Banach-Tarski Paradox/Lemma 3

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Lemma for Banach-Tarski Paradox

Relation Definition

Let $\approx$ denote the relation between sets in Euclidean space of $3$ dimensions defined as follows:

$X \approx Y$

if and only if:

there exists a partition of $X$ into disjoint sets:
$X = X_1 \cup X_2 \cup \cdots \cup X_m$
and a partition of $Y$ into the same number of disjoint sets:
$Y = Y_1 \cup Y_2 \cup \cdots \cup Y_m$

such that $X_i$ is congruent to $Y_i$ for each $i \in \set {1, 2, \ldots, m}$.


Let $X_1 \subseteq Y \subseteq X$.

Let$X \approx X_1$.

Then:

$X \approx Y$


Proof

Recall:

Lemma 2

Let $X$ and $Y$ be disjoint unions of $X_1, X_2$ and $Y_1, Y_2$ respectively.

Let $X_i \approx Y_i$ for each $i \in \set {1, 2}$.

Then:

$X \approx Y$

$\Box$


Let:

\(\ds X\) \(=\) \(\ds X^1 \cup X^2 \cup \cdots \cup X^n\) where superscripts are used for indexing
\(\ds X_1\) \(=\) \(\ds X_1^1 \cup X_1^2 \cup \cdots \cup X_1^n\)

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:

\(\ds X_0\) \(=\) \(\ds X\)
\(\ds X_1\) \(=\) \(\ds f \sqbrk X\) where $f \sqbrk X$ denotes the image of $X$ under $f$
\(\ds X_2\) \(=\) \(\ds f \sqbrk {X_1}\)
\(\ds \) \(\cdots\) \(\ds \)

and:

\(\ds Y_0\) \(=\) \(\ds Y\)
\(\ds Y_1\) \(=\) \(\ds f \sqbrk Y\)
\(\ds Y_2\) \(=\) \(\ds f \sqbrk {Y_1}\)
\(\ds \) \(\cdots\) \(\ds \)

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:
\(\ds Z\) \(\approx\) \(\ds f \sqbrk Z\)
\(\ds X\) \(=\) \(\ds Z \cup \paren {X \setminus Z}\)
\(\ds Y\) \(=\) \(\ds f \sqbrk Z \cup \paren {X \setminus Z}\)

and by Lemma $2$:

$X \approx Y$

$\blacksquare$


Sources

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