Proof
(related to Proposition: Complex Cauchy Sequences Vs. Real Cauchy Sequences)
In the following \((c_n)_{n\in\mathbb N}\) is a complex sequence with \(c_n=a_n+i\cdot b_n\) for some real numbers \(a_n,b_n\). We will prove two inclusions:
\("\Rightarrow"\)
- Assume, \((c_n)_{n\in\mathbb N}\) is a complex Cauchy sequence.
- Then, for each \(\epsilon > 0\), there exists an \(N\in\mathbb N\) with
\[ | c_n-c_m | < \epsilon\quad\quad \text{ for all }n,m\ge N.\]
- Because for every complex number \(x\in\mathbb C\), we have \(|\Re(x)|\le |x|\) and \(|\Im(x)|\le |x|\), it follows for all \(m,n\ge N\):\[\begin{array}{c}
| a_n-a_m |=|\Re(c_n -c_m)|\le | c_n-c_m | < \epsilon,\\
| b_n-b_m |=|\Im(c_n -c_m)|\le | c_n-c_m | < \epsilon.\\
\end{array}\]
- Thus, both real sequences \((a_n)_{n\in\mathbb N}\) and \((b_n)_{n\in\mathbb N}\) are real Cauchy sequences.
\("\Leftarrow"\)
- Assume, the real sequences \((a_n)_{n\in\mathbb N}\) and \((b_n)_{n\in\mathbb N}\) are real Cauchy sequences.
- Then, for each \(\epsilon > 0\), there exist indices \(N_a,N_b\in\mathbb N\) with \[ | a_n-a_m | < \frac\epsilon2\quad\quad \text{ for all }n,m\ge N_a\] and \[ | b_n-b_m | < \frac\epsilon2\quad\quad \text{ for all }n,m\ge N_b.\]
- Set \(N:=\max(N_a,N_b)\).
- Then, for all \(n,m\ge N\), it follows from the triangle inequality for the distance in the metric space of complex numbers that
$$\begin{align}| c_n-c_m |&= |a_n+ib_n -a_m-ib_m|\nonumber\\
&=|(a_n-a_m)+i(b_n-b_m)|\nonumber\\
&\le |a_n-a_m|+|b_n-b_m|\nonumber\\
& < \frac\epsilon2+\frac\epsilon2\nonumber\\
&=\epsilon.\nonumber\end{align}$$
- Therefore, \((c_n)_{n\in\mathbb N}\) is a complex Cauchy sequence.
∎
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
