The Real And Complex Number Systems

N
|fx f0| |fy 2k fy 2k1 | |fy 2k1 fy 2k2 |
k1
2N by (*)
which implies that
|fx| 2N |f0|
2 1 |x| |f0| since |x| N 1
2 |x| 2 |f0|.
Similarly for x 0. So, we have proved that |fx| |x| for all x.
4.56 In a metric space S, d, letA be a nonempty subset of S. Define a function
f A : S R by the equation
f A x infdx, y : y A
for each x in S. The number f A x is called the distance from x to A.
(a) Prove that f A is uniformly continuous on S.
(b) Prove that clA x : x S and f A x 0 .
Proof: (a) Given 0, we want to find a 0 such that as dx 1 , x 2 , x 1 , x 2 S,
we have
|f A x 1 f A x 2 | .
Consider (x 1 , x 2 , y S)
dx 1 , y dx 1 , x 2 dx 2 , y, anddx 2 , y dx 1 , x 2 dx 1 , y
So,
infdx 1 , y : y A dx 1 , x 2 infdx 2 , y : y A and
infdx 2 , y : y A dx 1 , x 2 infdx 1 , y : y A
which implies that
f A x 1 f A x 2 dx 1 , x 2 and f A x 2 f A x 1 dx 1 , x 2
which implies that
|f A x 1 f A x 2 | dx 1 , x 2 .
Hence, if we choose , thenwehaveasdx 1 , x 2 , x 1 , x 2 S, wehave
|f A x 1 f A x 2 | .
That is, f A is uniformly continuous on S.
(b) Define K x : x S and f A x 0 , we want to show clA K. We prove it
by two steps.
() Letx clA, then Bx; r A for all r 0. Choose y k Bx;1/k A, then
we have
infdx, y : y A dx, y k 0ask .
So, we have f A x infdx, y : y A 0. So, clA K.
() Letx K, then f A x infdx, y : y A 0. That is, given any 0, there
is an element y A such that dx, y . Thatis,y Bx; A. So,x is an adherent
point of A. Thatis,x clA. So, we have K clA.
From above saying, we know that clA x : x S and f A x 0 .
Remark: 1. The function f A often appears in Analysis, so it is worth keeping it in mind.