$\alpha < \beta$ implies that $\gamma+\alpha<\gamma+\beta$ and $\alpha+\gamma\le\beta+\gamma$ for ordinals This is an exercise from Kunen's book.
Show that $\alpha < \beta$ implies that $\gamma+\alpha<\gamma+\beta$ and $\alpha+\gamma\le\beta+\gamma$.  Given an example to show that the "$\le$" cannot be replaced by "$<$". Also show:
$$\alpha \le \beta \rightarrow \exists!\delta (\alpha+\delta=\beta).$$
What I've tried: I could give an example such that show the "$\le$" cannot be replaced by "$<$": $1+\omega = 2+\omega=\omega$, where $\alpha=1$, $\beta=2$ and $\gamma=\omega$. I could also see these two inequalities is obviously right, but I cannot how to use the set language to show. For the last question, I want let $\delta=\{\gamma:\alpha+\gamma\le\beta\}$ to prove. Am I right?
Could anybody help me? Thanks ahead:)
 A: 
$\alpha<\beta$ $\Rightarrow$ $\gamma+\alpha<\gamma+\beta$

$\alpha<\beta$ means: if $A$ has ordinal type $\alpha $ and $B$ has ordinal type $\beta$, then $A$ is isomorphic to some initial segment $B_a$ of the set $B$.
Using embedding $f \colon A\to B$ can you obtain similar embedding for the sets of ordinal types $\gamma+\alpha$ and $\gamma+\beta$?
Note: If you work with von Neumann's definition of ordinals, you could use $A=\alpha$, $B=\beta$. 
(For this construction, you have $\alpha\le\beta$ $\Leftrightarrow$ $\alpha\subseteq\beta$ and $\alpha<\beta$ $\Leftrightarrow$ $\alpha\subsetneqq\beta$; moreover you know that $\alpha$ is an initial segment of beta.)

A different solution. Suppose we already have shown the second part. Then $\alpha<\beta$ implies that there is $\delta>0$ such that $\beta=\alpha+\delta$. And then $\gamma+\beta=\gamma+\alpha+\delta>\gamma+\alpha$.

Transfinite induction:
Starting point of induction is $\beta=\alpha+1$. (The smallest ordinal greater than $\alpha$.) We have $\gamma+\alpha<\gamma+\alpha+1$.
Inductive step - successor ordinal: We combine $\gamma+\alpha<\gamma+\beta$ and $\gamma+\beta<\gamma+\beta+1$.
Inductive step - limit ordinal: If $\beta=\sup_{i\in I}\beta_i$ and $\gamma+\alpha<\gamma+\beta_i$ for each $i\in I$, then $\gamma+\alpha<\gamma+\beta$.
(This was induction on the well-ordered set $\{\beta\text{ is an ordinal}; \beta>\alpha\}$.)

$\alpha\le\beta$ $\Rightarrow$ $(\exists\delta) \beta=\alpha+\delta$

You know that $\alpha\subseteq\beta$. What about trying ordinal type of the well-ordered subset $\beta\setminus\alpha$ for $\delta$?
(If you work with the naive definition of ordinals and their inequality, you'll have $B\setminus A$ instead; where $A$ and $B$ have ordinal types $\alpha$ and $\beta$, respectively.)

Transfinite induction: Similar as the first part. We will use $\alpha+\sup_{i\in I}\beta_i=\sup_{i\in I}(\alpha+\beta_i)$. (Right-continuity of the addition.) 

$\alpha\le\beta$ $\Rightarrow$ $\alpha+\gamma\le\beta+\gamma$

One simple solution seems to be: Fix $\alpha$, $\beta$ and show this by transfinite induction on $\gamma$.
It is clear for $\gamma=0$.
For successor ordinals you need basically to know that $\alpha\le\beta$ $\Rightarrow$ $\alpha+1\le\beta+1$.
The inductive step for limit ordinals requires showing that $\alpha_i\le \beta_i$ $\Rightarrow$ $\sup_{i\in I}\alpha_i \le \sup_{i\in I}\beta_i$. 
(Which is easy if you're working with von Neumann's definition - the supremum is just union of ordinals.)

The first two parts seem to be doable by transfinite induction, too (if someone prefers such solution).
