Show that if $z_n \to z$ then $\exists M \in R $ such that $|z_n|\leq M\; \forall n \in N$

Question

Show that if $z_n \to z$ then $\exists M \in R $ such that $|z_n|\leq M\; \forall n \in N$

Let $z_n = x_n + \iota y_n$ and $z=x+\iota y$

$z_n \to z$

$\rightarrow x_n \to x $ and $y_n \to y$

because $\langle x_n \rangle$ and $\langle y_n \rangle$ are sequences of real numbers, $\exists M_1$ and $M_2$ such that $|x_n|\leq M_1$ and $|y_n|\leq M_2 \; \; \forall n \in N$

$|z_n|= |x_n + \iota y_n| \leq |x_n| + |y_n| \leq M_1 + M_2 = M \; \; forall n$

Is this proof correct?

My question is different, the sequence is in complex numbers and I have used the result directly quoted in the link

Answer 1

I guess this is a duplicate but :

$z_n\to z$ to for $n\geq n_0$, $|z_n-z|\leq 1$, so by the inverse triangular identity of $|\cdot|$ in $\mathbb C$, $|z_n|\leq |z|+1$ for $n\geq n_0$. So, $z_n\leq \max\{z_n|n\leq n_0\} + |z| + 1<\infty$ sor $(z_n)_n$is bounded.

As JustDroppedIn said, this is just a property of norms in normed vector spaces, or more generally, of metrics in metric spaces.