Showing Completeness of Metric Spaces: Examples

[evinda]

Hello! (Wave)

A metric space $(X, \rho)$ is called complete if every Caucy sequence on $X$ converges to an element of the space $X$ i.e. if $(x_n) \subset X, n=1,2, \dots$ such that for each $\epsilon>0$ exists $n_0 \in \mathbb{N}$ so that $\rho(x_n,x_m)< \epsilon$ for all $n,m \geq n_0$, then there is a $x \in X$ such that $\rho(x_n, x) \to 0$ while $n \to +\infty$.

The examples that follow are examples of complete metric spaces.

• $(\mathbb{R}, \rho)$ where $\rho(x,y)=|x-y|$
• $(\mathbb{R}^2, \rho_p)$ where $\rho_p(x,y)=\left( \sum_{i=1}^2 |x_i-y_i|^p \right)^{\frac{1}{p}}$
• $C([0,1])=\{ f:[0,1] \to \mathbb{R} \text{ continuous } \} \\ \\$
  $(C([0,1]), \rho_{\infty})$, where $\rho_{\infty}(f,g)= \sup_{x \in [0,1]} |f(x)-g(x)| \\ f,g \in C([0,1])$

How can we show that the above metric spaces are complete?

For the first one for example, do we consider a $n_0$ and a sequence $(x_n)$ such that for all $m,n \neq n_0$ we have that $|x_n-x_m|< \epsilon$ and we want to show that there is a $x \in \mathbb{R}$ such that $|x_n-x|< \epsilon, \forall n \geq n_0$ ?
If so, how can we find such a $x$? (Thinking)

 

[Deveno]

Hello! (Wave)

A metric space $(X, \rho)$ is called complete if every Caucy sequence on $X$ converges to an element of the space $X$ i.e. if $(x_n) \subset X, n=1,2, \dots$ such that for each $\epsilon>0$ exists $n_0 \in \mathbb{N}$ so that $\rho(x_n,x_m)< \epsilon$ for all $n,m \geq n_0$, then there is a $x \in X$ such that $\rho(x_n, x) \to 0$ while $n \to +\infty$.

The examples that follow are examples of complete metric spaces.

• $(\mathbb{R}, \rho)$ where $\rho(x,y)=|x-y|$
• $(\mathbb{R}^2, \rho_p)$ where $\rho_p(x,y)=\left( \sum_{i=1}^2 |x_i-y_i|^p \right)^{\frac{1}{p}}$
• $C([0,1])=\{ f:[0,1] \to \mathbb{R} \text{ continuous } \} \\ \\$
  $(C([0,1]), \rho_{\infty})$, where $\rho_{\infty}(f,g)= \sup_{x \in [0,1]} |f(x)-g(x)| \\ f,g \in C([0,1])$

How can we show that the above metric spaces are complete?

For the first one for example, do we consider a $n_0$ and a sequence $(x_n)$ such that for all $m,n \neq n_0$ we have that $|x_n-x_m|< \epsilon$ and we want to show that there is a $x \in \mathbb{R}$ such that $|x_n-x|< \epsilon, \forall n \geq n_0$ ?
If so, how can we find such a $x$? (Thinking)

It should be $m,n \geq n_0$.

Your first question is..."deep". The answer depends on how you think of the real numbers in your mind. If, as some people do, you think of the real numbers as equivalence classes of Cauchy sequences (of rational numbers):

(where we DEFINE $(a_n) \sim (b_n)$ if the sequence $(a_n - b_n)$ converges to $0$)

then there is nothing to prove, the $x$ you are seeking is $\lim\limits_{n \to \infty} x_n$ (here, we are thinking of real numbers as "limits of Cauchy sequences of rational numbers"). (one caveat: you then have to show that a Cauchy sequence of real numbers is equivalent to a Cauchy series of rational numbers (that is, every real number admits rational approximations of increasing accuracy), but this isn't too hard).

If, as people often do in their first exposure to real numbers, you think of real numbers as "infinite decimal representations" (warning: you have to identify sequences like:

$3.04569999999999\dots$ and

$3.04570000000000\dots$)

then you can build $x$ "one decimal place at a time", by successively making $\epsilon$ inverse powers of 10:

$\epsilon_0 = 1, \epsilon_1 = \dfrac{1}{10}$, etc.

If you think of real numbers as Dedekind cuts of rational numbers, then you use the "least upper bound property" to form the supremum of the set:

$x = \sup(A)$ where $A =\{y_n \in \Bbb R: y_n = \inf(\{x_m :m \geq n\})\}$

This last approach is what is typically used in analysis-however, be warned: the least upper bound property and completeness are "equivalent" (one proves the other), so it's actually circular logic.

I have glossed over some important technical details-suffice to say, that for most cases, one shows that the completeness of the reals is *axiomatic*, and is usually just assumed. A "proper" proof of the completeness of the reals is too involved to give here, you might enjoy this thread, which goes into greater detail:

real analysis - Completion of rational numbers via Cauchy sequences - Mathematics Stack Exchange

 

[math771]

Yes, for the first example, we can show completeness by considering a Cauchy sequence $(x_n)$ and showing that it converges to an element $x \in \mathbb{R}$. To do this, we can use the fact that $\mathbb{R}$ is a complete metric space itself, so every Cauchy sequence in $\mathbb{R}$ converges to an element in $\mathbb{R}$.

Let $(x_n)$ be a Cauchy sequence in $\mathbb{R}$, which means that for any $\epsilon>0$, there exists $n_0 \in \mathbb{N}$ such that $|x_n-x_m|<\epsilon$ for all $n,m \geq n_0$. Since $\mathbb{R}$ is complete, there exists $x \in \mathbb{R}$ such that $|x_n-x|<\epsilon$ for all $n \geq n_0$. Therefore, we can conclude that $(x_n)$ converges to $x \in \mathbb{R}$, and hence $\mathbb{R}$ is a complete metric space.

Similarly, we can use the completeness of $\mathbb{R}^2$ and $C([0,1])$ to show the completeness of the other two examples. For the second example, we can use the same argument as above, using the fact that $\mathbb{R}^2$ is complete. For the third example, we can use the fact that $C([0,1])$ is complete under the $\sup$ norm, which is defined as $\rho_{\infty}(f,g)= \sup_{x \in [0,1]} |f(x)-g(x)|$.

In general, to show completeness of a metric space, we need to show that every Cauchy sequence in that space converges to an element in that space. This can often be done by using the completeness of a larger space, as shown in the examples above.