Mapping from Set to Ordinate of Cartesian Product is Injection

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 $j_t: S \to S \times T$ be the mapping from $S$ to $S \times T$ defined as:

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

Then $j_t$ is an injection.

Proof

It has been shown in Correspondence between Set and Ordinate of Cartesian Product is Mapping that $j_t$ is a mapping.

Now it is to be shown that $j_t$ is injective, that is:

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

We have that:

\(\ds \map {j_t} {s_1}\)

\(=\)

\(\ds \map {j_t} {s_2}\)

\(\ds \leadsto \ \ \)

\(\ds \tuple {s_1, t}\)

\(=\)

\(\ds \tuple {s_2, t}\)

Definition of $j_t$

\(\ds \leadsto \ \ \)

\(\ds s_1\)

\(=\)

\(\ds s_2\)

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 $6$