19.1.1.2 Properties of metric spaces
Problem:
Is the mentioned function \( \rho(x, y) \) a metric in the given set? Fill in the table.
\begin{tabular}{|c|c|c|c|c|c|}
\hline Set & \begin{tabular}{l}
\( \quad \mathbb{R} \) \\
Of real \\
numbers \\
\( x, y \)
\end{tabular} & \begin{tabular}{l}
\( \mathbb{R}^{3} \) \\
Of 3 dimension vectors \\
\( x=\left(x_{1}, x_{2}, x_{3}\right) \) \\
\( y=\left(y_{1}, y_{2}, y_{3}\right. \)
\end{tabular} & \begin{tabular}{l}
\( \qquad \quad S(\mathbb{R}) \) \\
Of all possible \\
sequences of real \\
numbers \\
\( x=\left(x_{1}, x_{2}, \ldots, x_{n}, \ldots\right) \) \\
\( y=\left(y_{1}, y_{2}, \ldots, y_{n}, \ldots\right) \)
\end{tabular} & \begin{tabular}{l}
\( \qquad F_{c}([a ; b], \mathbb{R}) \) \\
Of countinuous \\
functions \\
\( x(t):[a ; b] \rightarrow \mathbb{R} \) \\
\( y(t):[a ; b] \rightarrow \mathbb{R} \)
\end{tabular} & \begin{tabular}{l}
\( \qquad M \) \\
Of the cells \( x, y \) of \\
the chessboard \( 8 \times \) \\
8
\end{tabular} \\
\hline Function \( \rho(x, y)=\cdots \) & \( \cos |x-y| \) & \( \sqrt{\left(x_{1}-y_{1}\right)^{2}+\left(x_{2}-y_{2}\right)^{2}} \) & \( \sup _{n \in \mathbb{N}}\left|x_{n}-y_{n}\right| \) & \( \int_{a}^{b}|y(t)-x(t)| d t \) & \begin{tabular}{l}
The minimum \\
number of the \\
knight moves on the \\
way from \( x \) to \( y \)
\end{tabular} \\
\hline Colclusion: & & & & & \\
\hline \begin{tabular}{l}
Function \( \rho \) doesn't set a metric in the given set, since it's \\
not defined for all pairs \( (x, y) \)
\end{tabular} & & & & & \\
\hline \begin{tabular}{l}
Function \( \rho \) doesn't set a metric in the given set, since it \\
isn't non-negative
\end{tabular} & & & & & \\
\hline \begin{tabular}{l}
Function \( \rho \) doesn't set a metric in the given set, since it \\
isn't non-degenerate
\end{tabular} & & & & & \\
\hline \begin{tabular}{l}
Function \( \rho \) doesn't set a metric in the given set, since it \\
doesn't have symmetry property
\end{tabular} & & & & & \\
\hline \begin{tabular}{l}
Function \( \rho \) doesn't set a metric on the given set, since it \\
doesn't satisfy the triangle inequality
\end{tabular} & & & & & \\
\hline Function \( \rho \) doesn't set a metric on the given set. & & & & & \\
\hline
\end{tabular}