If $X$ is a finite set and $Y \subset X$. Prove that $Y$ is finite. I am having trouble solving a question of section 3.6 (cardinality of sets) from Tao's Analysis I book. 
I have to prove the following statement:
My attempt (the boldline's refers to the points where I need help):
Since $X$ is a finite set, let $\#X=n$ and let 
$N_n:=\{i \in N:\forall 1\leq i \leq n\}$. Since $\#X=n$, $\exists$ a bijection $f:X \mapsto N_n $. Now, let $G:=\{f(x):x \in Y\} $. Since $G \subset N_n$ $  \exists g_1 \in G$ s.t $g_1=min(G)$ ( I am not bothering with proving this). Now, consider $G_2:=G-\{g_1\}$, if $G_2\ne \emptyset $, there exists $g_2 \in G_2 $ s.t $g_2=min(G_2)$.
And now I need help:
Going on like this I will get to a point where $G_m=\emptyset $.
I don't know how to 
technically write this argument, can someone help me?
Carrying on...
Now, let the set $G':=\{g_i: \forall 1\leq i \leq n'\}$. Note that $n'=m-1$. So (maybe I have to write some other steps here, please help), $\exists$ a bijection $g:Y \mapsto N_n' $, then Y is finite and $\#Y=n'$. 
And now, I have to proof that $n' \leq n$. 
Did I already prove this? If don't, I can prove this by contradiction, right?
So, that's it. I appreciate any help.
 A: What is a function? In set theory it is a set $f$ of ordered pairs such that if $(x,y)$ and $(x,y')$ belong to $f$ then $y=y'.$
Now $f:N_n\to X$ is a bijection iff $f^{-1}:X\to N_n$ is a bijection. So for $Y\subset X,$ the set $\{(y,f^{-1}(y):y\in Y\}$ is a bijection from $Y$ to$ M=\{f^{-1}(y):y\in Y\}.$ 
So, for $Y\ne \phi,$(because $Y=\phi$ is a trivial case), if you can show there is a bijection $g:M\to N_m$ for some natural number $m$ then the composite function $f g^{-1}$ is a bijection from $N_m to Y.$
So you have only the problem of showing that any subset of $N_n$ is finite. For this,you can employ induction on $n.$
The case $n=1$ is obviously true.
Suppose $n>1$ and it is true for $n-1.$ Now for $ Y\subset N_n,$ let $Y'=Y\backslash \{n\}.$ Then $Y'\subset N_{n-1}$ so there is a bijection $f':Y'\to N_m$ for some $m.$
If $n\not \in Y$ then $Y'=Y$ and you are done, because f' is then a bijection from  $Y$ to $N_m.$
Or if $n\in Y,$ then $f=f'\cup \{(n,m+1)\}$ is a bijection from $Y$ to $N_{m+1}.$
A: I recommend checking out Enderton's Elements of Set Theory (page 133) and/or Halmos' Naive Set Theory (page 53)
