Here $\alpha$ and $\beta$ are ordinals. I need to prove that if $\alpha+\beta=\beta$, then $\alpha\le \beta$.

My textbook just gives that that if $\alpha\le \beta$, then exists ordinal $\gamma$ such that $\alpha+\gamma=\beta$.

I am completely new to transfinite arithmetic, could someone please give a proof? Thanks in advance!


closed as off-topic by Andrés E. Caicedo, C. Falcon, Daniel W. Farlow, Claude Leibovici, user223391 Apr 24 '17 at 4:03

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  • $\begingroup$ Isn't $\alpha$ order-isomorphic to an initial segment of $ \alpha+\beta$, so that $\alpha\le\alpha+\beta$? $\endgroup$ – Lord Shark the Unknown Apr 22 '17 at 18:48
  • $\begingroup$ @LordSharktheUnknown It makes sense to me. But could you please expand a little bit, say, how to formally prove that they are order isomorphic and how can I thus deduce that $α≤α+β$? $\endgroup$ – non-abelian group of order 9 Apr 22 '17 at 18:56

Let $E$ and $F$ be ordered sets, and let $E\oplus f$ be the ordered set on the disjoint union $E\times\{0\}\cup F\times\{1\}$ where the order is taken to coincide with that of $E$ on $E$, that of $F$ on $F$, and where any element of $E$ is smaller than any element of $F$.

When $E$ and $F$ are well ordered, you can prove that $E\oplus F$ is as well (consider a non empty subset, and then discuss whether it intersects $E$ or not).

So if $\alpha$ and $\beta$ are ordinals, $\alpha\oplus\beta$ is order isomorphic with a unique ordinal. Now what I will prove is that $\alpha\oplus \beta \simeq \alpha + \beta$, and that the (unique) order isomorphism sends any element of $\alpha$ to itself. Once we have this, since $\alpha$ is obviously an initial segment of $\alpha\oplus\beta$, $\alpha$ will be an initial segment of $\alpha+\beta$, and thus $\alpha\leq \alpha+\beta$, and in your case since $\alpha+\beta =\beta$, this will give $\alpha\leq \beta$, as wanted.

To prove the claim $\alpha\oplus\beta\simeq \alpha+\beta$ we proceed by induction on $\beta$, fixing $\alpha$, on the following property : "there is an order isomorphism from $\alpha\oplus\beta$ to $\alpha+\beta$".

The property obviously holds for $\beta =0$, and the definition of addition makes the induction step $\beta\to \beta+1$ immediate.

Now for the limit case, we can argue as follows : let $\beta$ be a limit ordinal and assume the property holds for $\gamma<\beta$. Notice that for $\gamma<\beta$, $\alpha\oplus\gamma \subset \alpha\oplus\beta$, and in fact the LHS is an initial segment of the RHS. Fix for $\gamma<\beta$ an isomorphism $f_\gamma : \alpha\oplus \gamma \to \alpha+\gamma$ (this doesn't use the axiom of choice since order isomorphisms between well ordered sets are unique).

Whenever $\gamma\leq\delta<\beta$, the restriction of $f_\delta$ to $\alpha\oplus\gamma$ is an order preserving bijection with an initial segment of an ordinal, so an ordinal, so it is the unique ordinal with which $\alpha\oplus\gamma$ is isomorphic, i.e. $\alpha+\gamma$ by the induction hypothesis. Therefore, by unicity, the restriction of $f_\delta$ to $\alpha\oplus\gamma$ is $ f_\gamma$.

Therefore $f:=\displaystyle\bigcup_{\gamma<\beta}f_\gamma$ is a function, whose domain is $\alpha\oplus\beta$ (why ?) and whose range is included in $\displaystyle\bigcup_{\gamma<\beta}\alpha+\gamma = \alpha +\beta$.

Moreover $f$ is obviously order preserving and injective. Now if you let $\kappa\in \alpha+\beta$, there exists $\gamma<\beta$ such that $\kappa\in \alpha+\gamma$, because $\beta$ is a limit ordinal and by definition of the ordinal sum, and so $\kappa\in range(f_\gamma)\subset range(f)$, and so $f: \alpha\oplus\beta \to \alpha +\beta$ is an order preserving bijection, an isomorphism.

This completes the induction proof. Notice that it would be easy to prove at the same time that this isomorphism fixes $\alpha$, which proves my claim.

EDIT : I'm adding a precision on a point, as asked in the comments. Let $\delta$ and $\gamma$ be two ordinals, such that $\delta$ is an initial segment of $\gamma$. If $\delta$ is not a proper initial segment, then $\delta=\gamma$ and so $\delta\leq \gamma$. For the case where $\delta$ is a proper initial segment, I'll have to prove a general claim : Let $E$ be a well-ordered set, and $F$ a proper initial segment. Then there exists $x\in E$ such that $F= S_x := \{y\in E, y<x\}$. For this you can consider $x:= min\{z\in E, \forall y\in F, y<z\}$. This set is nonempty since $F$ is an initial segment and is not equal to $E$, and so $x$ is well-defined. You can easily see that then $F=S_x$.

So in our case, there is $\kappa\in \gamma$ such that $\delta = S_\kappa = \{y\in \gamma\mid y<\kappa\}$. Then you have to remember two things : on ordinals, $<$ is the same thing as $\in$, and $\kappa\subset\gamma$, by transitivity. Therefore $\delta = S_\kappa = \{y\in \gamma\mid y\in \kappa\} = \gamma\cap\kappa = \kappa$. And so, since $\kappa\in \gamma$, $\delta\in \gamma$, and so $\delta<\gamma$

  • $\begingroup$ It is much clearer! May I please ask how does $\alpha$ is a initial segment of $\alpha+\beta$ implies $\alpha\le \alpha+\beta$? $\endgroup$ – non-abelian group of order 9 Apr 23 '17 at 1:23
  • $\begingroup$ I've added a precision in my answer, you can look at it $\endgroup$ – Max Apr 23 '17 at 8:26

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