25. Every Metric Space can be Isometrically Embedded in a Complete Metric Space - II

We outline a second proof that every Metric Space can be Isometrically Embedded in a Complete Metric Space. Much of the material on this Page is taken directly from section 5.1 of the text.

25.1 Definition:

Let MATH be a sequence of continuous functions for a metric space $\QTR{Large}{(M,d)}$ to a metric space MATH . The sequence is said to converge uniformly to MATH if for any MATH0 , there is a n such that MATH for all MATH and i MATHn .

We write MATH.

25.2 Theorem:

In the setting of 25.1, $\QTR{Large}{f}$ is continuous. That is the limit of a uniformly convergent sequence of continuous functions is continuous.

Proof:

We need to prove the continuity of $\QTR{Large}{f}$ at each point MATH . Given MATH0 , Choose n such that MATH for all MATH and i MATHn. Next choose MATH0, such that for MATH we have MATH

then for MATH

MATH

MATH

25.3 Example:

The following example shows that something like uniform convergence is necessary. seen

Let MATH0 for MATH

MATH1 for MATH0

MATH1 for -MATH0

and

MATH0 for MATH0

$\QTR{Large}{f(}$0$\QTR{Large}{)=\ }$1

note that pointwise MATH

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The text discusses the following material in a more generalitly.

25.4 Definition:

Given a Metric Space $\QTR{Large}{(M,d)}$ we let MATH be the set of bounded real valued continuous functions on $\QTR{Large}{M}$. For MATH , define

MATH

25.5 Theorem:

MATH is a complete Metric Space.

Proof:

A Good Review Exercise.

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Solution:

The essence of the proof is that definition allows you to work point by point.

To show

MATHjust note that

MATH for every $\QTR{Large}{x}$. Hence

MATH

MATH

That is comple is even more straight forward. The definition tells us that a "$\QTR{Large}{D}$ " Cauchy sequence is a Cauchy sequence for each $\QTR{Large}{x}$ and that for any MATH0 the same n works for all $\QTR{Large}{x}$.

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25.5 Theorem:

There is an isometric embedding MATH MATH .

Proof:

Fix MATH and for all MATH define

MATH

we need to verify that

  1. MATHis continuous.

  2. MATHis an isometric embedding.

  3. MATHis bounded.

Proof:

Proving 2. first, suppose MATH, then

MATH

and

MATH

hence

MATH

Moreover,

MATH, choose $\QTR{Large}{x=v}$

but by the 17.1.4 form of the triangle inequality, for all MATH , MATH

Note that 3. follows from the same inequality with $\QTR{Large}{v=a.}$

To prove 1. it suffices to check that for all MATH, MATH is continuous in $\QTR{Large}{x}$ .

A Good Review Exercise.

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Solution:

To stress that $\QTR{Large}{y}$ is fixed we will write MATHfor MATH We need to show that for any MATH and for any MATH0 we can find a MATH0 such that

MATH

but

MATH

so choose MATH.