Math Help Forum - View Single Post

ThePerfectHacker · #10 June 10th, 2007, 09:25 PM

1)This problem was inspired from a proof I saw of a certain theorem. The proof was elegantly proven by enclosing a bounded set into a square and creating this nested sequence of squares. So I decided to make this a problem of the week.

The trick is to think of Cauchy Sequences. That is what Rebesques was suggesting. Doing this the ordinary way, i.e. convergent sequences might be difficult.

We can think of these sequence of points as complex numbers, if you want to. But that is not necessary. Let $L$ be the length of the square. Then the maximum distance between any two points in $S_1$ is $\frac{L\sqrt{2}}{2}$ , i.e. the diagnol. The length of the side of $S_2$ is $\frac{L}{2}$ and hence the maximum distance between any two points in $S_2$ is $\frac{L\sqrt{2}}{2^2}$ . By induction the maximum distance between any two points in $S_n$ is $\frac{L\sqrt{2}}{2^n}$ .
Let $s_1,s_2,s_3,...$ be the sequence of centers of these squares. It is necessary and sufficient to show that $(s_n)$ is a Cauchy sequence. Let $\epsilon >0$ then it is possible to find such an $N>0$ so that $\frac{L\sqrt{2}}{2^n} < \epsilon$ for all $n>N$ because $\lim \frac{L\sqrt{2}}{2^n} = 0$ . But if $n,m>N$ then $s_n,s_m$ all lie in the $S_{N+1}$ square because $S_{k+1} \subset S_k$ . But by above arguments the maximum distance between any two points in $S_{N+1}$ is $\frac{L\sqrt{2}}{2^{N+1}} < \epsilon$ . And so if $n,m>N$ we have $|s_n-s_m| < \epsilon$ . Hence $(s_n)$ is convergent (eventhough we have no idea what it converges to).
$\mathbb{Q}uad \ \mathbb{E}ra \ \mathbb{D}emonstrandum$ .
-----
2)I found this nice problem in some book. The following is my argument. First if $U = \{ \}$ the proof is immediate. And if $S \mbox{ or }T = \{ \}$ then the proof is immediate again. Having desposed the trivial cases let us consider a non-empty subset $U$ of $\mathbb{R}$ which is closed. And we can safely say that $S,T$ are non-empty subsets of $U$ . We will argue by contradiction: assume that both are not closed. Hence, there exists $s_1,s_2\in S \mbox{ and }t_1,t_2\in T$ with the property that $s_1s_2 \not \in S \mbox{ and }t_1t_2 \not \in T$ . But since these are disjoint subsets with the condition that $S\cup T = U$ and that $U$ is closed we must have $s_1s_2 \in U \mbox{ and }t_1t_2 \in U$ thus $s_1s_2 \in T \mbox{ and }t_1t_2 \in S$ . We have shown that $t_1,t_2,s_1s_2 \in T \mbox{ and }s_1,s_2,t_1t_2 \in S$ . Now we use the condition that the product of any three elements in again in the set and hence: $t_1t_2(s_1s_2) \in T \mbox{ and }s_1s_2(t_1t_2)\in S$ . But that means $S\cap T \not = \{ \}$ which is a contradiction. Which means one of these sets must be closed.
$\mathbb{Q}uad \ \mathbb{E}ra \ \mathbb{D}emonstrandum$ .
-----
3)This was fully answered before and a link was given. The nice thing about this problem is that ordinary induction does not work. This is a good time to learn Strong Induction.