1. ## Sequence

I need help with the following problem in analysis:

Find a sequence of closed intervals I_1 I_2 ... I_n ... whose end points are rational numbers and such that I_n = {e}.

2. Originally Posted by tarheelborn
I need help with the following problem in analysis:

Find a sequence of closed intervals I_1 I_2 ... I_n ... whose end points are rational numbers and such that I_n = {e}.
The rational numbers are dense in the reals.
$\displaystyle \left( {\exists p_1 \in \mathbb{Q}} \right)\left[ {e - 1 < p_1 < e} \right]\;\& \;\left( {\exists q_1 \in \mathbb{Q}} \right)\left[ {e < q_1 < e + 1} \right]$.

Let $\displaystyle \delta _1 = \max \left\{ {p_1 ,e - \frac{1}{2}} \right\}\;\& \,\varepsilon _1 = \min \left\{ {q_1 ,e + \frac{1}{2}} \right\}$.
$\displaystyle \left( {\exists p_2 \in \mathbb{Q}} \right)\left[ {\delta _1 < p_2 < e} \right]\;\& \;\left( {\exists q_2 \in \mathbb{Q}} \right)\left[ {e < q_2 < \varepsilon _1 } \right]$

If $\displaystyle n>1$
Let $\displaystyle \delta _n = \max \left\{ {p_n ,e - \frac{1}{n+1}} \right\}\;\& \,\varepsilon _n = \min \left\{ {q_n ,e + \frac{1}{n+1}} \right\}$.
$\displaystyle \left( {\exists p_{n+1} \in \mathbb{Q}} \right)\left[ {\delta _n < p_{n+1} < e} \right]\;\& \;\left( {\exists q_{n+1} \in \mathbb{Q}} \right)\left[ {e < q_{n+1} < \varepsilon _n } \right]$

See what you can do with the sequence $\displaystyle I_n=\left[p_n,q_n\right]$.

3. I feel certain that {e} is a set with one element, though the problem doesn't specifically state that is the case. However, our text does use the normal symbol as the empty set, so I imagine with the context of this section added, we are talking about a set with one element. Thank you so much!

4. Originally Posted by tarheelborn
I feel certain that {e} is a set with one element, though the problem doesn't specifically state that is the case. However, our text does use the normal symbol as the empty set, so I imagine with the context of this section added, we are talking about a set with one element. Thank you so much!

5. Now I am really confused... I think that e (the exponential function) is supposed to be contained in each of the I_n's, and it is the only element that is contained in all of the I_n's. But I am not sure how to define a set of closed intervals such that the intersection of those intervals is e. Thank you.

6. So would this work? Let I_1 = {e}, I_2 = {e, 1}. I_3 = {e, 1, 2}, ..., I_n= I_n-1 U {n-1}? Thanks.

7. Originally Posted by tarheelborn
Now I am really confused... I think that e (the exponential function) is supposed to be contained in each of the I_n's, and it is the only element that is contained in all of the I_n's. But I am not sure how to define a set of closed intervals such that the intersection of those intervals is e. Thank you.
Well I have given you a sequence of nested closed intervals with rational endpoints.
The intersection of these intervals is the set $\displaystyle \{e\}$.
But using what I gave you, you still must prove it all.

8. Originally Posted by tarheelborn
So would this work? Let I_1 = {e}, I_2 = {e, 1}. I_3 = {e, 1, 2}, ..., I_n= I_n-1 U {n-1}? Thanks.
Of course NOT.
For one, $\displaystyle I_1=\{e\}$ is not a closed interval, much less with rational endpoints.
The whole point of this question is this: e is not a rational number.
BUT e is the limit of an increasing sequence of rational numbers and the limit of an decreasing sequence of rational numbers.

9. I really appreciate your help and I can almost understand your example, but I believe it is several levels above that of my class. Thanks, though!

10. OK, now something is sinking in. I apologize for my denseness and I do appreciate you.

11. Well I hope so.