1. Nested sets

Q:

Consider the infinite expansion of $\sqrt{2}$ $=$ $
1.414213562373095048801688724209698078569671875376 9
· · · = 1.x_{1} x_{2} . . . x_{n} . . . .$

Denote by $r_{n}$ the truncation of the above representation after the first $n$ decimal digits. So $r_{n} = 1.x_{1}x_{2} . . . x_{n}$ ; for example $r_{1} = 1.4$, $r_{2} = 1.41$, $r_{3} = 1.141$, and so on.

(a) Consider the indexed family of sets $\{A_{n}\}_{n\in\mathbb{N}}$, where $A_{n} = (-\infty, r_{n} )$, open intervals of $\mathbb{R}$. Find $\bigcup_{n\in{\mathbb{N}}}A_{n}$ and $\bigcap_{n\in{\mathbb{N}}}A_{n}$. Justify your answers.

b) Consider the indexed family of sets $\{B_{n}\}_{n\in\mathbb{N}}$, where $A_{n} = [\frac{1}{n}, r_{n} )$, open intervals of $\mathbb{R}$. Find $\bigcup_{n\in{\mathbb{N}}}B_{n}$ and $\bigcap_{n\in{\mathbb{N}}}B_{n}$. Justify your answers.

A:

a) $\bigcup_{n\in{\mathbb{N}}}A_{n}=(-\infty,\sqrt{2})$ and $\bigcap_{n\in{\mathbb{N}}}A_{n}=(-\infty,1.4]$.

b) $\bigcup_{n\in{\mathbb{N}}}B_{n}=(0,\sqrt{2})$ and $\bigcap_{n\in{\mathbb{N}}}B_{n}=[1,1.4)$.

I do not know how to prove my answers. I have tried, and can't seem to reach a conclusion. Could someone please help me with either a or b? Is there a systematic whay to approach these kinds of problems? How do I use the Archimedean principle?

I didn't really get this stuff down in my foundations course and now its my first HW question for real analysis course .

2. Here is my attempt to prove part (a)

To show $\bigcup_{n\in{\mathbb{N}}}A_{n}=(-\infty,\sqrt{2})$, first let $x\in (-\infty,\sqrt{2})$. This means $x<\sqrt{2}$. Since $\lim_{x \to\infty} r_n=\sqrt{2}$, there exists $m\in \mathbb N$ such that $x (otherwise $r_n \leq x \forall n\in \mathbb N$ implies that $\sqrt{2} \leq x$, which is a contradiction). So, $x\in A_m \subseteq \bigcup_{n\in{\mathbb{N}}}A_{n}$. It follows that $(-\infty,\sqrt{2}) \subseteq \bigcup_{n\in{\mathbb{N}}}A_{n}$. To show there other inclusion, let $x\in \bigcup_{n\in{\mathbb{N}}}A_{n}$. This means $x\in A_n$ for some $n\in \mathbb N$. That is, $x. Since $r_n<\sqrt{2}, x\in (-\infty,\sqrt{2})$. Thus, equality follows.

I believe the second equality should be $\bigcap_{n\in{\mathbb{N}}}A_{n}=(-\infty,1.4)$. That is, not including 1.4. The method of proof is the same as above, i.e., show inclusion in both directions. First, let $x\in \bigcap_{n\in{\mathbb{N}}}A_{n}$. This implies $x, in particular, $x. So, $x\in (-\infty,1.4)$. To show the other direction of the inclusion, let $x\in (-\infty,1.4)$ and note that $x<1.4\leq r_n \forall n\in \mathbb N$. This implies $x\in \bigcap_{n\in{\mathbb{N}}}A_{n}$. This proves the equality.

I hope this helps a bit.

3. [INDENT]
Originally Posted by Monkey D. Johnny
Here is my attempt to prove part (a)

I believe the second equality should be $\bigcap_{n\in{\mathbb{N}}}A_{n}=(-\infty,1.4)$. That is, not including 1.4. The method of proof is the same as above, i.e., show inclusion in both directions. First, let $x\in \bigcap_{n\in{\mathbb{N}}}A_{n}$. This implies $x, in particular, $x. So, $x\in (-\infty,1.4)$. To show the other direction of the inclusion, let $x\in (-\infty,1.4)$ and note that $x<1.4\leq r_n \forall n\in \mathbb N$. This implies $x\in \bigcap_{n\in{\mathbb{N}}}A_{n}$. This proves the equality.
Why is 1.4 not included?

