S is nonempty for any  &gt

1. Some Examples

The set of all boundary points of a set S ⊂ Rⁿ is denoted by ∂S. By definition, a point p is a boundary point of a set S if B(p, ) ∩ S and B(p, ) ∩ (Rⁿ\ S) is nonempty for all  > 0. By definition, if B(p, ) ∩ S is nonempty for any  > 0, then p ∈ cl(S). Similarly, if B(p, ) ∩ (Rⁿ\ S) is nonempty for any  > 0, p ∈ cl(Rⁿ\ S). By definition,

∂S = cl(S) ∩ cl(Rⁿ\ S).

Example 1.1. Show that p(0, 0) is an accumulation point of D = {(x, y) : x > 0, y > 0, x 6= y}.

Proof. We have to show that B(p, )∩D is nonempty for all  > 0. Let  > 0. Choose p^= (4 10, 3

10).

Since  > 0, 4

10 > 0 and 3

10 > 0 and 4 10 > 3

10. The point p_ belongs to D. Moreover, d(p, p_) =

s

 4 10

2

+ 3 10

2

=  2 < ,

which shows p∈ B(p, ). Both B(p, ) and D contains p, we find p ∈ B(p, ) ∩ D. In other words,

B(p, ) ∩ D is nonempty. 

Example 1.2. Let B = {x²+ y²< 1}. Show that ∂B is S = {x²+ y²= 1}.

Let p(0, 0) be the origin. Let us prove that ∂B = S. We want to show that S ⊂ ∂B and ∂B ⊂ S.

Let us prove S ⊂ ∂B first. To do this, we prove that S is contained in cl(B) ∩ cl(R²\ B).

Let q(x0, y0) be a point on S. Then x²₀+ y₀²= 1. Let  > 0. Choose a point q=

(1 + 

2)x0, (1 +  2)y0

 . Then

d(q, q_) =  2

q

x²₀+ y₀²=  2 < , which shows that q∈ B(p, ). Moreover,

d(p, q_) = (1 +  2) > 1.

Hence q∈ R²\ B. We find that both B(q, ) and R²\ B both contains q. Hence q∈ B(q, ) ∩ (R²\ B).

Hence B(q, ) ∩ (R²\ B) is nonempty for any  > 0. We see that q ∈ cl(R²\ B).

When  > 1, choose q = p ∈ B. Then d(q, q^) = d(p, q) = 1 <  which shows that q ∈ B(q, ).

Since q_= p ∈ B, we find that B and B(q, ) both contain q. In other words, q_∈ B(q, )∩B whenever

 > 1. Thus B(q, ) ∩ B is nonempty whenever  > 1. When 0 <  < 1, we choose q=

(1 − 

2)x0, (1 −  2)y0

 . Then

d(p, q) = (1 −  2) < 1, which gives q∈ B. Moreover,

d(q, q_) =  2 < 1.

Thus q∈ B(q, ). We find that both B and B(q, ) contain q. In other words, q∈ B(q, ) ∩ B. As a consequence, B(q, ) ∩ B is nonempty whenever 0 <  < 1. Combining with the previous result, we find B(q, ) ∩ B is nonempty for every  > 0. This implies that q ∈ cl(B). Thus

q ∈ cl(B) ∩ cl(R²\ B).

1

2

This implies that S ⊂ ∂B.

Now let us prove the converse. Let q ∈ ∂B. If q ∈ B, then q is an interior point of B. We can choose  > 0 so that B(q, ) ⊂ B. In this case, B(p, ) ∩ R²\ B is empty and hence p is not on the boundary of B. Hence p 6∈ B, and thus d(p, q) ≥ 1. If d(p, q) > 1, then we can show that p is an interior point of R²\ B and hence it is never on ∂B. The only possibilty is d(p, q) = 1. Thus q ∈ S.