Proving the usual distance metric in $\mathbb{R}$ is complete

Question

If we allow the metric to be $d(x,y)=|x-y|$, we must prove that this is complete. Now, I have proven all properties of a metric space. However, I don't particularly now where to begin to prove that this is complete.

I understand that we need every Cauchy sequence in this space to converge inside the space itself. Now, I know the definition of Cauchy convergence, but I don't particularly know how to even begin by showing that this is complete. I have much harder metrics to prove but I would like to try to understand this simpler space before I move onto more difficult ones.

Since we know that the output of this distance will always be some constant in $\mathbb{R}$, and since we are working with $\mathbb{R}$, then it must always converge to some real number, and thus its complete?

Answer 1

Let $(x_n)$ be a Cauchy sequence in $\mathbb{R}$. We note that Cauchy sequences are bounded and hence any subsequence will also be bounded. We also note that every sequence of the real numbers has a monotone subsequence. By the Monotone Convergence Theorem, we have that the subsequence convergence. We have shown an arbitrary Cauchy sequence converges. Hence $\mathbb{R}$ is complete with the distance metric.

Remarks:

• We know that Cauchy sequences are bounded because taking $\epsilon=1$ gives us $\exists N\in\mathbb{N}:\forall m,n\geq N\implies |x_n-x_m|<1$ and so $|x_m|\leq |x_N|+|x_m-x_N|\leq |x_N|+1$ (by the Triangle inequality)

• To show that every sequence in $\mathbb{R}$ has a monotone subsequence:

Suppose $(x_n)$ is a real sequence. Let $E=\{n\mid x_n\geq x_m \forall m\geq n\}$. If $E$ is infinite, then each element of $E$ can be collected and is a monotone decreasing subsequence. Else, if $E$ is finite, we can construct a increasing subsequence. We consider terms of the subsequence that are not elements of $E$, say $x_{n+1}$. This means that $\exists n_2>n_1: x_{n_2}>x_{n_1+1}$. Label this point $x_{K_1}$. Then $x_{n_2}\notin E$ and so similarly to what we just did, we note that $\exists n_3>n_2: x_{n_3}>x_{n_2}$. We label this point $x_{K_2}$. We can of course keep doing this.

You can then make a subsequence $x_{K_1}<\ldots<x_{K_p}$ which is of course increasing.

So, if $E$ is finite, we have a increasing monotone subsequence and if $E$ is infinite, we have a decreasing monotone subsequence. Hence regardless of whether $E$ is finite or infinite, we are ensured the existence of a monotone subsquence.

Answer 2

If I am understanding the question correctly, you must suppose the metric holds in $R$ and suppose some sequence $p_1,..., p_n$ converges to $p$ and thus is Cauchy. And then show by the metric that $p \in R$.

Suppose $p_1,...,p_n$ be a sequence in $R$ that converges to some point $p$. Since $p_1,...,p_n$ converges, it is Cauchy, by theorem. Also by definition of convergent, for some arbitrary $r \in R$, there exist an natural number $N$ s.t. $d(pn, p) < r$, $\forall n>N$. And since $d(pn, p) = |pn - p|$ we have, $|pn - p| < r$ and thus $p > p_n + r$ or $p > p_n - r$.

And since $r \in R$, in either case $p \in R$ and the metric space is complete.

Answer 3

Suppose $\{x_n\}$ is a Cauchy sequence in $\mathbb{R}$. We note that Cauchy sequences are bounded. The Bolzano-Weierstrass Theorem tells us that every infinite sequence in $\mathbb{R}$ has a convergent subsequence, so $\lim_{j\to\infty}x_{n_j}=x$.

Cauchy sequences are bounded since for $\epsilon=1$, we have $\exists N\in\mathbb{N}:\forall m,n\geq N\implies |x_n-x_m|<1$ and it then follows from the triangle inequality that $|x_m|\leq |x_N|+|x_m-x_N|\leq |x_N|+1$

The limit of a subsequence is the limit of the Cauchy sequence because $\forall\epsilon>0\exists N\in\mathbb{N}:\forall m,n\geq N\implies|x_m-x_n|<\frac{\epsilon}{2}$

We note that $\lim_{j\to\infty}x_{n_j}=x$ means that $\exists N_1\in\mathbb{N}: \forall j\geq N_1\implies |x_{n_j}-x|<\frac{\epsilon}{2}$

Now, $\forall\epsilon>0\exists N,N_1\in\mathbb{N}:\forall n,n_{j}\geq N,\forall j\geq N_1$, we have $|x_{n_j}-x|\leq |x_n-x_{n_j}|+|x_{n_j}-x|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

Hence an arbitrary Cauchy sequence is convergent in $\mathbb{R}$. Hence $\mathbb{R}$ is complete.

Answer 4

This is one of those theorems whose proof really depends on how real numbers were introduced in the course. You could take the completion of the rationals with respect to the usual metric and show the resulting space is complete and homeomorphic to $\mathbb{R}$ (this is the construction due to Cantor).

If you just want to use the property that every bounded set has a least upper bound, take a Cauchy sequence $\{a_n\}$, then look at the sequence $b_k = \inf\limits_{n\geq k}(a_n)$. This is non-decreasing and bounded from above, since $\{a_n\}$ is Cauchy (and therefore bounded). Hence it has a least upper bound $L$ and in fact converges to $L$. Verify that $a_n \to L$.