Answer
(a) \([1, 2]\)
(b) \([1, 1.25]\)
(c) Not mutually disjoint
(d) \([1, 2]\)
(e) \([1, 1+\tfrac{1}{n}]\)
(f) \([1, 2]\)
(g) \(\{1\}\)
Work Step by Step
We are given a family of **closed intervals**:
\[
R_i = \left[1,\; 1 + \frac{1}{i} \right], \quad \text{for all positive integers } i.
\]
As \(i\) increases, the length of the interval \(R_i\) shrinks, approaching \([1,1]\), but always contains 1 and is entirely within \([1,2]\).
Let’s now answer each part.
---
## **(a)** \(\displaystyle \bigcup_{i=1}^{4} R_i\)
We compute:
- \(R_1 = [1, 2]\)
- \(R_2 = [1, 1.5]\)
- \(R_3 = [1, 1.\overline{3}]\)
- \(R_4 = [1, 1.25]\)
All of these are **contained in** \(R_1 = [1,2]\), so the union is:
\[
\boxed{\bigcup_{i=1}^{4} R_i = [1, 2]}
\]
---
## **(b)** \(\displaystyle \bigcap_{i=1}^{4} R_i\)
The intersection is the overlap of all four intervals. That is, the interval from:
- left endpoint: always \(1\), and
- right endpoint: the **smallest** among \(1+\frac{1}{i}\)
So the smallest right endpoint is:
\[
1 + \frac{1}{4} = 1.25
\]
Thus the intersection is:
\[
\boxed{\bigcap_{i=1}^{4} R_i = [1, 1.25]}
\]
---
## **(c)** Are \(R_1, R_2, R_3, \dots\) mutually disjoint?
**No**.
They all **include the point 1** (the left endpoint). So they **overlap**, and are **not** mutually disjoint.
\[
\boxed{\text{No, they are not mutually disjoint. All intervals share the point }1.}
\]
---
## **(d)** \(\displaystyle \bigcup_{i=1}^{n} R_i\)
The largest interval among \(R_1, \dots, R_n\) is \(R_1 = [1, 2]\), and all others are subsets of it.
So:
\[
\boxed{\bigcup_{i=1}^{n} R_i = [1, 2]}
\]
---
## **(e)** \(\displaystyle \bigcap_{i=1}^{n} R_i\)
Each interval is of the form \([1,\;1+\frac{1}{i}]\), so the smallest right endpoint is from \(R_n\).
Thus:
\[
\boxed{\bigcap_{i=1}^{n} R_i = \left[1,\; 1 + \frac{1}{n} \right]}
\]
---
## **(f)** \(\displaystyle \bigcup_{i=1}^{\infty} R_i\)
As \(i \to \infty\), the right endpoint \(1 + \frac{1}{i} \to 1\), but **each** \(R_i\) extends a little beyond 1.
So collectively the union includes:
\[
[1, 2)
\quad\text{(not including 2, but approaching it via } R_1 = [1,2])
\]
**Note**: 2 is only included in \(R_1\), so it *is* in the union.
So the union is exactly:
\[
\boxed{\bigcup_{i=1}^{\infty} R_i = [1, 2]}
\]
---
## **(g)** \(\displaystyle \bigcap_{i=1}^{\infty} R_i\)
As \(i\to\infty\), \(1 + \frac{1}{i} \to 1\), so the right endpoints shrink toward 1. That means the infinite intersection is just:
\[
\boxed{\bigcap_{i=1}^{\infty} R_i = \{1\}}
\]
Because every \(R_i\) contains 1, but the right ends get smaller and smaller.
