Show that $X^2 \cap \le_X $ is a well ordered set. Prove that if $X \subset \Bbb{N},$ then the order induced by $\le$ in $X$ is a well-ordered set. By induced order I mean $X^2 \cap \le_X .$
The excercise cites this other problem: If $R$ is an order in $A,$ and $B \subset A$, then $B^2 \cap R $ is an order in B. It is called the induced order.   
My main issue is that I don't know what $X^2 \cap \le_X $ means. It could be that the set refers to the pairs $\{(k,k), (k+1, k+1), ... , (k+m, k+m)\} $ where $k+i, i\in \Bbb{N} $. Or maybe the set are the ordered pairs $(m,n)$ with $(m,n)$ in the cartesian product $X^2$. And as $(\Bbb{N}, \le)$ is a well-ordered set, $(a,a)$ could be a least element. But I guess It would all depend in how the actual order is defined.  
Thanks in advance
 A: The set-theoretic definition of a binary relation on a set $X$ is a subset of $X\times X.$ So $a\leq b$ means there is a set called $\leq,$ and the ordered pair $(a,b)$ belongs to $\leq.$ In set-theory, if $R$ is a binary relation on the set $X$ then  the set which in non-set-theoretic notation is denoted by $\{(a,b)\in X^2: aRb\}$ is denoted by $R.$
What  you meant to ask ( after you fix the typos) was to show that $X^2\cap \leq_{\mathbb N}$ is a well-order on $X\subset \mathbb N$ where $\leq_{\mathbb N}$ is the usual order on $\mathbb N.$ In non-set theoretic language, to prove that if $X\subset \mathbb N$ then the usual order $\leq$ of $\mathbb N,$ restricted to $X,$ is a well-order on $X.$ 
BTW. The method of proof applies to any well-order on any set $N'$ and any $X\subset N'.$
Remark:This kind of notation in set-theory is also seen with regard to functions: In set-theory a function is the same thing as "its graph". 
A: Let's look at it step by step:
$$X^2 = \{ (m,n) \mid m,n \in X \}$$
is the set of all ordered pairs whose members are in $X$.
Now $\le$, the natural order of $\mathbb N$
$$
\begin{align*}
\le &= \{ (m,n) \mid m,n \in \mathbb N \wedge  m \le n \} \\
&\left( = (m,n) \mid m,n \in \mathbb N \wedge \exists k \in \mathbb N \colon m + k = n \} \right)
\end{align*}
$$
again is a set of ordered pairs of members of $\mathbb N$ - but not all of them. Only those which are listed in the correct order. Hence, for any $X \subseteq \mathbb N$
$$
\begin{align*}
\le \cap X^2 &= \{ (m,n) \mid m,n \in \mathbb N \wedge m \le n \} \cap \{(m,n) \mid m,n \in X \} \\
&= \{ (m,n) \mid m,n \in \mathbb N \wedge m \le n \wedge m,n \in X \} \\
&= \{ (m,n) \mid m,n \in X \wedge m \le n \}
\end{align*}
$$
again is a set of ordered pairs of members of $X$ - those that are listed in the correct order (of $\mathbb N$).
The verification that this is a wellorder is now very straightforward. Since $\le$ is reflexive, it follows immediately that $\le \cap X^2$ is reflexive as well. The same applies to transitivity, antisymmetry and wellfoundedness. I'll leave it to you to spell out the details.
