A concrete example of a proof of completness

Question

I'v been looking arround in some of my books, all say the same: "To prove a metric space is complete, show that every Caychy sequence is convergent to a pointin the space". Yet noone gives me an example of doing this, my mind can't connect the Cauchy and convergent sequences :/ Can you gime a concrete example?

Answer 1

Here is a concrete example: Starting from order completeness of ${\mathbb R}$ it has been proven to you that ${\mathbb R}$ is also complete as a metric space. Now it is an exercise for you to prove that ${\mathbb R}^n$ with the usual euclidean metric is a complete metric space as well.

"Connecting" Cauchy and convergent sequences:

Any convergent sequence in any metric space is automatically a Cauchy sequence in that space, for the following reason: If all $x_n$ are near the limit point $\xi$ when $n$ is large, then $x_n$ and $x_m$ are also near each other when both $n$ and $m$ are large, by the triangle inequality.

The essence of completeness is the converse: You can test the "Cauchyness" of a sequence by just looking at the given terms $x_n$, and don't have to know the prospective limit for this test. If the underlying metric space is complete then any sequence that passes the Cauchy test is automatically convergent.

Answer 2

Here a something elaborated example with a functional space. I hope this interests you.

We know $\mathbb R$ is complet with the usual distance of the absolue value. Let $E$ be a set and define the space $$\mathcal F=\{\text {bounded functions of E in } \mathbb R \} $$ $\mathcal F$ becomes a metric space with the distance $$d(f,g)=\sup_{x\in E}|f(x)-g(x)|\space\text {(left as an exercise)}$$ We prove that $\mathcal F$ becomes a complet metric space.

Let $\{f_n\}$ a Cauchy sequence in $\mathcal F$. One has by definition of the distance $d$ $$|f_n(x)-f_m(x)|\le d(f_n,f_m)$$ It follows that $\forall x\in E$ the sequence $\{f_n(x)\}$ is also Cauchy in the complet space $\mathbb R$ so this sequence converges to a point, say $f(x)$, of $\mathbb R$ This defines a function $f$ of $E$ in $\mathbb R$.

We should prove that $f$ is bounded and that $f_n\to f$ in the metric space $\mathcal F$.

$(1)$ $f$ is bounded.

For all $\epsilon \gt 0$ there is $N$ such that $n,m\ge N$ implies $d(f_m,f_n)\le \epsilon$ so $|f_m(x)-f_n(x)|\le \epsilon$ for all $x\in E$. Then for $x$ fixed and continuity of the absolue value we can make $m$ tends to $\infty$ which gives $|f(x)-f_n(x)|\le \epsilon$ for $n\ge N$.

Since $f_n$ is bounded $f_n(E)$ is contained in a ball centered, say,at $a_n$ and radius $r_n$. Now $|a_n-f_n(x)|\le |a_N-f_n(x)|+|f_n(x)-f(x)|$ which proves that $f$ is bounded (i.e. $f\in \cal F$).

$(2)$ $f_n\to f$ when $n\to \infty$

For all $\epsilon\gt 0$ the same integer $N$ of $(1)$ gives $|f(x)-f_N(x)|\le \epsilon$ for all $x\in E$ and all $n\ge N$ which shows that $d(f,f_N)\le \epsilon$ i.e. $f_n\to f$.