Thursday, 2 July 2015

real analysis - Why are Baire functions defined in terms of ordinals?



The Baire-$0$ functions, $\mathcal{B}_{0}$, are defined as the continuous real functions, and the Baire-$n$ functions, which I denote as $\mathcal{B}_{n}$, are defined as functions which can be expressed as pointwise limits of a sequence of functions $\{f_k\}_{k \geq 1}$ where each $f_k \in \bigcup_{0\leq i < n} \mathcal{B}_{i}$. According to Wikipedia, some authors require the additional condition that if $f \in \mathcal{B}_{n}$, then $f \notin \mathcal{B}_{i}$ for any $i

Wiki also states that Baire functions are defined in terms of ordinal numbers. In other words, we consider Baire functions of class $\alpha$ for any countable ordinal $\alpha$ instead of Baire functions of class $n$ where $n$ is a natural number (zero inclusive).



What exactly goes wrong if we work with natural numbers? Why do we need to use ordinals?




I don't know much about ordinals, so sorry if this is a naive question.


Answer



Every natural number is an ordinal, but not conversely. If I can define something for all ordinal numbers, that gets me further than if I just define it for all natural numbers.



Sometimes this doesn't matter, but in the Baire hierarchy (and the related Borel hierarchy) it really does. It's a good exercise to find a sequence of functions $f_0, f_1, f_2, f_3, ...$ such that




  • $f_i$ is of Baire class $i$,


  • The sequence $(f_i)_{i\in\mathbb{N}}$ converges pointwise to some function $g$, but



  • $g$ is not of any finite Baire class.




(HINT: Try to build an appropriate $g$ first, then figure out what the sequence of $f_i$s should be. We can build $g$ "in pieces": have $g\upharpoonright [0, 1]$ be Baire class $0$, $g\upharpoonright [1, 2]$ be Baire class $1$ but not Baire class $0$, $g\upharpoonright [2,3]$ be Baire class $2$ but not Baire class $1$, etc. - can $g$ be Baire class $n$ for any finite $n$?)



This function $g$ is intuitively "Baire class $\omega$;" getting a precise notion of this, and going beyond $\omega$, is accomplished by extending the definition of Baire classes to arbitrary (not just finite) ordinals.






Interestingly, it turns out that there is a limit to how far up the ordinals we have to go before we hit the point of diminishing (in fact, zero) returns: if $f$ is Baire class $\alpha$ for some ordinal $\alpha$, then $f$ is Baire class $\omega_1$ (where $\omega_1$ is the first uncountable ordinal) - in fact, $f$ is Baire class $\gamma$ for some countable $\gamma$. So while we can define "Baire class $\omega_{\omega_7+3}+32$," this doesn't get us anything that we didn't get at stage $\omega_1$ anyways. (The same thing happens with the Borel hierarchy.)



No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...