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Suppose $\alpha$ and $\beta$ are ordinals. If $\alpha>0$, prove that the class function $$\beta\mapsto\alpha\cdot\beta$$ is stricly increasing and continuous.

I have no clue what the answer is, it was on my final exam and I am curious what the answer is for my own knowledge, thanks for the help.

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Asking two questions with the same title is unnecessarily confusing, especially since that title doesn't contain much information anyway. Just ask the question in the title (and then ask it again, with more details, in the body). – Qiaochu Yuan May 4 '11 at 1:11
up vote 3 down vote accepted

Recall that $\alpha\beta$ is the order type of $\beta$ copies of $\alpha$, one after the other. If $\beta<\gamma$, then obviously ($\gamma$ copies of $\alpha$) is longer than only ($\beta$ copies). This shows that the function $\beta\mapsto \alpha\beta$ is strictly increasing.

The argument showing that the map is continuous is similar: If $\gamma>\alpha\delta$ for all $\delta<\beta$, then $\gamma\ge$ ($\beta$ copies of $\alpha$). [Or you can think of it this way: If $\gamma<\alpha\beta$, then $\gamma$ lies inside (or "just to the left of") one of the $\beta$ intervals of type $\alpha$ that make up $\alpha\beta$, say the $\delta$-th one. Then $\gamma<\alpha\delta$. This shows that $\alpha\beta$ is the sup of the $\alpha\delta$ for $\delta<\beta$.]

Of course, details may vary if the definition of ordinal multiplication you are using is different. (There are several equivalent possibilities.)

Let me expand a bit on why we argued as we did, since this is an issue that comes up with some frequency: The topology here is the order topology on ordinals. Note that $0$ and all successor ordinals are isolated: $\{\alpha+1\}=(\alpha,\alpha+2)$ and $\{0\}=(-\infty,1)$. On the other hand, any open neighborhood of $\alpha$ limit contains one of the form $(\beta,\alpha+1)$ for some $\beta<\alpha$.

To show that a function $f$ is continuous, we need to show that the preimage of any open set is open. It follows that it suffices to prove that if $f(\lambda)$ is limit, then for any $\beta<f(\lambda)$ there is a neighborhood $U$ of $\lambda$ contained in $(\beta,f(\lambda)+1)$. Of course, this is trivial if $\lambda$ itself is not a limit ordinal. On the other hand, if $\lambda$ is limit, we need to check that $\beta<f(\gamma)\le f(\lambda)$ for $\gamma<\lambda$ sufficiently large.

Suppose now, as in the question, that $f$ is strictly increasing. Then the condition above reduces to check, for $\lambda$ limit, that $f(\lambda)=\sup\{f(\gamma)\mid\gamma<\lambda\}$.

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can you elaborate on alphabeta is the sup of the alphasigma for sigma<beta? – Aaron Flood May 4 '11 at 3:15
i know what supremum means just wondering if you can give me a little extra info, thank you so much – Aaron Flood May 4 '11 at 3:15
alphabeta sup? :) – user641 May 4 '11 at 5:05

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