Correspondence between Set and Ordinate of Cartesian Product is Mapping

From ProofWiki
Jump to navigation Jump to search


Let $S$ and $T$ be sets such that $T \ne \O$.

Let $S \times T$ denote their cartesian product.

Let $t \in T$ be given.

Let $j_t \subseteq S \times \paren {S \times T}$ be the relation on $S \times {S \times T}$ defined as:

$\forall s \in S: \map {j_t} s = \tuple {s, t}$

Then $j_t$ is a mapping.


First it is to be shown that $j_t$ is left-total.

This follows from the fact that $j_t$ is defined for all $s$:

$\map {j_t} s = \tuple {s, t}$


Next it is to be shown that $j_t$ is many-to-one, that is:

$\forall s_1, s_2 \in S: \map {j_t} {s_1} \ne \map {j_t} {s_2} \implies s_1 \ne s_2$

We have that:

\(\ds \map {j_t} {s_1}\) \(\ne\) \(\ds \map {j_t} {s_2}\)
\(\ds \leadsto \ \ \) \(\ds \tuple {s_1, t}\) \(\ne\) \(\ds \tuple {s_2, t}\) Definition of $j_t$
\(\ds \leadsto \ \ \) \(\ds s_1\) \(\ne\) \(\ds s_2\) Definition of Ordered Pair

Hence the result.


Also see