Ordinal numbers addition property: $b<c$ implies $b+a \le c+a$

Question

I'm having trouble proving the following property of ordinal numbers.

If $a, b, c$ are ordinal numbers such that $b \lt c$, then $b+a \le c+a$.

I first started by assuming $g$ as an order isomorphism from a well-ordered set $B$ to $C_q$, which is a proper segment of $C$, such that $ord(B)=b$ and $ord(C)=c$. However, I don't know how to extend this situation to get the desired result. Any help?

Answer 1

Proceed as follows. All sets are well orders.

1) If $f: D\rightarrow D$ is order preserving, ($x<y \rightarrow f(x) <f(y)$) then $x\leq f(x)$

2) If $f: D\rightarrow E$ is order preserving, then $ord(D) \leq ord(E)$. (Hint: use trichotomy and 1) to show you cannot order preservingly map a set into an initial segment of itself.)

(I take your definition as $ord(D) \leq ord(E)$ means that $D$ is isomorphic to an initial segment of $E$.)

3)Now show your statement by finding an order preserving $f:B+A \rightarrow C+A$ by maping $B$ to $C$ as you have done, then using the identity on $A$.

Answer 2

I use that $a \le b$ for order types iff there is a strict order-preserving function from a set with order type $a$ to one with order type $b$.

As $a < b$ in particular $a \le b$, so there is a (strict) order preserving $f: a \rightarrow b$ (so for all $x,y \in a: a < b \rightarrow f(a) < f(b)$, where $<$ denotes the order on both $a$ and $b$. Assume that $a$ and $b$ are represented by disjoint well-ordered sets here (and $c$ is disjoint from both of them; such can always be arranged, of course). So $a + c$ is represented by the union of $a$ and $c$ in the sum order, and $b + c$ also by the union in the sum order.

Define $f': a \cup c \rightarrow b \cup c$ by $f'(x) = f(x)$ if $x \in a$, $f(x) = x$ for $x \in c$.

Claim: $x < y$ ($x,y \in a \cup c$) implies $f'(x) < f'(y)$: if $x,y$ are both in $a$ this follows from the property of $f$, if $x,y$ are both in $c$, this is clear as $f'$ is the identity on $c$, and if $x \in a$, $y \in c$ (the other way round cannot be, by the definition of the sum order: everything in $a$ is strictly before $c$), $f(x') \in b$, and $f'(y) = y \in c$, so again the definition on the sum order on $b \cup c$ says that $f'(x) < f'(y)$.

This implies that $a + c \le b + c$.