# Exponentiation and Set of Functions Notation.

Given arbitrary sets $A$ and $B$, the notation $A^{B}$ is mostly clear from context to mean $A^{B} = \{f : f : B \rightarrow A\}$.

However, when these sets are ordinal or cardinals, especially $\omega$, the notation is not consistent even among subfields of logic.

For example $2^\omega$ in one sense can denote ordinal exponentiation. Hence $2^\omega = \lim_{n < \omega} 2^n = \omega$.

However, you can also consider $2^{\aleph_0}$. By using the cardinal $\aleph_0$, some people may consider it clear that $2^{\aleph_0}$ denotes the cardinal of the set $\{f : \aleph_0 = \omega_0 = \omega \rightarrow 2\}$.

In the above paragraph $2^{\aleph_0}$ is a cardinal, (in ZFC) it is a special ordinal. However, it descriptive set theory, you may want to consider not the cardinal but Cantor Space (or Baire Space), i.e. the set of functions from $\omega \rightarrow 2$. When you want the set of functions as oppose to the ordinal, is there a notation for that.

In recursion theory, I have found that $2^\omega$ or $\omega^\omega$ most frequently refers to Cantor or Baire space, and not the ordinal or cardinal. In Mostovachis book, he uses $\text{}^\omega2$ to denote Cantor Space.

Does anyone know of any establish custom to distiguish between ordinal exponentiation, cardinality of the set of functions between ordinals, and the actual set of functions between ordinals. I was thinking perhaps the left right exponent like $\text{}^\omega2$ and $2^\omega$ could be used as distinction, but from reference to recursion theory and Moschivakis's book, it seems that this is not the case.

• Language nit: Sets are neither ordinals nor cardinals. You mean when $A$ and $B$ are taken to be ordinals or cardinals, not when the sets are "ordinals" or "cardinals." – Thomas Andrews Aug 8 '12 at 17:25
• $2^{\aleph_0}$ is cardinal regardless to the axiom of choice. In fact specifying $\aleph_0$ in the exponent means that it is implied that the interpretation is the cardinal of the power set of any set of size $\aleph_0$. Yes, cardinals make sense without the axiom of choice. – Asaf Karagila Aug 8 '12 at 17:33
• It's not clear that $2^\omega = \omega$ since there is one order-preserving map from $\omega$ to (the ordinal) $2$ that is greater than all other such maps, the maximal constant function. So $2^\omega = \omega + 1$ – Thomas Andrews Aug 8 '12 at 17:36
• @ThomasAndrews The third paragraph, I am referring explicitly to ordinal exponentiation. By definition (since $\omega$ is a limit ordinal), $2^\omega = \lim_{\alpha < \omega} 2^\alpha = \omega$. Refer to en.wikipedia.org/wiki/Ordinal_arithmetic#Exponentiation – William Aug 8 '12 at 17:53
• It's probably too much to expect an unambiguous notation that everyone would understand, but if you're going to develop one for yourself, you might want to take a hint from computer science and let "$A\to B$" itself mean the set of all functions with domain $A$ and codomain $B$. You can then decorate the arrow in various ways when you need to consider only special kind of functions (e.g. hom(e)omorphisms). – Henning Makholm Aug 8 '12 at 17:55

[Good] writers will usually try to ensure the context is set when setting the notation "consider Cantor space, denoted by $2^\omega$, ..." or something of this kind.
That been said, when $\aleph$ numbers are in the exponent this is always the cardinal version of exponentiation.
There is also a common notation of $^AB$ as the set of all functions from $A$ into $B$, in this aspect $^\omega\omega$ can be seen as the Baire space (as its underlying set is exactly this). Note also that $\left|^AB\right|=|B|^{|A|}$.
Mildly related: Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?