1. ## one-one function

Definition of 0ne-one function:

$\displaystyle f:A\rightarrow B$ is one-one if $\displaystyle f(a)=f(b)$ implies $\displaystyle a=b$.

Now let $\displaystyle A=\varnothing$ and $\displaystyle B=\varnothing$.

I know the logical statement $\displaystyle g: a\notin \varnothing \Rightarrow b\notin \varnothing$ is true, since were $\displaystyle P$ is true and $\displaystyle Q$ is true, $\displaystyle P \Rightarrow Q$ is true by vacuous proof, but it's not the definition of one-one function.

Question: How can we prove $\displaystyle g:\varnothing \rightarrow \varnothing$ using the definition?

$\displaystyle g$(blank)=$\displaystyle g$(blank) implies blank = blank or
$\displaystyle g(\qquad)=g(\qquad)$ implies $\displaystyle \qquad\qquad\qquad$ =$\displaystyle \qquad\qquad\qquad$ ?

2. Originally Posted by novice
Definition of 0ne-one function:

$\displaystyle f:A\rightarrow B$ is one-one if $\displaystyle f(a)=f(b)$ implies $\displaystyle a=b$.

Now let $\displaystyle A=\varnothing$ and $\displaystyle B=\varnothing$.

I know the logical statement $\displaystyle g: a\notin \varnothing \Rightarrow b\notin \varnothing$ is true, since were $\displaystyle P$ is true and $\displaystyle Q$ is true, $\displaystyle P \Rightarrow Q$ is true by vacuous proof, but it's not the definition of one-one function.

Question: How can we prove $\displaystyle g:\varnothing \rightarrow \varnothing$ using the definition?

$\displaystyle g$(blank)=$\displaystyle g$(blank) implies blank = blank or
$\displaystyle g(\qquad)=g(\qquad)$ implies $\displaystyle \qquad\qquad\qquad$ =$\displaystyle \qquad\qquad\qquad$ ?
I might be missing something, but my thought process would be: since there is no $\displaystyle a$ or $\displaystyle b$ for which $\displaystyle g(a)$ or $\displaystyle g(b)$ is defined, the statement

$\displaystyle g(a)=g(b) \implies a=b$.

is vacuously true because $\displaystyle g(a)=g(b)$ can never be true.

3. Originally Posted by undefined
I might be missing something, but my thought process would be: since there is no $\displaystyle a$ or $\displaystyle b$ for which $\displaystyle g(a)$ or $\displaystyle g(b)$ is defined, the statement

$\displaystyle g(a)=g(b) \implies a=b$.

is vacuously true because $\displaystyle g(a)=g(b)$ can never be true.
If we look at the set of functions $\displaystyle B^A$, we see that the is a function since $\displaystyle |B|^{|A|}=0^0=1$

The proof in my book shows this:

Given $\displaystyle A$ and $\displaystyle B$ are empty sets.
Let $\displaystyle b\in B$. A function $\displaystyle f:A \rightarrow B$ is a one-one function for every $\displaystyle a \in A$.

I am not convinced, and very annoyed by it since $\displaystyle a\notin A$ and $\displaystyle b\notin B$.

4. Originally Posted by novice
If we look at the set of functions $\displaystyle B^A$, we see that the is a function since $\displaystyle |B|^{|A|}=0^0=1$

The proof in my book shows this:

Given $\displaystyle A$ and $\displaystyle B$ are empty sets.
Let $\displaystyle b\in B$. A function $\displaystyle f:A \rightarrow B$ is a one-one function for every $\displaystyle a \in A$.

I am not convinced, and very annoyed by it since $\displaystyle a\notin A$ and $\displaystyle b\notin B$.
Yeah I don't understand the statement

Let $\displaystyle b\in B$.

Isn't that like writing

Let $\displaystyle 0 = 1$.

?