Cartesian Product is Set Product/Family of Sets

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\family {S_i}_{i \mathop \in I}$ be a family of sets.

For all $j \in I$, let $\pr_i: \displaystyle \prod_{j \mathop \in I} \family {S_j} \to S_i$ be the $i$th projection from $\displaystyle \prod_{j \mathop \in I} \family {S_j}$ to $S_i$.


Then $\struct {\displaystyle \prod_{j \mathop \in I} \family {S_j}, \family {\pr_i}_{i \mathop \in I} }$ is a set product.


Proof

Let $i \in I$.

Consider any set $X$ and any indexed family of mappings $\family {f_i: X \to S_i}_{i \mathop \in I}$.

Define $h: X \to \displaystyle \prod_{j \mathop \in I} \family {S_j}$ by:

$\forall x \in X: \map h x = \family {\map {f_i} x}_{i \mathop \in I}$

Then for all $x \in X$ and $i \in I$ we have:

$\map {\paren {\pr_i \mathop \circ h} } x = \map {\pr_i} {\family {\map {f_i} x}_{i \mathop \in I} } = \map {f_i} x$

So:

$\pr_i \mathop \circ h = f_i$

Thus $h$ is shown to exist.


Suppose there exists a mapping $k: X \to \displaystyle \prod_{j \mathop \in I} \family {S_j}$ such that:

$\forall i \in I: \pr_1 \mathop \circ k = f_i$

Let $x \in X$ and let:

$\map k x = \family {s_j}_{j \mathop \in I}$

for some $\family {s_j}_{j \mathop \in I} \in \displaystyle \prod_{j \mathop \in I} \family {S_j}$.

Then:

$\map {f_i} x = \map {\paren {\pr_i \mathop \circ k} } x = \map {\pr_i} {\family {s_j}_{j \mathop \in I} } = s_i$

and so:

$\map k x = \family {s_j}_{j \mathop \in I} = \family {\map {f_i} x}_{i \mathop \in I} = \map h x$

and so $k = h$.

So $h$ is unique.

Hence the result.

$\blacksquare$


Sources