Cfrgtkky

I have to proof that the function $f : X rightarrow Y$ is an injection if and only if $forall T subseteq X$, $f(Xsetminus T) subseteq Y setminus f(T)$.
I'm having some difficulties. First (1) I proof that if $f$ is an injection then $forall T subseteq X$, $f(Xsetminus T) subseteq Y setminus f(T)$, successively I'll prove the inverse implication (2).

(1):

I want to show that a generic element of $f(Xsetminus T)$ belongs to $Y setminus f(T)$ too, but I don't know how to continue, it is not apparent to me how to use the injection of $f$.

(2)

EDIT as suggested by Mark, i'll try the direction 2.

We know that $forall x in Xsetminus T$ and $forall t in T$, we have $xneq t $, because x belongs to X but it does not belong to T. Now , if I choose $y in f(X setminus T)$, there is $x in X setminus T$ : $ y = f(x)$, because y is an element of the image of $Xsetminus T$ through $f$. But we know also that $y in Ysetminus f(T)$, so $y notin f(T)$, this implies that $forall t in T$, $yneq f(t)$. This reduces to $f(t) neq f(x) $. Is this proof valid ?

So we want to show $f$ is injective iff

$$forall T subseteq X: f[Xsetminus T]subseteq Y setminus f[T]tag{*}$$

So let $f$ be injective, $T subseteq X$ and let $y in f[Xsetminus T]$, i.e.
$y=f(x)$ with $x notin T$. By definition of the function, $y in Y$, but I claim that also $y notin f[T]$, because otherwise $y=f(t)$ for some $t in T$, and as $x notin T$, we have that $x neq t$ but also $y=f(x)=f(t)$ contradicting that $f$ is injective. So $y in Ysetminus f[T]$, and $(ast)$ has been shown.

Suppose that $(ast)$ holds. Suppose $f$ were not injective, so that we have that $x_1 neq x_2$ in $X$ with $y:= f(x_1) = f(x_2)$. Then $f[Xsetminus {x_1}] = Y$ (as the only point we could miss is $y$ but this is also assumed by $x_2 in Xsetminus {x_1}$) but $Ysetminus {y}$ is a proper subset of $Y$, so that $(ast)$ fails for $T={x_1}$ (or $T={x_2}$ too).
This contradiction shows that $f$ is injective.

It's straight from the definitions. Let $yin f(Xsetminus T)$. We need to show that $yin Y$ and that $ynotin f(T)$. Obviously $yin Y$ because $f$ maps elements of $X$ to $Y$. Now assume that $yin f(T)$. Then there is an element $xin T$ such that $f(x)=y$. On the other hand $yin f(Xsetminus T)$ so there is also an element $zin Xsetminus T$ such that $f(z)=y$. Hence $f(x)=f(z)$ when $xne z$ which is a contradiction to $f$ being injective. So $yin Ysetminus f(T)$. Now can you prove the other direction?

Partial answer:

2) $forall T subset X$, $f(X$ $T) subset Y$ $f(T)$

$Rightarrow$ $f$ injective.

Let $x not = t$;

With $T=${$t$}:

$x in X$ {$t$} ; and

$f(x)in f(X$ {$t$}) $ subset Y$ $f(${$t$}).

Then

$f(x) not in f(${$t$}$)$, i.e. $f(x)not =f(t)$, injective.