1. ## one-one function

Definition of 0ne-one function:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The proof in my book shows this:

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

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

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

The proof in my book shows this:

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

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

Let $b\in B$.

Isn't that like writing

Let $0 = 1$.

?