Proof.

Convergent implies Cauchy

If \(x_n\rightarrow x\) then given \(\varepsilon >0\) choose \(N\) with \(\left |x_n-x\right |<\varepsilon /2\) for \(n\geq N\). For \(m,n\geq N\), \[\left |x_m-x_n\right | \leq \left |x_m-x\right | + \left |x-x_n\right | < \varepsilon .\] This direction holds in every metric space.

Cauchy implies convergent

This direction is special to \(\mathbb {R}\) and rests on its completeness axiom. A Cauchy sequence is bounded: taking \(\varepsilon =1\) gives an \(N\) with \(\left |x_n-x_N\right |<1\) for \(n\geq N\), so every term lies within \(\max \left \{\left |x_1\right |,\dots ,\left |x_{N-1}\right |,\left |x_N\right |+1\right \}\) of the origin. By the Bolzano–Weierstrass theorem a bounded real sequence has a convergent subsequence, say \(x_{n_k}\rightarrow x\).

A Cauchy sequence with a convergent subsequence converges to the same limit. Given \(\varepsilon >0\), choose \(N\) with \(\left |x_m-x_n\right |<\varepsilon /2\) for \(m,n\geq N\), and then \(n_k\geq N\) with \(\left |x_{n_k}-x\right |<\varepsilon /2\). For any \(n\geq N\), \[\left |x_n-x\right | \leq \left |x_n-x_{n_k}\right | + \left |x_{n_k}-x\right | < \varepsilon .\] □

Proof. Let \((x_n)\) be a Cauchy sequence in \(\mathbb {R}^2\) and let \(x_n=(x_1^{(n)},x_2^{(n)})\) since \((x_n)\) is Cauchy, for every \(\epsilon >0\) there is an \(N>0\) such that
\(d(x_n,y_m)=\sqrt {(x_1^{(n)}-y_1^{(m)})^2+(x_2^{(n)}-y_1^{(m)})^2}<\epsilon \) for all \(n,m>N\). Squaring, we obtain that \((x_1^{(n)}-y_1^{(m)})^2<\epsilon ^2\), so that \(|x_1^{(n)}-y_1^{(m)}|<\epsilon \) for all \(n,m>N\).
Similarly, \(|x_2^{(n)}-y_2^{(m)}|<\epsilon \) for \(n,m>N\). This shows that \((x_1^{(n)},x_2^{(n)},..........)\) and \((y_1^{(m)},y_2^{(m)},...........)\) are Cauchy sequences of real numbers and therefore convergent by the previous theorem say \(x_1^{(n)}\rightarrow x_1\) as \(n\rightarrow \infty \) and \(x_2^{(m)}\rightarrow x_2\) as \(m\rightarrow \infty \). Let \(x=(x_1,x_2)\). Then \(x\in \mathbb {R}^2\) and \(d(x_n,x)<\epsilon \) for \(n>N\). This proves that the sequence \((x_n)\) converges to x. Hence \(\mathbb {R}^2\) is complete.

We say that a sequence of real numbers is Cauchy if and only if it is convergent. This is not generally true in a metric space. However, we have the following result.
□

Proof. Suppose that \(x_n\rightarrow x\) as \(n\rightarrow \infty \). Then for every \(\epsilon >0\) there is an \(N>0\) such that \(d(x_n,x)< \dfrac {\epsilon }{2}\) for \(n>N\). From the triangle inequality, it follows that for \(n,m>N\), \begin {align*} d(x_m,x_n) &\leq d(x_m,x)+d(x,x_n)\\ &< \frac {\epsilon }{2} +\frac {\epsilon }{2}\\ &=\epsilon \end {align*}

Hence \((x_n)\) is a Cauchy sequence.

The previous two theorems provide examples of complete metric spaces. The following is an example of an incomplete metric space.

□

Proof. For every \(\epsilon >0\), the open ball \(B(\sqrt {2},\epsilon )\) contains a real number. Therefore there is a sequence \((x_n)\) of rational numbers converging to \(\sqrt {2}\). By the
previous theorem, we have that \((x_n)\) is a Cauchy sequence in \(\mathbb {Q}\) which does not converge in \(\mathbb {Q}\). Hence \((\mathbb {Q},d)\) is incomplete.

So far in this section we have defined a complete metric space, given examples and shown that every convergent sequence is Cauchy. We now turn to separable metric spaces.
□

Proof. If \(x_0\) is any real number , then there is a sequence \((x_n)\) of rational numbers such that \(x_n\rightarrow x_0\) as \(n\rightarrow \infty \). Therefore for any \(r>0\), the open ball \(B(x_0,r)\) always contains a rational number. Thus \(x_0\) is an an accumulation point of \(\mathbb {Q}\). Since \(x_0\) was an arbitrary real number, this means that \(\overline {\mathbb {Q}}=\mathbb {R}\).

□

Solution. From the previous, \(\mathbb {Q}\) is a countable dense subset of \(\mathbb {R}\).

Proof. A countable dense subset of \(\mathbb {C}\) is the set of all complex numbers whose real and imaginary parts are both rational.

□