Correspondence between Set and Ordinate of Cartesian Product is Mapping
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Theorem
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.
Proof
First it is to be shown that $j_t$ is left-total.
This follows from the fact that $j_t$ is defined for all $s$:
- $\map {j_t} s = \tuple {s, t}$
$\Box$
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.
$\blacksquare$
Also see
- Definition:Canonical Injection (Abstract Algebra) for an instance of this construct in the context of algebraic structures
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $6$