Need help in vetting my answers for the question #6 in CRM series book by MAA: Exploratory Examples for Real Analysis, By Joanne E. Snow, Kirk E. Weller. 
Consider the following two collections of sets:
$$\{I_n = [0,\frac1n] \ : n \in \mathbb{N} \}$$ $$\{J_n = (0,\frac1n) \ : n \in \mathbb{N} \}$$(a) Compute $\bigcap_{n=1}^{\infty} I_n$.
(b) Compute $\bigcap_{n=1}^{\infty} J_n$.
(c) Can you explain the differences in the two answers using any of the terms introduced in this lab (in chapter #1)?
(a) $I_1 = [0,1], I_2 = [0,\frac12], I_3 = [0,\frac13], I_4 = [0,\frac14], \cdots, I_{\infty} = [0,0] = 0$
$\bigcap_{n=1}^{\infty} I_n = [0,1]\cap[0,\frac12]\cap[0,\frac13]\cap[0,\frac14]\cdots \cap 0$
There need be found intersection of real points in all the given intervals, leading to finally single point $0$, which is the same as $I_{\infty}$.
(b) $J_1 = (0,1), J_2 = (0,\frac12), J_3 = (0,\frac13), J_4 = (0,\frac14), \cdots, J_{\infty} = (0,0)=\emptyset$
The last set is empty set ($\emptyset$), as .
$\bigcap_{n=1}^{\infty} J_n = (0,1)\cap(0,\frac12)\cap(0,\frac13)\cap(0,\frac14)\cdots \cap \emptyset $
$\bigcap_{n=1}^{\infty} J_n = \emptyset $
In all non-null sets, the boundary of the intervals are not included.
The intersection of the sets is null set, as there is nothing in common of any set with $\emptyset$.
(c) Unable to find any term in chapter 1 of the book, that describes something related to empty set intersection with other sets leading to null set again. The chapter describes supremum, maximum, Upper bound, infimum, minimum, Lower bound.
This can be checked by the google-book link for the same.
Doubt:
1. Am I correct about taking the interval $I_\infty$, that represents a single point?
2. Similarly, about $J_\infty$, which is empty set.
I don't like $I_\infty$ or $J_\infty$. It's not that your final answer is wrong, but these "last sets" in the collection are not defined in the question. You've invented them using your intuition. Your intuition happens to be consistent with the question, but you may not be so lucky next time.
(As an example of when your intuition might fail you, note that $\bigcap_{i=1}^\infty (-\frac{1}{n}, \frac{1}{n}) \neq (0, 0) = \emptyset$.)
Instead, use your intuition to conjecture that $\bigcap_{n=1}^\infty I_n = \{0\}$. To prove it, we do a classic subset proof. Start by supposing $x \in \{0\}$, i.e. $x = 0$. Then, show that $x \in \bigcap_{n=1}^\infty I_n$. By definition of intersection, this is equivalent to showing $x \in I_n$ for all $n$, which is true, since $0 \le 0 \le \frac{1}{n}$.
Conversely, suppose $x \in \bigcap_{n=1}^\infty I_n$, which is to say that $x \in I_n$ for all $n$, and hence further that $0 \le x \le \frac{1}{n}$. If we consider the constant sequence $x, x, x, \ldots$, then this means this constant sequence is squeezed between $0$ and $\frac{1}{n} \to 0$, hence the constant sequence tends to $0$. But, the constant sequence tends to $x$ as well, hence by uniqueness of limits, $x = 0$. So, $\bigcap_{n=1}^\infty I_n \subseteq \{0\}$.
Putting these two arguments together, we have proven that $\bigcap_{n=1}^\infty I_n = \{0\}$.
Try something similar with the $J_i$s.