Posted by Jason Polak on 11. January 2018 · Write a comment · Categories: analysis · Tags: ,

A total ordering $\lt$ on a set $S$ is called dense if for every two $x,y\in S$ with $x \lt y$, there exists a $z\in S$ such that $x\lt z\lt y$. A total ordering is said to be without endpoints if for every $x\in S$ there exists $y,z\in S$ such that $y \lt x\lt z$. In other words, a total order is without endpoints if there is no largest element and no smallest element.

Obviously, if a total order is dense or without endpoints, it must be a relation on an infinite set. One such example is the rational numbers with the usual order.

Theorem. [Cantor] Up to order isomorphism, there is exactly one countable dense totally ordered set without endpoints.

So the rational numbers is the only example of such a totally ordered set up to isomorphism. In particular, this means that $\Q$ and $\Q – \{0\}$ must be order isomorphic. Can you find the isomorphism?

If we take Cantor’s theorem and remove the word ‘countable’, then the new statement is no longer true! For example, Suppose we take the real numbers $\R$. Then we claim that $\R$ is not order isomorphic to $\R – \{0\}$.

Indeed, suppose $\varphi:\R\to\R-\{0\}$ is an order isomorphism. Consider the sequence $a_0,a_1,\dots$ defined by $\varphi(a_i) = 2^{-i}$. Since $\varphi(a_i) \gt \varphi(0)$, we see that $a_0,a_1,\dots$ is a bounded strictly decreasing sequence with limit $\lim a_i = a$.

Because $\varphi$ is an order isomorphism, we conclude $0 \gt\varphi(a)$. Now let $z\in \R$ be such that $0 \gt \varphi(z)\gt \varphi(a)$. Then $z\gt a$ and so $z \gt a_n$ for some $n$. But this means that $\varphi(z) \gt 0$, which is a contradiction! Hence no order isomorphism $\R\to \R-\{0\}$ exists.

Posted by Jason Polak on 24. December 2017 · Write a comment · Categories: analysis

Series hold endless fascination. To converge or not to converge? That is the question.

Let’s take the series $1 + 1/2 + 1/3 + \cdots$. It’s called the harmonic series, and it diverges. That’s because it is greater than the series
$$1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + \cdots = 1 + 1/2 + \cdots$$

The harmonic series diverges rather slowly, however. In fact, by comparing with the integral of $1/x$, we see that $1 + 1/2 + \cdots + 1/N$ can never be more than $\log(N) + 1$. For example, the sum of the first two hundred million terms is about 19.691044.

On the other hand, the sum of the reciprocals of the squares $1 + 1/4 + 1/9 + 1/16 + \cdots$ converges, which can be seen by comparing it to the integral of $1/x^2$ from one to infinity. In fact, $1 + 1/4 + 1/9 + \cdots = \pi^2/6$ as proved by Leonhard Euler. Here’s a question for you: does $\sum_{n=1}^\infty 1/n^s$ converge or diverge for $1 < s < 2$?

Even though the sum of reciprocals of squares converges, the sum of reciprocals of primes $1/2 + 1/3 + 1/5 + 1/7 + 1/11 + \cdots$ diverges. One could say by the convergence-divergence metric that the primes are more numerous than the squares. Also, in a similar vein to the harmonic series, the sum $1/2 + 1/3 + 1/5 + \cdots + 1/p – \log\log p$ is bounded.

Let’s go back to that harmonic series: $1 + 1/2 + 1/3 + 1/4 + \cdots$. Take this series, and delete every term whose denominator has the digit “9” somewhere in its decimal expansion. The resulting series converges! Can you prove it?

Posted by Jason Polak on 23. December 2017 · Write a comment · Categories: analysis · Tags: ,

Did you know that the closed interval $[0,1]$ cannot be partitioned into two sets $A$ and $B$ such that $B = A + t$ for some real number $t$? Of course, the half-open interval $[0,1)$ can so be partitioned: $A = [0,1/2)$ and $t = 1/2$. Why is this? I will leave the full details to the reader but I am sure they can be reconstructed without much difficulty using the following sketch:

Assume such a partition can be so made, and assume without loss of generality that $t \gt 0$. Then we must have $[0,t)\subseteq A$ and $(1-t,1]\subseteq B$. This shows that $t \lt 1/2$. Now, by assumption, $B = A + t$. Therefore, $[t,2t)\subseteq B$ since $[0,t)\subseteq A$ and similarly $(1-2t,1-t]\subseteq A$. Because $A$ and $B$ are disjoint, this implies that $t \lt 1/4$. We can continue to play this game, which shows that $t$ is strictly less than $1/(2n)$ for any $n$ and hence $t = 0$. This shows that such a partition cannot be made.

Pretty good. But did you know that the closed interval $[0,1]$ cannot be partitioned into two nonempty, disjoint open sets? Neither can any interval, whether open, closed, or half-open. In the language of topology, intervals of real numbers are connected. Proof?

Posted by Jason Polak on 13. June 2013 · Write a comment · Categories: analysis, elementary · Tags: , ,

A class of fractals known as Mandelbrot sets, named after Benoit Mandelbrot, have pervaded popular culture and are now controlling us. Well, perhaps not quite, but have you ever wondered how they are drawn? Here is an approximation of one:

xframe-200-30

From now on, Mandelbrot set will refer to the following set: for any complex number $ c$, consider the function $ f:\mathbb{C}\to\mathbb{C}$ defined by $ f_c(z) = z^2 + c$. We define the Mandelbrot set to be the set of complex numbers $ c\in\mathbb{C}$ such that the sequence of numbers $ f_c(0), f_c(f_c(0)),f_c(f_c(f_c(0))),\dots$ is bounded.
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