@XanderHenderson I think arguments involving transfinite induction can be a little subtle, especially if you are proving that a statement is true for all ordinals. Then, your proof will likely informally use proper classes rather than sets, and convincing yourself that the proof can be translated into the language of ZFC can be a little painful.
Even in the case of more down to earth kinds of induction/recursion, there can be subleties. For example, consider the statement: if $R$ is a ring in which there is no strictly ascending chain of ideals $I_1\subsetneq I_2\subsetneq I_3\subsetneq\dots$, then every ideal of $R$ is finitely generated.
The usual argument would be as follows: suppose $R$ has an ideal $I$ which is not finitely generated. Pick an element $x_0\in I$. Then, $\{x_0\}$ does not generate $I$, so we can pick an element $x_1\in I\setminus\langle x_0\rangle$. Then, $\{x_0,x_1\}$ does not generate $I$, so we can pick an element $x_2\in I\setminus\langle x_0,x_1\rangle$. Continuing in this manner, we obtain a strictly ascending chain of ideals; contradiction.
I feel that the gap between this informal argument (that would almost unanimously be accepted as correct by mathematicians), and how the formal proof goes is quite large. For one thing, the "continuing in this manner" hides a lot of metamathematical subtleties.
Proofs can't be infinitely long, and so we certainly cannot "continue" in any literal sense of the word. Turning this into a formal proof would require you to pick a well-ordering of $R$, or perhaps pick a choice function on $\mathcal P(R)\setminus\{\varnothing\}$, and it would probably also require the use of the recursion theorem, which is also not completely trivial.
@Thorgott: Indeed, and even Zorn's lemma can be proved without transfinite induction. The first proof of Zorn's lemma I read was in Halmos's book Naive Set Theory, and at that stage in the book Halmos hadn't even introduced ordinals. The result was a proof that I personally found to be horribly unintuitive and technical.
I want to show that $$\beta(Y, Y') = \begin{cases} 0 & \text{if $|Y \cap Y'|$ is even} \\ 1 & \text{otherwise} \end{cases}$$ is linear in the first argument ($\beta(Y + Y', Z) = \beta(Y, Z) + \beta(Y', Z)$) where $+$ is the symmetric difference ($Y + Z = (Y \cup Z) \setminus (Y \cap Z)$).
While $$\beta(Y, Z) + \beta(Y', Z) = \begin{cases} 0 & \text{if $\Big(|Y \cap Z|$ and $|Y' \cap Z|$ are even$\Big)$ or both are odd } \\ 1 & \text{otherwise} \end{cases}$$
Now we somehow need to match these two
Any ideas?
Assume that both $|Y \cap Z|$ and $|Y' \cap Z|$ are even. Can we say anything about $|Y|, |Y'|, |Z|$?
Hi, I have some questions about recreational math. How to get started, where to find the resources, etc? Also happy to hear from anyone with experience in math competition. How do you find university math/recreational math?
@Simd Fractal geometry and analysis on fractals. I also do a little bit of $p$-adic analysis, and am probably slightly more knowledgable than most on random walks and Brownian motion.
@ModularMindset Gross. Don't get your disgusting nonsense involved in my beautiful analysis.
@Simd Again, it isn't really my area. But anything involving the distribution of primes is probably going to require a better working knowledge of number theory. There is also the issue of things like the prime number theorem being asymptotic, and not really saying a ton about any finite collection of numbers.
@ModularMindset In the usual setting Euclidean / Cartesian setting the heat equation is given by $\lambda u_t = \Delta x$, n'est-ce pas? Consider this on some open subset of $\mathbb{R}^n$, and assume Dirichlet boundary conditions (i.e. solutions vanish at the boundary). Solutions to this equation can be obtained via the spectrum of the Dirichlet operator.
@XanderHenderson $$\frac{\partial T}{\partial t}=k\nabla^{\alpha} T $$ like this? where the RHS involves the fractional laplacian where $\alpha$ is the fractal dim.?
My phd advisor's big idea was that you can associate a kind of "geometric zeta function" to a bounded set in $\mathbb{R}^n$---this function "sees" the geometry of the set. The Mellin transform of this zeta function (under certain hypotheses) is a spectral zeta function---the poles of the spectral zeta function are the spectrum of the Dirichlet operator.
My own work sought to push that idea to metric measure spaces with fairly general hypotheses (basically, as long as the Assouad dimension is finite, a lot of the theory pushes through nicely).
