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By definition of exponent operator on ordinals, we have $$0^\omega=\lim_{\xi\to\omega}0^\xi$$

However, Note that $0^\xi$ is not increasing, so if we still let $\lim_{\xi\to\omega}0^\xi=\sup\{0^\xi|\xi<\omega\}$ then it is followed by $$0^\omega=\sup\{1,0,0,\ldots\}=1$$ An incredible result.

Nevertheless if we use the definition of limit superior $$\overline{\lim}_{\eta\to\beta}\alpha^\eta:=\inf_{\eta < \beta}\sup_{\eta \le\xi<\beta}\alpha^\xi$$ and limit inferior $$\underline{\lim}_{\eta\to\beta}\alpha^\eta:=\sup_{\eta < \beta}\inf_{\eta \le\xi<\beta}\alpha^\xi$$ then we get $$\lim_{\xi\to\omega}0^\xi=\underline{\lim}_{\xi\to\omega}0^\xi=\overline{\lim}_{\xi\to\omega}0^\xi=0$$

Conceivable, but this is not the end of the story. Let's consider about $m^n$ with $0<m,n< \omega$. Since exponent operator is continuous in the second slot, this sentence following must hold: $$m^n=\lim_{\xi\to n}m^\xi=\underline{\lim}_{\xi\to n}m^\xi=\overline{\lim}_{\xi\to n}m^\xi=m^{n-1}$$ fail when $1<m$.

So can we find a perfect definition of limit on ordinals?


By the way, if we deal the Ordinal class as a discrete topology space, then $\{\alpha\}$ is a neighborhood of $\alpha$, hence $\{\alpha\}$ must in every filterbase which converge to $\alpha$. So if we let $$\overline{\lim}_{\eta\to\beta}\alpha^\eta:=\inf_{\eta \le \beta}\sup_{\eta \le\xi\le\beta}\alpha^\xi$$ and $$\underline{\lim}_{\eta\to\beta}\alpha^\eta:=\sup_{\eta \le \beta}\inf_{\eta \le\xi\le\beta}\alpha^\xi$$

Then $$\alpha^\beta=\lim_{\xi\to \beta}\alpha^\xi=\underline{\lim}_{\xi\to \beta}\alpha^\xi=\overline{\lim}_{\xi\to \beta}\alpha^\xi$$ hold for every $(\alpha,\beta) \in \mathbb O^2$

But actually it cannot be a definition since $$\inf_{\eta \le \beta}\sup_{\eta \le\xi\le\beta}\alpha^\xi:=\inf\{\sup\{\alpha^\xi|\eta \le\xi\le\beta\}|\eta \le \beta\}$$ but $\{\alpha^\xi|\eta \le\xi\le\beta\}$ contains $\alpha^\beta$!

Update: Edited title. The eventual question is as title illustrates.

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You wrote $\lim_{\xi\to\omega}0^\xi=\sup\{0^\xi|\xi<\omega\}$. This is true for non-decreasing sequences, so you cannot deduce this from the fact that the sequence $(0^\xi)$ is non-increasing. – Martin Sleziak Jun 10 '12 at 12:50
@MartinSleziak:Indeed, but what's the exact definition of ordinal limit? – Popopo Jun 10 '12 at 13:50
You keep changing your question and its title. At some point you should figure out one question, then ask a new follow up question. Also the definition of limit in what context? – Asaf Karagila Jun 10 '12 at 15:14
@AsafKaragila:OK, thank you for your advisement. – Popopo Jun 10 '12 at 15:47
By the way,Wiki says that for every countable sequence of sets $$\overline{\lim}_{n\to \infty}X_n=\bigcap_{n=1}\bigcup_{m=n}^{\infty}X_m$$ and $$\underline{\lim}_{n\to \infty}X_n=\bigcup_{n=1}\bigcap_{m=n}^{\infty}X_m$$. And finally $$\lim_{n\to \infty}X_n=\overline{\lim}_{n\to \infty}X_n=\underline{\lim}_{n\to \infty}X_n$$ if this equation holds. Since ordinal numbers are all sets, it can be applied by this kind of limit. – Popopo Jun 10 '12 at 16:17
up vote 3 down vote accepted

You asked in your comment:

Indeed, but what's the exact definition of ordinal limit?

Let me try to address this, although this is not an answer to your original question.

The notion of limit of of transfinite sequence is defined only for limit ordinals see e.g. Wikipedia

If $\alpha$ is a limit ordinal and $X$ is a set, an $\alpha$-indexed sequence of elements of $X$ is a function from $\alpha$ to $X$. This concept, a transfinite sequence or ordinal-indexed sequence, is a generalization of the concept of a sequence. An ordinary sequence corresponds to the case $\alpha=\omega$.

Some authors consider only increasing sequences, see e.g. Kechris: Classical Descriptive Set Theory p.349 or Sierpinski: Cardinal and ordinal numbers, p.287 or Kuratowski, Mostowski: Set Theory: p.231. In the case of increasing (non-decreasing) sequence we have $\lim_{\xi\to\alpha} \beta_\xi=\sup\{\beta_\xi; \xi<\alpha\}$.

However, it makes sense to define limit of any transfinite sequence $(\beta_\xi)_{\xi < \alpha}$, if $\alpha$ is a limit ordinal. (We do not have to use only monotone sequence.)

