Indexed family of sets defined by a half-open interval

Question

Suppose $A = \{A_n : n \in\mathbb N\}$ is an indexed family of sets with N as the index set, where $A_n$ is the half-open interval $\left[\frac1n, 2 − \frac1n\right)$ for each $n \in\mathbb N$.

(a) Write $A_1$, $A_2$ and $A_3$.

(b) Find the intersection of $A_n$

(c) Find the union of $A_n$

I'm attempting to understand this, but the half-open interval is throwing me off. For (a) I have $A_1$ = $\{1\}$, $A_2$ = $\left[\frac12, \frac32\right]$and $A_3$ = $\left[\frac13, \frac53\right]$ and I believe the answer to (b) and (c) on a closed interval would be $[1]$ and $(0, 2)$ respectively, but it's the first week of class and the text only offers a paragraph on the matter. Does the interval being half-open have any effect on $A_1$, $A_2$, and $A_3$, and how does it change the intersection and union.

Answer 1

The fact that the intervals are half open does affect the result. Your answers for $A_1, A_2$ and $A_3$ are not correct.
The fact that the interval in open at right means that the right endpoint does not belong to the set. For instance $A_1=\emptyset$ since by tge very definition $A_n=\{x\in \mathbb{R} | \frac{1}{n} \leq x < 2-\frac{1}{n}\}$ and of course there are no points which are both greater or equal than $1$ and strictly smaller than $1$. You must correct $A_2$ and $A_3$ also.

Answer 2

obviously $$A_1\subset A_2\subset A_3\subset...$$therefore $$\bigcap_{n=1}^{\infty}A_n=A_1$$For finding the union first note that any $\epsilon>0$ belongs to set $\{A_N,A_{N+1},A_{N+2},...\}$ where $N>\dfrac{1}{\epsilon}$. With the same argument for sufficiently small positive $\epsilon$, $2-\epsilon$ belongs to $\{A_N,A_{N+1},A_{N+2},...\}$ for some $N$. This is clear that $\{0,2\}$ doesn't belong to any $A_n$, therefore $$\bigcup_{n=1}^{\infty}A_n=(0,2)$$