If A is a set with $n$ elements and $B \subseteq A$ is a set with $1$ element, then $A-B$ is a set with $n-1$ elements.
I read a hint that said this:
Define $\Bbb N_{n-1}\to (A-C)$ as (roughly) the restriction of $f$ to a subset of $\Bbb N_n$ you have to choose.
But why this is a bijection, there is another way to prove this?
Could someone help me please, thanks for your time and help.
Let $N$ be the cardinal of $A$.
Consider a function $\quad f : \Bbb N_{N} \rightarrow A \quad$ built as it follows:
It exists a $k \in \Bbb N_{N}$ as that $\quad f(n) \in B \quad \forall n \geq k$
and $\quad \forall p,q \in \Bbb N_{N} \quad f(p) \neq f(q)$
We can show it is a bijection. It is injective by definition.
Let's suppose there is an element $a \in A$ as that $a \notin Im(f)$. By construction $Im(f) \subset A$. Also for all the $N$ elements of $\Bbb N_{N}$ we have a different image that means $card(Im(f)) = N$ which means $Im(f) = A$ and therefore $a$ can not exist.
$f$ is surjective so it is bijective.
Suppose $k$ was different from $N$. Then it must be strictly smaller than $N$. We have that the set $\Bbb N_{N}- \Bbb N_{k-1} = \{k, ..., N\}$ is bijective to $\Bbb N_{t}$ where $t=N-k + 1$. (It is clear $t$ is strictly bigger than one)
With the bijection from $\Bbb N_{t}$ to $\Bbb N_{N}- \Bbb N_{k}$
$g: x \mapsto x+k$
That would mean $Im(f \circ g) = B$ has cardinal $t$. As it is clear that $f \circ g: \Bbb N_{t} \rightarrow Im(f \circ g) $ is a bijection. Which enters in contradiction with the fact that $B$ has one element.
So we have $k = N$ and so we can say $B$ is not included in the image of the restriction of $f$ to $\Bbb N_{N-1}$. That means the image of $f$ restricted to $\Bbb N_{N-1}$ is a subset of $A-B$.
Lets prove the other implication to show that $A-B = Im(f_{|\Bbb N_{N-1}})$. Let $a \in A-B$ as that $a \notin Im(f_{|\Bbb N_{N-1}})$ due to the fact that $f$ is bijective that would mean $a = f(N)$ as $a \in A$. And then $a \in B$ which contradicts the belonging of $a$ to $A-B$.
So we have $A-B$ is the image of $f_{|\Bbb N_{N-1}}$. And therefore its cardinal is $N-1$.