Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In most Descriptive Set Theory books, the rationale for working with the Baire space ($\mathbb{N}^{\mathbb{N}}$) as opposed to the real line ($\mathbb{R}$) is that the connectedness of the latter causes 'technical difficulties'.

My question is, what are these technical difficulties, and why does Descriptive Set Theory (normally?) stick to zero-dimensional Polish spaces?

Thanks in advance.

share|improve this question

1 Answer 1

up vote 18 down vote accepted

There are several useful properties of $\mathbb{N}^\mathbb{N}$:

  • $\mathbb{N}^\mathbb{N}$ is homeomorphic to $\mathbb{N}^\mathbb{N}\times\mathbb{N}^\mathbb{N}$. So we can view each element of $\mathbb{N}^\mathbb{N}$ as a code for a pair of elements, and the decoding maps are continuous. This is not the case for $\mathbb{R}$.
  • $\mathbb{N}^\mathbb{N}$ is not connected. Because $\mathbb{R}$ is connected, there are no nonconstant continuous functions from $\mathbb{R}$ to $\{0,1\}$; you have to move up a few levels in the Borel hierarchy to get such functions. This doesn't matter if you just care about the functions being Borel, but when you're looking at the actual levels of the Borel hierarchy it's more convenient to start with continuous functions rather than starting a little higher up.
  • It's easy to construct an element of $\mathbb{N}^\mathbb{N}$: you just have to construct a sequence of natural numbers. On the other hand, the representations of real numbers (Cauchy sequences, Dedekind cuts) are not as straightforward to work with. That makes proofs technically more difficult without making them more interesting. For example, compare the diagonalization proof that $\mathbb{N}^\mathbb{N}$ is uncountable with the diagonalization proof that $\mathbb{R}$ is uncountable.

There are also a few reasons that the use of $\mathbb{N}^\mathbb{N}$ does not result in a loss of generality:

  • For any uncountable complete separable metric space (c.s.m.s.) $X$, there is a bijection between $X$ and $\mathbb{N}^\mathbb{N}$ that is both Borel measurable and has a Borel measurable inverse. So if the property we are studying is preserved by Borel isomorphisms, we can just replace an uncountable c.s.m.s. $X$ with $\mathbb{N}^\mathbb{N}$.

  • Every c.s.m.s. is a continuous image of $\mathbb{N}^\mathbb{N}$. In fact, for any c.s.m.s. $X$ there is a closed subset $C$ of $\mathbb{N}^\mathbb{N}$ and a continuous bijection from $C$ to $X$. So if we are studying a property preserved by continuous maps, we can work with $\mathbb{N}^\mathbb{N}$ or with its closed subspaces without losing generality.

Those types of reasons are why it's safe to stick with $\mathbb{N}^\mathbb{N}$ most of the time: the goal is to study an arbitary c.s.m.s. (including $\mathbb{R}$) but for most purposes there's no loss of generality in studying $\mathbb{N}^\mathbb{N}$.

share|improve this answer
    
Any particular reason you prefer the name "c.s.m.s." to "Polish space"? Got something against Poland? :) –  Nate Eldredge Mar 11 '11 at 14:52
    
I don't have a preference between the two, I just happened to write c.s.m.s. this time. My own area is reverse math, where we always work with spaces that come equipped with a complete metric, rather than just completely metrizable spaces. So that may show through in my choice of words. –  Carl Mummert Mar 11 '11 at 15:16
4  
There is an additional technical advantage that is worth adding to Carl's list. Namely, many arguments in descriptive set theory use what we call the "effective" theory, that relates the underlying topological notions to computable counterparts. Key theorems in the field (e.g., the Glimm-Effros dichotomy) were originally proved using the effective theory. The natural setting to work with the required computability-theoretic notions is ${\mathbb N}^{\mathbb N}$. (Turing machines are identified with numbers, and their outcomes with sequences, i.e., points of ${\mathbb N}^{\mathbb N}$.) –  Andres Caicedo Mar 11 '11 at 19:41
    
Thanks, Andres. –  Carl Mummert Mar 11 '11 at 20:09
1  
It also comes up in relation to the computability issue that Andres Caicedo mentioned. There is a general heuristic that any computable function will be continuous. This is an obstacle to doing computability theory on the reals, again because there are only two continuous functions from $\mathbb{R}$ to $\{0,1\}$. So, for example, the function that takes a real and tells whether it's nonnegative can't be computable as a function on the reals, because it isn't continuous. There are various ways to work around this when we want to work with reals, but's easier to just work with Baire space. –  Carl Mummert Mar 12 '11 at 16:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.