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Christmas Riddle

Posted by Mathoman, Friday 25 December 2009 at 13:57 - Riddles And Exercises - Tags

X-mas is the time to crack nuts. At least that's what normal people do. But mathematicians enjoy cracking very peculiar nuts, like the following one. Show that the equation below is true for every positive integer.

$\sum_{k=0}^n\left\(\begin{array}{c}2n+1\\2k+1\end{array}\right\)(2k+1)\:=\:2^{2n-1}(2n+1).$

As always in maths, the slogan is short is beautiful, i.e., you should find a solution that is as short and elegant as possible, without using much computations. For this exercise ou may get inspiration from Santa Claus who has to distribute gifts into ChristmasStockings...

An exercise about groups

Posted by Mathoman, Sunday 20 December 2009 at 12:45 - Riddles And Exercises - Tags

A topological group is a set G with a group structure and a topology such that the binary law

$G \times G \rightarrow G ,\;\; (x,y) \rightarrow xy,$

and the map of the inverse

$G \rightarrow G ,\;\; x \rightarrow x^{-1},$

are continuous. In other words both structures, the algebraic and the topological one, are linked in a natural way. So the following question is of interest.

Question

Are there topological groups which are isomorphic as groups and homeomorphic as topological spaces but not isomorphic as topological groups?

Answer

Yes. The proof goes in three stages:

Let G and H be dense subsets of $\mathbb{R}$ and $f :\, G \rightarrow H$ a monotonic bijection. Then f is a homeomorphism.

Without restriction of generality we can suppose that f is increasing. We prove that it is continuous. Let $x_0\in G$ and $\epsilon>0$ . Since H is dense in $\mathbb{R}$ we have $H\cap\,]f(x_0)-\epsilon,f(x_0)[\,\neq\emptyset$ . Thus there exists $y_1\in H\cap\,]f(x_0)-\epsilon,f(x_0)[\,$ . Similarly there exists $y_2\in H\cap\,]f(x_0),f(x_0)+\epsilon[\,$ . Since f is onto we can write $y_k=f(x_k)$ with $x_k\in G, k=1,2$ . Let $\delta=\min(x_0-x_1,x_2-x_0)$ . Then for all x in G
$\begin{align*}x_0-\delta<x<x_0+\delta \;\;\;\Longrightarrow\;\;\;& f(x_0-\delta)<f(x)<f(x_0+\delta)\\ \Longrightarrow\;\;\;&y_1=f(x_1)\leq f(x)\leq f(x_2)=y_2\\ \Longrightarrow\;\;\;&f(x_0)-\epsilon<f(x)<f(x_0)+\epsilon\,, \end{align*}$
which proves that f is continuous in $x_0$ . The proof of the continuity of the inverse function $f^{-1}$ is the same.
Let G and H be countable and dense subsets of $\mathbb{R}$ . Then they are homeomorphic.

First we write
$\begin{align*} G&=\{x_0,x_1,x_2,\ldots\}\;\;\;\;\;(*)\,,&H&=\{y_0,y_1,y_2,\ldots\}\;\;\;\;(**)\,. \end{align*}$
We will now list G and H in a new way, and The aim is to make a monotonic bijection (then it is automatically a homeomorphism). For this we proceed as follows.
- k=0. We take $x'_0=x_0,\;y'_0=y_0$
- k=1. We take $x'_1=x_1$ . To choose $y'_1$ we look at the order of $x'_0$ and $x_1'$ .
  If $x'_1<x'_0$ then we take as $y'_1$ any element in H which is smaller than $y'_0$ .
  If $x'_1>x'_0$ then we take as $y'_1$ any element in H which is greater than $y'_0$ .
- k=2. We take as $y'_2$ the first element of $H\setminus\{y'_0,y'_1\}$ listed in (**). To choose $x'_2$ we look at the order of $y'_0,y'_1,y'_2$ .
  If $y'_2$ is smaller than $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is smaller than $x'_0$ and $x'_1$ .
  If $y'_2$ is greater than $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is greater than $x'_0$ and $x'_1$ .
  If $y'_2$ is between $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is between $x'_0$ and $x'_1$ .
- k=3. We take as $x'_3$ the first element of $G\setminus\{x'_0,x'_1,x'_2\}$ listed in (*). To choose $y'_3$ we look at the order of $x'_0,x'_1,x'_2,x'_3$ . There are 24 possible orders for these four numbers.
  If $x'_3<x'_0<x'_1<x'_2$ we take as $y'_3$ any element in H which is smaller than $y'_0,y'_1,y'_2$ .
  If $x'_2<x'_3<x'_0<x'_1$ we take as $y'_3$ any element in H which is bewteen $y'_2$ and $y'_0$ .
  And so forth.
The topological groups $G=\mathbb{Q}+\mathbb{Q}\sqrt2$ and $H=\mathbb{Q}+\mathbb{Q}\sqrt3$ fulfill our wishes.

After what has been said before, G and H are homeomorphic as topological spaces. Clearly they are isomorphic groups. But they are not isomorphic topological groups. In fact, suppose that there exists an isomorphism of topological groups $f :\, G \to H$ . One proves easily by induction that f(n)=nf(1) for every integer n, and then that f(r)=rf(1) for every rational r. Therefore by continuity
$f(\sqrt2)=\sqrt2f(1),\;\;\;\;\;\lightning$
a contradiction in $H=\mathbb{Q}+\mathbb{Q}\sqrt3$ .