Definition:Concatenation (Topology)

From ProofWiki
Jump to navigation Jump to search


Definition

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $c_1, c_2: \left[{0 \,.\,.\, 1}\right]^n \to S$ be maps.

Let $c_1$ and $c_2$ both satisfy the concatenation criterion $c \left({\partial \left[{0 \,.\,.\, 1}\right]^n}\right) = x_0$.


Then the concatenation $c_1 * c_2$ is defined as:

$\left({c_1 * c_2}\right) \left({t_1, t_2, \ldots, t_n}\right) = \begin{cases} c_1 \left({2t_1, t_2, \ldots, t_n}\right) & : t_1 \in \left[{0 \,.\,.\, 1/2}\right] \\ c_2 \left({2t_1-1, t_2, \ldots, t_n}\right) & : t_1 \in \left[{1/2 \,.\,.\, 1}\right] \end{cases} $

where $\left({t_1, \ldots, t_n}\right)$ are coordinates in the $n$-cube.

By Continuous Mapping on Finite Union of Closed Sets, $c_1*c_2$ is continuous.

This resulting map is continuous, since:

  • $2 \left({\dfrac 1 2}\right) = 1$ and $2 \left({\dfrac 1 2}\right) - 1 = 0$;
  • anywhere any coordinate of $\hat t$ is either $1$ or $0$, $\left({c_1*c_2}\right) \left({\hat t}\right) = x_0$.

The resulting map also clearly satisfies the concatenation criteria itself.


Also see


Special cases


Linguistic Note

The word concatenation derives from the Latin com- for with/together and the Latin word catena for chain.

However, the end result of such an operation is not to be confused with a (set theoretical) chain.