Prove that the cardinality of the set X without element a = the cardinality of X - 1.
I know that |X\{a}|=|X|-1 but I'm not sure how to prove this. I was thinking of defining an enumeration of f: [n] -> x and g: [n-1] -> x{a}. But I'm not sure how to define a function g(i) such that n doesn't have to map to a.
If $A$ is finite and $|A| =n$ then there is bijection $f: \{1,2,.....,n\} \to A$.
Let $g: \{1,2,3,.....,n-1\} \to A\setminus \{a\}$ be defined as
$g(k)=\begin{cases}f(k)&f^{-1}(a) > k\\f(k+1)&f^{-1}(a) \le k\end{cases}$
It's easy to show it's a bijection.
But if $A$ is infinite..... I'm not sure where or if it is ever written that for infinite sets $|A| - 1 = |A|$.
But the find a bijection between $A$ and $A-\{a\}$ for infinite sets is well known.
Claim: we can find a countable subset of $A$.
Let $a_1 = a$. As $|A| > 1$ there is an $b\in A$ so that $b\ne a$. Let $a_2 = b$.
Via induction. If $\{a_1,....,a_k\} \subset A$. If $ \{a_1,....,a_k\} = A$ then $|A| = k$ so there is a $c \in A$ so that $c\not \in \{a_1, ....,a_k\}$. Let $a_{k+1} = c$. By induction a set $B=\{a_i|i\in \mathbb N\}\subset A$ exists.
Define $f:\mathbb A \to \mathbb A-\{a\}$ via. $f(x)=\begin{cases}f(a_{i+1})&\text{if }x = a_i \in B\\x&\text{otherwise}\end{cases}$.
Easy to show $f$ is a bijection.