If $A$ is finite with $|A|=n \geq 1$ and $x \in A$, I need to prove that $A - \{x\}$ is finite with $|A-\{x\}| = n-1$, where $|\cdot|$ is the cardinality.
I know that there are proofs for $|A-\{x\}| = n-1$ which rely on the result $A = (A-\{x\})\cup \{x\} $, but I was wondering if there was a way to prove this without relying on it (as relying on it would require me to prove it first).
Instead, I was thinking of proving it the following way:
Since $|A|= n \geq 1$, $A \neq \emptyset$, so $\exists$ at least one element $x \in A$.
Suppose that $A = \{x\}$, then $|A|=1$ and $A - \{x\} = \emptyset$, which is certainly finite, and $|A-\{x\}|=|\emptyset|=0 = |A|-1$.
Suppose that $|A|=n>1$, $x \in A$. Then, there exists a bijection $f:A \to \mathbb{N}_{n}$.
Now, to go any further with bullet point 2, I need to define a function $g:A-\{x\} \to \mathbb{N}_{n-1}$ and show that it is a bijection, but I'm not sure what function $g$ to choose. I know that I can use $f$ to help me define $g$, but how? What choice of $g$ will allow me to prove this result this way without needing to rely on $A = (A-\{x\})\cup \{x\} $?
Thank you.
Suppose that $f:A \to \Bbb N_n$ is a bijection. We can find an "obvious" formula for a bijection $\tilde f:(A - \{x\}) \to \Bbb (N_n - \{f(x)\})$. Then, find an explicit bijection $g:(\Bbb N_n - \{f(x)\}) \to \Bbb N_{n-1}$. It would follow that $g \circ \tilde f$ gives us the desired bijection from $A \setminus \{x\}$ to $\Bbb N_{n-1}$.