Recall that the Cantor set is defined as the intersection where

and is obtained by removing from each closed interval that makes up its open middle third, so

,

,

etc. Each is the union of closed intervals, each of length .

Let’s prove that is the interval . (Cf. Abbott, Understanding analysis, Exercise 3.3.6.)

1.

The usual proof consists in showing inductively that for all . This is easy: Note first that

,

where

and

.

This equality is verified by induction. Using this, we can use induction again to verify that, indeed, for all .

We clearly have that . To prove the converse, for each and each , pick such that . The sequence of is bounded, so it has a convergent subsequence . The corresponding subsequence has itself a convergent subsequence . One argues that their limit values belong to , because they belong to each , since these sets are nested and closed. Finally, it follows immediately that as well.

2.

A very elegant different argument is obtained by using an alternative characterization of : Note that each can be written in base three as

where each is , , or . By induction, one easily verifies that iff it admits such an expansion with . It follows that iff it admits an expansion where no is .

Given , we have , so we can write where the ternary expansion of has only s and s (so ), and the expansion of has only s and s: If

,

we can set where unless , in which case as well, and similarly where unless , in which case as well.

We then have that , and both and are in .

This construction has the further advantage of making clear that the typical admits continuum many () representations as sum of two members of : If we can split (where the expansions of only have s and s), we can set

.

This gives us as many representations as subsets of .

3.

The related problem of describing appears to be much more complicated. See here and here.

Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

The description below comes from József Beck. Combinatorial games. Tic-tac-toe theory, Encyclopedia of Mathematics and its Applications, 114. Cambridge University Press, Cambridge, 2008, MR2402857 (2009g:91038). Given a finite set $S$ of points in the plane $\mathbb R^2$, consider the following game between two players Maker and Breaker. The players alternat […]

Yes. This is a consequence of the Davis-Matiyasevich-Putnam-Robinson work on Hilbert's 10th problem, and some standard number theory. A number of papers have details of the $\Pi^0_1$ sentence. To begin with, take a look at the relevant paper in Mathematical developments arising from Hilbert's problems (Proc. Sympos. Pure Math., Northern Illinois Un […]

It is easy to see without choice that if there is a surjection from $A$ onto $B$, then there is an injection from ${\mathcal P}(B)$ into ${\mathcal P}(A)$, and the result follows from Cantor's theorem that $B

Only noticed this question today. Although the selected answer is quite nice and arguably simpler than the argument below, none of the posted answers address what appeared to be the original intent of establishing the inequality using the Arithmetic Mean-Geometric Mean Inequality. For this, simply notice that $$ 1+3+\ldots+(2n-1)=n^2, $$ which can be easily […]

First of all, $f(z)+e^z\ne 0$ by the first inequality. It follows that $e^z/(f(z)+e^z)$ is entire, and bounded above. You should be able to conclude from that.

Yes. The standard way of defining these sequences goes by assigning in an explicit fashion to each limit ordinal $\alpha$, for as long as possible, an increasing sequence $\alpha_n$ that converges to $\alpha$. Once this is done, we can define $f_\alpha$ by diagonalizing, so $f_\alpha(n)=f_{\alpha_n}(n)$ for all $n$. Of course there are many possible choices […]

I disagree with the advice of sending a paper to a journal before searching the relevant literature. It is almost guaranteed that a paper on the fundamental theorem of algebra (a very classical and well-studied topic) will be rejected if you do not include mention on previous proofs, and comparisons, explaining how your proof differs from them, etc. It is no […]

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