I know that the boundary of a topologic set $E$ is given by $$ \partial E=\overline{E} \backslash \overset{\circ}{E} $$
Assume that I know that for a subset of $\mathbb{R}^2$, if $\left|z\right|<1$ then a defined sequence $\left(z_n\right)_{n \in \mathbb{N}}$ converges, if $\left|z\right|>1$ then the same sequence $\left(z_n\right)_{n \in \mathbb{N}}$ diverges and if $\left|z\right|= 1$ we cannot conclude whether it converges or not. How can I show that the boundary of the set defined as the $z$ for which the sequence $\left(z_n\right)_{n \in \mathbb{N}}$ converges is exactly given by the set of $z$ which satisfies $\left|z\right|= 1$ ?
I've tried to use the definition but it led me nowhere, any help ?
Define $A$ as the set of $z$ for which the given sequence converges. Because the sequence converges for all $z$ with $|z| < 1$ and diverges for all $z$ with $|z| > 1$, we know that
$$\{z \in \mathbb{R}^2 \mid |z| < 1\} \subseteq A \subseteq \{z \in \mathbb{R}^2 \mid |z| \leq 1\}.$$
Now we see that $\{z \in \mathbb{R}^2 \mid |z| < 1\}$ is an open set contained in $A$, so by definition we see that also $\{z \in \mathbb{R}^2 \mid |z| < 1\} \subseteq \mathring{A}$. Furthermore, the largest open set contained in $\{z \in \mathbb{R}^2 \mid |z| \leq 1\}$ is $\{z \in \mathbb{R}^2 \mid |z| < 1\}$, so from this it follows that also $\mathring{A} \subseteq \{z \in \mathbb{R}^2 \mid |z| < 1\}$. Now combining these two yields $\mathring{A} = \{z \in \mathbb{R}^2 \mid |z| < 1\}$.
In a similar way, we see because $A \subseteq \{z \in \mathbb{R}^2 \mid |z| \leq 1\}$, and $\{z \in \mathbb{R}^2 \mid |z| \leq 1\}$ is closed, that $\bar{A} \subseteq \{z \in \mathbb{R}^2 \mid |z| \leq 1\}$. Also $\bar{A}$ is a closed set containing $\{z \in \mathbb{R}^2 \mid |z| < 1\} \subseteq A$, and the smallest such set is $\{z \in \mathbb{R}^2 \mid |z| \leq 1\}$, so we have that $\bar{A} \supseteq \{z \in \mathbb{R}^2 \mid |z| \leq 1\}$. We conclude that $\bar{A} = \{z \in \mathbb{R}^2 \mid |z| \leq 1\}$.
Now it follows by the definition of the border of $A$ that $$\partial A = \{z \in \mathbb{R}^2 \mid |z| \leq 1\} \setminus \{z \in \mathbb{R}^2 \mid |z| < 1\} = \{z \in \mathbb{R}^2 \mid |z| = 1\},$$ as was to be shown.