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1

Show that the collection of all finite strings of letters A-Z is countable, but the collection of all infinite (i.e. unending) sequences of letters is uncountable.

Take $\mathcal S$ to be the collection of all finite strings of A-Z. We can write this as $\mathcal S = \bigcup_{k=1}^{\infty} \left\{ A, \dots, Z \right\}^k$. Since $\mid A-Z \mid < \mid \mathbb N \mid$, we can define a one-to-one function $f: A-Z \rightarrow \mathbb N$. Since $\mathcal S \rightarrow \bigcup_{k=1}^\infty \mathbb N^k$, we know that $\mathcal S$ is countable.

Now consider $\mathcal S_\infty$ to be the collection of infinitely long sequences of strings of A-Z. We have shown that $T = \left\{ 0,1 \right\}^{\infty}$ is uncountable and $\mathcal S_\infty \subset T$. Thus, $\mathcal S_\infty$ must be uncountable. p

2

Let $A_n$ be a sequence of subsets of $\Omega$. Show that $\lim \inf A_n = \left\{ \omega: \lim_{n\rightarrow \infty} \mathbb I_{A_n}(\omega) = 1 \right\}$. What is $\left\{ \omega: \lim_{n\rightarrow \infty} \mathbb I_{A_n}(\omega) = 0 \right\}$? Also characterize the set $\left\{ \omega: \mathbb I_{A_n}(\omega) \text{ does not converge} \right\}$.

Proof

Take $\omega \in \lim \inf A_n$. By definition, $\omega \in \bigcup_{n=1}^{\infty} \left[ \bigcap_{m=n}^\infty A_m \right]$. Take $B_n = \bigcap_{m=n}^\infty A_m$. Notice that as $n$ increases, so does $B_n$. Then, there exists $n_0$, $\omega \in B_{n_0}$. Then, since $B_n$ is increasing, for all $l \geq n_0$ we know $\omega \in B_l$, or $ \mathbb I_{B_l}(\omega) = 1$. Thus, $\omega \in \left\{ \lim_{n\rightarrow \infty} \mathbb{I}_{A_n}(\omega = 1) \right\}$. So, $\lim \inf A_n \subset \left\{ \omega: \lim_{n\rightarrow \infty} \mathbb I_{A_n}(\omega) = 1 \right\}$.

Now take $\omega \in \left\{ \omega: \lim_{n\rightarrow \infty} \mathbb{I}_{A_n}(\omega) = 1 \right\}$. Thus, ther exists some $n_0$ such that for $l > n_0$, $\omega\in A_n$. Then,

\[\omega \in \bigcap_{m = n_0}^\infty A_m \Rightarrow \omega \in \bigcup_{n=1}^\infty \bigcap_{m = n_0}^\infty A_n = \lim \inf A_n.\]

So, $\lim \inf A_n \supset \left\{ \omega: \lim_{n\rightarrow \infty} \mathbb I_{A_n}(\omega) = 1 \right\}$.

Thus, $\lim \inf A_n = \left\{ \omega: \lim_{n\rightarrow \infty} \mathbb I_{A_n}(\omega) = 1 \right\}$.

Notice that $\left\{ \omega: \lim_{n\rightarrow \infty} \mathbb I_{A_n}(\omega) = 0 \right\}$ means taht at some stage onward, $\omega$ is not in any $A_n$. That is, at some point $\omega \in A_n^C$. By the same argument as above,

\[\left\{ \omega: \lim_{n\rightarrow \infty} \mathbb I\_{A\_n}(\omega) = 0 \right\} = \lim \inf A_n^C.\]

If $\left\{ \omega: \mathbb I_{A_n}(\omega) \text{ does not converge} \right\}$, then $\omega \in A_n$ infinitely many times AND $\omega \in A_n^C$ infinitely many times. That is,

\[\omega \in \lim \sup A_n \cap \lim \sup A_n^C.\]

3

Let $\Omega = \left\{ 1,2,3,4,5,6 \right\}$. Find the smallest $\sigma$-field containing the class $\left\{ \left\{ 1,2,3\right\}, \left\{ 2,5,6\right\} \left\{ 3,6\right\} \right\}$.

