Set Equivalence Less One Element
Theorem
Let $S$ and $T$ be sets which are equivalent:
- $S \sim T$
Let $a \in S$ and $b \in T$.
Then:
- $S \setminus \set a \sim T \setminus \set b$
where $\setminus$ denotes set difference.
Proof
As $S \sim T$, there exists a bijection $f: S \to T$.
Let $g: \paren {S \setminus \set a} \to \paren {T \setminus \set b}$ be the mapping defined as follows:
$\forall x \in S \setminus \set a: \map g x = \begin{cases} \map f x: \map f x \ne b \\ \map f a: \map f x = b \end{cases}$
First it is shown that $g$ is an injection.
Let $x, y \in S \setminus \set a$ and $x \ne y$.
The following cases apply:
$(1): \quad \map f x \ne b$ and $\map f y \ne b$
By definition of $g$:
- $\map f x = \map g x$
and:
- $\map f y = \map g y$
It follows from the injectivity of $f$ that:
- $\map g x \ne \map g y$
$(2): \quad \map f x = b$
By definition of $g$:
- $\map g x = \map f a$
We have that:
- $y \in S \setminus \set a$
Therefore by definition of set difference:
- $y \ne a$
So by the injectivity of $f$:
- $\map f y \ne \map f a = \map g x$
Also by the injectivity of $f$:
- $\map f y \ne b$
Thus, by the definition of $g$:
- $\map g y = \map f y \ne \map f a = \map g x$
$(2): \quad \map f y = b$
The proof is the same as that of case $2$, with the roles of $x$ and $y$ reversed.
$\Box$
Next it is shown that $g$ is a surjection.
Let $y \in T \setminus \set b$.
By definition of surjectivity of $f$:
- $\exists x \in S: \map f x = y$
It is to be shown that:
- $\exists x' \in S, x \ne a: \map f {x'} = y$
$(1): \quad \map f a = b$
- $\map f x = y \ne b$
Then by the rule of transposition:
- $x \ne a$
so let:
- $x' = x$
$(2): \quad \map f a \ne b$
By the rule of transposition:
- $\map {f^{-1} } b \ne a$
where $\map {f^{-1} } b$ denotes the preimage of $b$ under $f$.
- Case $(2.1): \quad \map f a = y$
By the definition of $g$:
- $\map g {\map {f^{-1} } b} = y$
so let:
- $x' = \map {f^{-1} } b$
- Case $(2.2): \quad \map f a \ne y$
- $\map f x = y \ne \map f a$
Then by the rule of transposition:
- $x \ne a$
so let:
- $x' = x$
Thus:
- $\forall y \in T \setminus \set b: \exists x \in S \setminus \set a: \map g x = y$
and so $g$ is a surjection.
$\Box$
Thus $g$ is both an injection and a surjection, and is by definition a bijection.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.2$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 6$: Finite Sets: Lemma $6.1$