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) ∈ R2: (x, y) 6= (0, 0)}
is an open subset of R2.
Proof. The statement is true. The proof is given as follows. Let U = {(x, y) ∈ R2: (x, y) 6=
(0, 0)}. By definition,
U = R2\ {(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) ∈ R2: x2+ 4y2≤ 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
x2+ 4y2<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) ∈ R2: 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, ) ∩ Uc= ∅.
(4) ( ) The point p(0, 0) is an isolated point of {(x, 0) ∈ R2: 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 ∩ B0((0, 0), ). Thus S ∩ B0((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) ∈ R2: 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) ∈ R2: 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 Si0 is an empty set for all i = 1, 2, 3. Use the fact that a set is closed if and only if it contains A0. You find Siis closed (since it contains Si0= ∅.)
