Banach-Tarski Paradox/Lemma 3

Relation Definition
Let $X_1 \subseteq Y \subseteq X$.

Let$X \approx X_1$.

Then:
 * $X \approx Y$

Proof
Recall:

Lemma 2
Let:

such that $X^i$ is congruent to $X_1^i$ for each $i \in \set {a, 2, \ldots, n}$.

Let us choose a congruence:
 * $f^i: X^i \to X_1^i$

for each $i \in \set {a, 2, \ldots, n}$.

Let $f$ be the bijection of $X$ to $X_i$ which agrees with $f^i$ on each $X^i$.

Now let:

and:

Let:
 * $Z = \ds \bigcup_{n \mathop = 0}^\infty \paren {X_n \setminus Y_n}$

Then:
 * $f \sqbrk Z$ and $X \setminus Z$ are disjoint, and:

and by :
 * $X \approx Y$