We can start by creating a partition by taking intersections of the given sets and their complements.

\[\begin{align} \text{set} & & \text{ complement} \\ \left\{ 1,2,3 \right\} & & \left\{ 4,5,6 \right\} \\ \left\{ 2,5,6 \right\} & & \left\{ 1,3,4 \right\} \\ \left\{ 3,6 \right\} & & \left\{ 1,2,4,5 \right\} \end{align}\]

Take $C_1 = \left\{ 1,2,3\right\}$, $C_2 = \left\{ 2,5,6\right\}$, and $\left\{ 3,6\right\}$. Our partition is

\[\begin{align} \mathcal P = C_1^C \cap C_2^C \cap C_3^C & = \left\{ 4 \right\} \\ C_1^C \cap C_2^C \cap C_3 & = \left\{ 5 \right\}\\ C_1^C \cap C_2 \cap C_3^C & = \varnothing\\ C_1^C \cap C_2 \cap C_3 & = \left\{ 1 \right\}\\ C_1 \cap C_2^C \cap C_3^C & = \left\{ 6 \right\}\\ C_1 \cap C_2^C \cap C_3 & = \left\{ 3 \right\}\\ C_1 \cap C_2 \cap C_3^C & = \left\{ 2 \right\}\\ C_1 \cap C_2 \cap C_3 & = \varnothing \end{align}\]

The partitions contains all singletons of elements in $\Omega$. From this, we can generate all 64 element of $\Omega$ by unioning them together. So the smallest $\sigma$-field is the whole power set, $\mathcal P(\Omega)$.

4

Let $\Omega = \left\{ 1,2,3 \right\}$, $\mathcal F_1 = \left\{ \varnothing, \left\{ 1 \right\}, \left\{ 2,3, \right\}, \Omega \right\}$, $\left\{ \varnothing, \left\{ 1,2 \right\}, \left\{ 3 \right\}, \Omega \right\}$. Show that both $\mathcal F_1$ and $\mathcal F_2$ are $\sigma$-fields, but $\mathcal F_1 \cup \mathcal F_2$ is not a $\sigma$-field.

Let’s start with $\mathcal F_1$.

1. $\varnothing, \Omega \in \mathcal F_1$ ✅
2. $\left\{ 1 \right\} \cup \left\{ 2,3 \right\} = \Omega \in \mathcal F_1$ ✅
3. • $\varnothing^C = \Omega \in \mathcal F_1$
   • $\left\{ 1 \right\}^C = \left\{ 2,3 \right\} \in \mathcal F_1$
   • $\left\{ 2,3 \right\}^C = \left\{ 1 \right\} \in \mathcal F_1$
   • $\Omega^C = \varnothing \in \mathcal F_1$ ✅

Thus, $\mathcal F_1$ is a $\sigma$-field. Now we will look at $\mathcal F_2$

1. $\varnothing, \Omega \in \mathcal F_2$ ✅
2. $\left\{ 1,2 \right\} \cup \left\{ 3 \right\} = \Omega \in \mathcal F_2$ ✅
3. • $\varnothing^C = \Omega \in \mathcal F_2$
   • $\left\{ 1,2 \right\}^C = \left\{ 3 \right\} \in \mathcal F_2$
   • $\left\{ 3 \right\}^C = \left\{ 1,2\right\} \in \mathcal F_2$
   • $\Omega^C = \varnothing \in \mathcal F_2$ ✅

Thus, $\mathcal F_2$ is a $\sigma$-field. Now let’s look at $\mathcal F_1 \cup \mathcal F_2$.

\[\mathcal F_1 \cup \mathcal F_2 = \left\{ \varnothing, \left\{ 1 \right\}, \left\{ 3 \right\}, \left\{ 2,3 \right\}, \left\{ 1,2 \right\}, \Omega \right\}\]

1. $\varnothing, \Omega \in \mathcal F_1 \cup \mathcal F_2$ ✅
2. $\left\{ 1 \right\} \cup \left\{ 3 \right\} = \left\{ 1,3 \right\} \notin \mathcal F_1 \cup \mathcal F_2$ ❌

Since the union of two elements of $\mathcal F_1 \cup \mathcal F_2$ is not contained in $\mathcal F_1 \cup \mathcal F_2$, it is not a $\sigma$-field.

5

Call a subset $A$ of $\mathbb{R}$ periodic is $x \in A$ and $n \in \mathbb N$ implies that $x \pm n \in A$. Show that the class of all periodic sets forms a $\sigma$-field.

Take $\mathcal A$ to be the class of all periodic sets.

Notice that if $A = \Omega$ then for any $x \in A$, $x \pm n \in \mathcal A$. If $A = \varnothing$, then $A$ is trivially periodic.

Take $A_1, A_2, \dots, \in \mathcal A$ and $x \in \bigcup_{i=1}^{\infty} A_i$. So, $x \in A_k$ for some $k$. Then $x\pm n \in\bigcup_{i=1}^{\infty} A_i $ for $n \in \mathbb N$. Thus, $\bigcup_{i=1}^{\infty} A_i$ is periodic.

Take $z \in A^C$ for periodic $A$. For sake of contradiction, assume $z \pm n \notin A^C$. Then $z \pm n \in A$. But this would mean that $(z \pm n) \mp n = z \in A$. This contradicts $z \in A^C$, so $A^C$ must be periodic.

Thus, $\mathcal A$ is a $\sigma$-field.