A transfinite sequence transfinite sequence $(\beta_\xi)_{\xi < \alpha}$ converges to an ordinal $\beta\ne0$ if, for every ordinal number $\gamma<\beta$, there exists an ordinal number $\eta < \alpha$ such that $\gamma <\beta_\xi\le\beta$ whenever $\eta <\xi < \alpha$.

$$(\forall \gamma <\beta) \quad (\exists \eta < \alpha) \quad (\forall \xi) \qquad (\eta<\xi<\alpha \Rightarrow \gamma < \beta_\xi \le \beta) $$

A transfinite sequence $(\beta_\xi)_{\xi < \alpha}$ converges to $0$ if it is eventually equal to zero. $$(\exists \eta < \alpha) \quad (\forall \xi) \qquad (\eta<\xi<\alpha \Rightarrow \beta_\xi =0) $$

If you are familiar with nets, you can notice that this is the same as limit of the transfinite sequence $(\beta_\xi)_{\xi< \alpha}$ considered as a net in the order topology on $\lambda$, where $\lambda$ is any ordinal larger than all $\beta_\xi$'s.

The basis for the order topology on some totally ordered set $X$ consists of sets of the form $(-\infty,b)=\{x\in X; x<b\}$, $(a,\infty)=\{x\in X; a<x\}$ and $(a,b)=\{x\in X; a<x<b\}$ for $a<b$.

It is not difficult to notice, that if $\beta>0$ then neighborhood basis of $\beta$ (as an element of some $\lambda>\beta$ with order topology) consists of all sets of the form $(\gamma,\beta+1)$ where $\gamma>\beta$. Precisely these sets were used in the above definition.

Neighborhood basis at $0$, cannot be described in that way, that's why I had to treat the case $\beta=0$ separately. (In this case, we can take the neighborhood basis consisting of the single set $(-\infty,1)=\{0\}$.) Thanks for letting me know that the definition I suggested originally wasn't working.

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Thank you for your attention. I slept on it a day, but finally found out that this is still not the correct definition. Let $\alpha=7$, then for any ordinal $\beta$ and $\gamma<\beta$, we can choose $\eta=6$. Then the whole sentence holds since there is no $\xi$ between $\eta$ and $\alpha$. In conclusion any $(\beta_\xi)_{\xi<7}$ converges to every ordinal number$\ldots$ – Popopo Jun 11 '12 at 14:02
You've noticed that in my answer I stressed the phrase only for limit ordinals, right? In your example: $\alpha=7$ is not a limit ordinal. – Martin Sleziak Jun 11 '12 at 14:04
Oh, I'm sorry. But for limit ordinals, let $\beta=0$, since there is no $\gamma<0$, this sentence is still true. Therefore $(\beta_\xi)_{\xi<\alpha}$ must converges to $0$ whenever $\alpha$ is a limit ordinal$\ldots$ – Popopo Jun 11 '12 at 14:10
You're correct about the problem with $\beta=0$. I don't see any more elegant way out of this than treating the case $\beta=0$ separately. Sorry, I should have been more careful when I wrote down that definition. – Martin Sleziak Jun 11 '12 at 14:26
I slept on it a day again, and found out this is actually exact definition for limit of ordinal function in limit ordinal case. Thank you for your patient solution. – Popopo Jun 12 '12 at 9:24

The answer $0^\omega = 0$ is correct, and intended. The "sup" definition is a mistake someone made by overlooking this single case.

The "limit" definition for ordinals must be $\lim_{\xi \to n} f(\xi) = f(n)$ in case $n$ is not a limit ordinal. This is the same as we use in any topological space, the limit at an isolated point.

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:Thank you. But when I deal with it as a isolated point, it would occur in the definition sentence $\ldots$ – Popopo Jun 10 '12 at 13:40

The problem is that the ordinal topology is not the discrete topology. It is the order topology, namely an open set is an interval.

It is easy to see that successor ordinals are indeed isolated points, but limit ordinals are exactly those ordinals which are not isolated.

The definition of the exponentiation is somewhat topological and therefore when considering a limit ordinal we consider this a limit in the topological sense.

What may confuse you is that if $f\colon\mathrm{Ord\to Ord}$ then $\lim\limits_{\xi\uparrow\delta}f(\xi)=\sup\{f(\xi)\mid\xi<\delta\}$ if and only if $f$ is non-decreasing.

As you noted $0^0=1$ but $0^1=0$, therefore the function is decreasing and we cannot interchange the $\sup$ and $\lim$.

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If $\beta$ isn't a limit ordinal, we can't really talk about limits of functions $f(\gamma)$ as $\gamma$ approaches $\beta$. After all, $\beta$ is isolated, and so we can't approach it arbitrarily closely.

If we are dealing with a non-decreasing sequence of ordinals, there are two possibilities: (1) The sequence is eventually constant, so has a greatest element, which will be the limit of the sequence. (2) The sequence is not eventually constant, so has no greatest element, so its supremum (and limit) will be a limit ordinal, specifically the least limit ordinal greater than every ordinal in the sequence.

If we are dealing with non-increasing sequences, they must be eventually constant, because every non-$\emptyset$ set of ordinals has a least member. This really precludes any kind of "convergence from above", so we will always be dealing with convergence from below (as Martin expresses explicitly in his answer).

Ordinal exponentiation is more accurately recursively defined as follows:

$\alpha^0=1$; $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$; and for limit $\lambda$, $\alpha^\lambda=\sup\{\alpha^\beta:0<\beta<\lambda\}$.

We exclude the $\beta=0$ case from the supremum precisely to avoid having $0^\lambda=1$ for all limit $\lambda$.

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