Answer
a. \(\displaystyle \bigcup_{i=0}^4 D_i = [-4,4].\)
b. \(\displaystyle \bigcap_{i=0}^4 D_i = \{\,0\}.\)
c. \(D_0, D_1, D_2,\dots\) are **not** mutually disjoint (they all share 0).
d. \(\displaystyle \bigcup_{i=0}^n D_i = [-n,n].\)
e. \(\displaystyle \bigcap_{i=0}^n D_i = \{\,0\}.\)
f. \(\displaystyle \bigcup_{i=0}^\infty D_i = \mathbb{R}.\)
g. \(\displaystyle \bigcap_{i=0}^\infty D_i = \{\,0\}.\)
Work Step by Step
We have a family of sets
\[
D_i \;=\;\{\,x\in\mathbb{R}: -i \le x \le i\,\}
\quad\text{for nonnegative integers }i.
\]
Concretely:
- \(D_0 = \{\,0\}\).
- \(D_1 = [-1,1]\).
- \(D_2 = [-2,2]\).
- … and so on, growing as \(i\) increases.
We want to analyze various unions and intersections of these sets.
---
## (a) \(\displaystyle \bigcup_{i=0}^4 D_i\)
Since
\[
D_0 \;\subset\; D_1 \;\subset\; D_2 \;\subset\; D_3 \;\subset\; D_4,
\]
the union over \(i=0,1,2,3,4\) is simply the largest of these intervals, namely
\[
D_4 = [-4,4].
\]
Hence
\[
\boxed{\bigcup_{i=0}^4 D_i = [-4,4].}
\]
---
## (b) \(\displaystyle \bigcap_{i=0}^4 D_i\)
Again using
\[
D_0 = \{\,0\},\quad D_1=[-1,1],\quad D_2=[-2,2],\;\dots
\]
we see that \(D_0\subset D_1\subset D_2 \dots\), so the **smallest** set among them is \(D_0=\{\,0\}\). Since an intersection is contained in each set, and \(D_0\) is already just \(\{0\}\), the intersection must be \(\{0\}\). Concretely,
- \(D_0 \cap D_1 = \{0\}\cap[-1,1]=\{0\}\).
- Intersecting with larger intervals will not remove 0, so it stays \(\{0\}\).
Thus
\[
\boxed{\bigcap_{i=0}^4 D_i = \{\,0\}.}
\]
---
## (c) Are \(D_0, D_1, D_2, \dots\) **mutually disjoint**?
**No.** “Mutually disjoint” would mean that no two distinct \(D_i\) share a point. But in fact,
- \(0\in D_0\)
- \(0\in D_1\) (since \([-1,1]\) includes 0)
- \(0\in D_2\), etc.
All these sets share at least the point \(0\). Hence they are **not** mutually disjoint.
---
## (d) \(\displaystyle \bigcup_{i=0}^n D_i\)
For a fixed \(n\),
\[
D_0 \;\subset\; D_1 \;\subset\; \dots \;\subset\; D_n,
\]
so the union over \(i=0\) to \(n\) is simply \(D_n\), i.e.
\[
\boxed{\bigcup_{i=0}^n D_i = [-n,\,n].}
\]
---
## (e) \(\displaystyle \bigcap_{i=0}^n D_i\)
Similarly, among \(D_0,\dots,D_n\), the smallest set is \(D_0=\{0\}\). Hence their intersection is again just
\[
\boxed{\bigcap_{i=0}^n D_i = \{\,0\}.}
\]
---
## (f) \(\displaystyle \bigcup_{i=0}^\infty D_i\)
As \(i\) grows, \([-i,i]\) expands to cover arbitrarily large intervals on the real line. In the limit, the union of all these intervals is
\[
\bigcup_{i=0}^\infty [-i,i] \;=\; \mathbb{R}.
\]
Formally, for any real \(x\), you can pick \(i\) large enough so that \(-i \le x \le i\). Thus \(x\in D_i\) for some (indeed infinitely many) \(i\). Therefore
\[
\boxed{\bigcup_{i=0}^\infty D_i = \mathbb{R}.}
\]
---
## (g) \(\displaystyle \bigcap_{i=0}^\infty D_i\)
Finally, we look at the infinite intersection. Because \(D_0 = \{0\}\) is part of the family, any element in the intersection must be in \(\{0\}\). That forces the intersection to be
\[
\boxed{\bigcap_{i=0}^\infty D_i = \{\,0\}.}
\]
*(If we had started from \(i=1\) instead of \(i=0\), the intersection would be \(\bigcap_{i=1}^\infty [-i,i] = \bigcap_{i=1}^\infty [-i,i] = \{x\in\mathbb{R}: -i\le x\le i \text{ for all }i\ge1\} = \{\,0\}\) anyway, but more obviously the presence of \(D_0\) = \(\{0\}\) forces everything to collapse to \(\{0\}\) immediately.)*
