# R2: (x, y is an open subset of R2

## 全文

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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.

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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= ∅.)

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