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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?

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Samantha Smith
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  • You should edit the previous question, rather than asking a new one. – Asaf Karagila Apr 23 '13 at 19:02
  • Sorry about that, didn't really know the procedure and thought that by asking it again more people would see it – Samantha Smith Apr 23 '13 at 19:05
  • By editing the previous question you make it reappear on the top of the front page. – Asaf Karagila Apr 23 '13 at 19:07
  • How exactly is ordinal addition defined for you? (I couldn't find your previous question on this subject. Did you delete it?) – Cameron Buie Apr 23 '13 at 20:46
  • What do you mean Cameron Buie? What are the different ways addition can be defined in this case? Sorry, this is my first time learning this stuff and its still really confusing to me – Samantha Smith Apr 23 '13 at 20:50
  • There is also no need to post edits to the question when you want someone to add a comment. You should leave a comment on the answer itself; if there is no reply within a day or so, you could post a new question linking to the answer and requesting for clarification. – Asaf Karagila Apr 24 '13 at 14:15
  • Alright I'll do that from now on. – Samantha Smith Apr 24 '13 at 14:41

2 Answers2

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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.

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Asaf Karagila
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  • Thanks a lot for the example. I read about the three steps on Wikipedia, but its great seeing a concrete example that makes sense. Thanks again! – Samantha Smith Apr 23 '13 at 20:59
  • Samantha, no problems. I know that I had a lot of troubles when I was learning this topic for the first time. – Asaf Karagila Apr 23 '13 at 21:11
  • I think I've almost digested this answer completely, but there is one step that I'm still fuzzy on and that's the base case. On Wikipedia they say to show that the property being proved $P$ holds in the zero case (ie $P(0)$ is true), but in this proof you say its enough to show that $0 \in C$ are these one in the same? – Samantha Smith Apr 24 '13 at 13:05
  • Yes, this is exactly what it means. $\cal C$ is exactly the class of ordinals for which $P$ holds. – Asaf Karagila Apr 24 '13 at 13:27
  • Ahh, so your defining a class for which $P(0)$ is already true, that's pretty neat. – Samantha Smith Apr 24 '13 at 13:33
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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.

hot_queen
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  • Is this a valid proof? I guess I'm looking for other peoples opinion to see if this is right or not, because if it is then it is a great simplification – Samantha Smith Apr 24 '13 at 00:02
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    What matters here is what you "think" not "other people's opinion". – hot_queen Apr 24 '13 at 00:12
  • Well I'm not really sure. It makes sense to me, but I'm just a beginner so I could be overlooking something, so I just wanted to get the opinion of some other people to see if it makes sense to them as well. – Samantha Smith Apr 24 '13 at 00:17
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    One can define ordinal addition by "transfinite induction" in which case all theorems about ordinal addition will need a proof by "transfinite induction". I am a big fan of this tool because it is very powerful in certain situations: For example, show that there is a Vitali subset of reals of outer measure exactly 7. But here it is an overkill. In my opinion, sum of two linear orders is just putting them next to each other and then your statement is obvious once you realize that for any two well ordered sets x and y, either x is an initial segment of y or y is an initial segment of x. – hot_queen Apr 24 '13 at 00:22
  • Or, we can define ordinal addition in either way and then prove that the two ways are equivalent. Then we have both tools at our disposal! Huzzah. – Asaf Karagila Apr 24 '13 at 14:16