Ordinals Addition Property

Question

I’m sill trying to figure out how to prove that the following property of ordinals holds

Let $x, y, z \in$ On where On is the class of ordinals.

If $x < y$ then $\forall z$ we have that $z + x < z + y$

I asked this same question earlier and some people suggested to use transfinite induction, but I don’t see how this helps me. Can someone help with this problem or provide an example of how transfinite induction works?

Edit:

Can someone please comment on the base case of Asaf's answer? Why is it enough to show that $0 \in C$ and not that the property holds in the zero case, or are these the same?

Answer 1

The idea of transfinite induction is pretty much like the one in mathematical inductions on the natural numbers:

Let $\cal C$ be a non-empty class of ordinals which has the properties:

1. $0\in\cal C$;
2. If $\alpha\in\cal C$, then $\alpha+1\in\cal C$;
3. If $\delta$ is a limit ordinal, and $\delta\subseteq\cal C$ then $\delta\in\cal C$.

Then $\cal C=\sf Ord$.

Now we define $\cal C$ to be the class of those $y$ such that for all $x<y$, and for all $z$, $z+x<z+y$. We will show that $\cal C$ has the three properties and then we can conclude that it holds for any two ordinals.

Step 1: $0\in\cal C$.

Well, since there is no ordinal strictly smaller than $0$ it holds vacuously that $0\in\cal C$.

Step 2: Successor case.

Suppose that $y\in\cal C$, we want to show that for every $x<y+1$ and for every $z$, $z+x<z+(y+1)$. Let $z$ be any ordinal, and $x<y+1$.

• If $x=y$ then it is obvious, because $z+y<(z+y)+1=z+(y+1)$.
• If $x<y$ then by the assumption that $y\in\cal C$ we have $z+x<z+y<z+y+1$ (where the last inequality follows from the previous bullet).

Therefore if $y\in\cal C$, so is $y+1$.

Step 3: Limit case.

If $y$ is a limit ordinal, and for every $y'<y$ we have that $y'\in\cal C$, let us show that $y\in\cal C$ as well. Let $z$ be any ordinal, and recall the definition of $z+y$: $$z+y=\sup\{z+y'\mid y'<y\}.$$ Suppose $x<y$ then there is some $y'<y$ such that $x<y'$, since $y'\in\cal C$ we have $z+x<z+y'<\sup\{z+y'\mid y'<y\}=z+y$.

Therefore $y\in\cal C$ in the limit case as well.

Answer 2

A simpler way to see this: Since x is less than y, x is a proper initial segment of y. Also z + x is z followed by x etc., so obviously z + x is a proper initial segment of z + y.