Exercise 1.4

Question

% The scalar field C:

ewcommand\usualSet[1]{{\mathbf #1}}
\def\C{\usualSet{C}}

extit{Let be %
%
  $B = \{(z_{1}, z_{2}) \in \C^2: |z_{1}| \leq |z_{2}| \}$. %
%
Show that $B$ is balanced but that its interior is not.
}

Cordier · Accepted Answer

% The scalar field C:

ewcommand\usualSet[1]{{\mathbf #1}}
\def\C{\usualSet{C}}

\begin{proof}
It is obvious that the nonempty set $B$ contains the origin $(0,0)$. %
Additionally, its interiror $B^\circ$ is nonempty as well. %
Indeed, the following set %
%
\begin{align}
  \{(z_1, z_2) \in \C^2: |1- z_1| + |2- z_2] < 1/2 \} \subset B %
\end{align}
%
is a neighborhood of $(1, 2) \in B$. %
Moreover, $B$ is balanced, since
\begin{align}
  |\alpha z_1|  = |\alpha| |z_1| \leq  |\alpha| |z_2| = |\alpha z_2| %
  \quad (|\alpha| \leq 1)
\end{align}
%
for all $(z_1, z_2)$ in $B$. %
%
Nevertheless, the nonempty set $B^\circ$ is not balanced, what we now %
establish by showing that $(0, 0) 
otin B^\circ$. %
%
To do so, assume, to reach a contradiction, %
that the origin has a neighborhood %
%
\begin{align}
  U 	riangleq \{(z_1, z_2) \in \C^2: |z_1| + |z_2] < r \} \subset B %
\end{align}
%
for some positive $r$. Clearly, $U$ contains $(r/2, 0)$,  %
and that special case $(r/2, 0) \in B$ now contradicts the definition of $B$. %
So ends the proof.
\end{proof}
% END

Exercise 1.4

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Exercise 1.4

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