Can a proof by contradiction be done for ( $\Rightarrow$)? I keep trying by assuming there exists an x not in the intersection, but I can't form a solid contradiction. I keep thinking, if I choose n+1 and take the limit, the limit will still be $\sqrt{2}$ and thus, no such x can exist since, but I can't seem to formalize this idea and make it work for me.

Aslo, for part b, how would I use the Archimedean principle?

4. Originally Posted by Danneedshelp
Q:

Consider the infinite expansion of $\sqrt{2}$ $=$ $
1.414213562373095048801688724209698078569671875376 9
• • • = 1.x_{1} x_{2} . . . x_{n} . . . .$

Denote by $r_{n}$ the truncation of the above representation after the first $n$ decimal digits. So $r_{n} = 1.x_{1}x_{2} . . . x_{n}$ ; for example $r_{1} = 1.4$, $r_{2} = 1.41$, $r_{3} = 1.141$, and so on.
(a) Consider the indexed family of sets $\{A_{n}\}_{n\in\mathbb{N}}$, where $A_{n} = (-\infty, r_{n} )$, open intervals of $\mathbb{R}$. Find $\bigcup_{n\in{\mathbb{N}}}A_{n}$ and $\bigcap_{n\in{\mathbb{N}}}A_{n}$. Justify your answers.
b) Consider the indexed family of sets $\{B_{n}\}_{n\in\mathbb{N}}$, where $A_{n} = [\frac{1}{n}, r_{n} )$, open intervals of $\mathbb{R}$. Find $\bigcup_{n\in{\mathbb{N}}}B_{n}$ and $\bigcap_{n\in{\mathbb{N}}}B_{n}$. Justify your answers.

A:

a) $\bigcup_{n\in{\mathbb{N}}}A_{n}=(-\infty,\sqrt{2})$ and $\bigcap_{n\in{\mathbb{N}}}A_{n}=(-\infty,1.4]$.

b) $\bigcup_{n\in{\mathbb{N}}}B_{n}=(0,\sqrt{2})$ and $\bigcap_{n\in{\mathbb{N}}}B_{n}=[1,1.4)$.

I do not know how to prove my answers. I have tried, and can't seem to reach a conclusion. Could someone please help me with either a or b? Is there a systematic whay to approach these kinds of problems? How do I use the Archimedean principle?
First, both of those answers are correct.

Second, I cannot imagine what you mean by ‘proving’ these answers.
It really is about knowing how to use union and intersection.

BTW: Here is an easy way: $r_n = \frac{{\left\lfloor {10^n \sqrt 2 } \right\rfloor }}{{10^n }}$.

5. Originally Posted by Plato
First, both of those answers are correct.

Second, I cannot imagine what you mean by ‘proving’ these answers.
It really is about knowing how to use union and intersection.

BTW: Here is an easy way: $r_n = \frac{{\left\lfloor {10^n \sqrt 2 } \right\rfloor }}{{10^n }}$.
Well, as the question states, my teacher wants us to "justify" our answers. He went over something similar last semester in my foundations class. I guess he wants to show there is no other possible answer, by showing every arbitrary value we choose are within the union or intersection.

6. 1.4 is not included because if it were it would be in $A_n$ for all n, in particular, n=1. But $1.4 \notin A_1$.

I think the Archimedean principle is needed to show that $(0,\sqrt{2})$ is a subset of the union of the $B_n's$: Let x be an element of $(0,\sqrt{2})$, then there exists $m\in \mathbb N$ such that 1/m<x by the Archimedean principle. Next, we know that $(\exists m'\in \mathbb N)(x, by the same argument used in part (a). Lastly, take $M=max(m,m')$ and note that $x\in B_M$. Hence x is an element of the union of the $B_n's$.