Answer
(a) \((1, 2)\)
(b) \((1, 1.25)\)
(c) Not mutually disjoint
(d) \((1, 2)\)
(e) \(\left(1,\; 1 + \tfrac{1}{n}\right)\)
(f) \((1, 2)\)
(g) \(\varnothing\)
Work Step by Step
We are given the family of **open intervals**:
\[
S_i = \left(1,\; 1 + \frac{1}{i}\right), \quad \text{for all positive integers } i.
\]
These intervals all:
- start just above 1 (not including 1),
- have decreasing length as \(i\) increases,
- and "shrink" toward \((1, 1]\), i.e., becoming "smaller right neighborhoods" of 1.
---
## (a) \(\displaystyle \bigcup_{i=1}^{4} S_i\)
The intervals are:
- \(S_1 = (1, 2)\)
- \(S_2 = (1, 1.5)\)
- \(S_3 = (1, 1.\overline{3})\)
- \(S_4 = (1, 1.25)\)
These are nested and increasing toward \((1,2)\), so their union is:
\[
\boxed{\bigcup_{i=1}^4 S_i = (1, 2)}
\]
---
## (b) \(\displaystyle \bigcap_{i=1}^{4} S_i\)
We take the smallest of the intervals:
- right endpoints: \(2, 1.5, 1.\overline{3}, 1.25\)
So intersection is:
\[
\left(1,\; 1.25\right)
\]
\[
\boxed{\bigcap_{i=1}^4 S_i = (1, 1.25)}
\]
---
## (c) Are \(S_1, S_2, S_3, \dots\) mutually disjoint?
**No**.
All intervals start at 1 (excluded) and stretch beyond, with overlapping regions.
For example:
- \(S_1 = (1, 2)\),
- \(S_2 = (1, 1.5)\),
- Their overlap: \((1, 1.5)\)
So they're **not** disjoint.
\[
\boxed{\text{No, they are not mutually disjoint. They all overlap near }1.}
\]
---
## (d) \(\displaystyle \bigcup_{i=1}^{n} S_i\)
Since \(S_1 = (1, 2)\) contains all the rest, the union is:
\[
\boxed{\bigcup_{i=1}^n S_i = (1, 2)}
\]
---
## (e) \(\displaystyle \bigcap_{i=1}^{n} S_i\)
Smallest right endpoint is from \(S_n\):
\[
\boxed{\bigcap_{i=1}^n S_i = \left(1,\; 1 + \frac{1}{n}\right)}
\]
---
## (f) \(\displaystyle \bigcup_{i=1}^{\infty} S_i\)
As \(i \to \infty\), \(1 + \frac{1}{i} \to 1\), so the intervals get tighter near 1, but all are **contained** in \(S_1 = (1,2)\). And collectively, their union is:
\[
\boxed{\bigcup_{i=1}^{\infty} S_i = (1, 2)}
\]
---
## (g) \(\displaystyle \bigcap_{i=1}^{\infty} S_i\)
Now, as \(i \to \infty\), the right endpoint \(1 + \frac{1}{i} \to 1\). So each interval is a smaller open interval approaching 1 from the right.
That means:
- For any \(x > 1\), eventually \(x\) is **not** in \(S_i\) for large enough \(i\).
- So no number lies in **all** of the \(S_i\).
Thus the infinite intersection is empty:
\[
\boxed{\bigcap_{i=1}^\infty S_i = \varnothing}
\]
