(p, ) ⊂ U. Then

B(p, ) ∩ U^{c}= ∅.

(4) ( ) The point p(0, 0) is an isolated point of
{(x, 0) ∈ R^{2}: x ∈ Q}.

Here Q is the set of all rational numbers.

Proof. The statement is false. Let S be the given set. For any > 0, (by the density
of Q) we can choose a rational number r^{} so that 0 < r < . Take p = (r, 0). Then
p ∈ S ∩ B^{0}((0, 0), ). Thus S ∩ B^{0}((0, 0), ) is always nonempty for any > 0. Thus (0, 0) is
not an isolated point; it is an accumulation point.

(5) ( ) The set

{(m, n) ∈ R^{2}: m, n ∈ Z, 0 < n < 4}

is a closed set. Here Z is the set of all integers.

1

2

Proof. The statement is true. Let S be the given set. For i = 1, 2, 3, we set Si= {(m, i) ∈
R^{2}: m ∈ Z}. Then

S = S1∪ S2∪ S3.

You need to show that S1, S2, S3 are closed sets. Since any finite union of closed sets is
closed, S is closed. To show that Si is a closed set, you show that S_{i}^{0} is an empty set for all
i = 1, 2, 3. Use the fact that a set is closed if and only if it contains A^{0}. You find Siis closed
(since it contains S_{i}^{0}= ∅.)