数学代写|MATH2400 Mathematical Analysis

Exercise $4 \mathrm{~b})$. If $x \in \hat{\mathcal{C}}$, then there exists a sequence $\left(x_n\right_{n=1}^{\infty}$ such that $x_n \notin \hat{\mathcal{C}}$ and $x_n \rightarrow x$ and $x_n \in I_n$ where $I_n$ is a subinterval in the complement of $\hat{\mathcal{C}}$ with $\left|I_n\right| \rightarrow 0$

Proof. Let $x \in \hat{\mathcal{C}}$. At each stage of the construction of $\hat{\mathcal{C}}$, there remain $2^{-1}$ disjoint closed intervals, and $x$ is in exactly one of these intervals. At the $k$ th stage, that closed interval has length
$$
\frac{1-\sum_{j=1}^k l_k 2^{k-1}}{2^{k-1}}
$$
since the numerator is the total length of remaining intervals, and the denominator is the number of remaining intervals. When we remove the open middle interval of length $l_k$ from that closed interval, let $x_k$ be the midpoint of that interval. Then
$$
\left|x_k-x\right| \leq \frac{1-\sum_{j=1}^k l_k 2^{k-1}}{2^{k-1}} \leq \frac{1}{2^{k-1}}
$$
Thus $x_k \rightarrow x$, and by construction $x_k \notin \hat{\mathcal{C}}$. We let $I_k$ be the interval of which $x_k$ is the midpoint, so $I_k \subset(\hat{\mathcal{C}})^c$. We know that $\left|I_k\right| \rightarrow 0$ since $\left|I_k\right|=l_k$ and by definition of $\hat{\mathcal{C}}$, we have
$$
\sum_{j=1}^k l_j 2^{j-1}<1
$$
for all $k$.

(Exercise 4c). $\hat{\mathcal{C}}$ is perfect, and contains no open interval.
Proof. Let $x \in \hat{\mathcal{C}}$. By part (b), there exists a sequence $\left(x_n\right)_{n=1}^{\infty}$ with $x_n \notin \hat{\mathcal{C}}$ and $x_n \rightarrow x$. Thus, for every $r>0$, there exists $N \in \mathbb{N}$ such that $n \geq N \Longrightarrow x_n \in B(x, r)$. Hence $x$ is not an isolated point, so $\hat{\mathcal{C}}$ has no isolated points. It is clear that $\hat{\mathcal{C}}$ is closed, since its complement is a union of open intervals by construction. Thus $\hat{\mathcal{C}}$ is perfect.

Now we show that $\hat{\mathcal{C}}$ contains no open interval. Suppose that $\hat{\mathcal{C}}$ contains an open interval $(a, b)$, and let $x \in(a, b)$. Then by part (b) we have a sequence $\left(x_n\right)$ with $x_n \notin \hat{\mathcal{C}}$ and $x_n \rightarrow x$. Let $r=\min (|x-a|,|x-b|)$. Then there exists $N \in \mathbb{N}$ such that $n \geq N$ implies $x_n \in B(x, r)$. But by construction of $r, B(x, r) \subset(a, b)$, so for $n \geq N, x_n \in(a, b)$. This is a contradiction, since $x_n \notin \hat{\mathcal{C}}$ and $(a, b) \subset \hat{\mathcal{C}}$. Thus $\hat{\mathcal{C}}$ contains no open interval.

(Exercise $4 \mathrm{~d}$ ). $\hat{\mathcal{C}}$ is uncountable.
Proof. Let $x \in \hat{\mathcal{C}}$. At the $k$ th stage of construction of $\hat{\mathcal{C}}$, we separate each closed subinterval into two subintervals. We define a funtion $f: \hat{\mathcal{C}} \rightarrow[0,1]$ by
$$
f(x)=\sum_{k=1}^{\infty} \frac{a_k}{2^k}
$$
where $a_k=0$ if $x$ is in a left subinterval at stage $k$, and $a_k=1$ if $x$ is in a right subinterval at stage $k$. We claim that $f$ is surjective. To see this, let $y \in[0,1]$. Then $y$ has a (not necessarily unique) binary expansion
$$
y=\sum_{k=1}^{\infty} \frac{b_k}{2^k}
$$
Then we use the $b_k$ to choose a subinterval remaining at each stage of the construction of $\hat{\mathcal{C}}$. Let $I_k$ be this chosen subinterval. For example, if $b_1=0$, we choose the left of the two subintervals left after the first removal of a middle interval of length $l_1$. Now we take the intersection over all such intervals, $\cap_{k=1}^{\infty} I_k$, and we know that this intersection is non-empty, since the endpoints of each $I_k$ are never removed. If we choose some $x \in \cap_k I_k$, then $f(x)=y$ by construction. Hence $f$ is surjective.

Since there is a surjective function $\hat{\mathcal{C}} \mapsto[0,1]$, we know that $\hat{\mathcal{C}}$ cannot have smaller cardinality than $[0,1]$. Hence $\hat{\mathcal{C}}$ is uncountable.

Proposition 0.12 (Exercise $5 \mathrm{~b}$ ). The above proposition need not hold when $E$ is closed and unbounded, or when $E$ is open and bounded.

Proof. First, we let $E$ be the closed and unbounded set $\mathbb{Q}$ lying in $\mathbb{R}$. We know that $m(\mathbb{Q})=0$. (One way to see this is by using Theorem 3.2 on the collection of singleton sets of rationals.) However, $\mathcal{O}_n={x: d(x, \mathbb{Q})<1 / n}$ is $\mathbb{R}$ for all $n$, since no matter how small $1 / n$, every real number is always within $1 / n$ of some rational. But $m(\mathbb{R})=\infty$, so $\lim _n m\left(\mathcal{O}_n\right)=\infty$

Now we construct an open, bounded set that doesn’t satisfy the above property. Let $\epsilon>0$. Consider the rationals in $[0,1]$, and order them $q_1, q_2, \ldots$. Around $q_k$, form an open ball of radius $2^{-k} \epsilon$. Then let
$$
E=\bigcup_{k=1}^{\infty} B\left(q_k, 2^{-k} \epsilon\right)
$$
$E$ is open, since it is a union of open sets, and $E$ is bounded within $[-1 / 2,3 / 2]$. We have an upper bound on the measure of $E$ by $m(E) \leq \sum_{k=1}^{\infty} 2^{-k+1} \epsilon=2 \epsilon$. Now consider $\mathcal{O}n$. For each $k$, we can see that $B\left(q_k, 2^{-k} \epsilon+1 / n\right) \subset \mathcal{O}_n$, since everything in $B\left(q_k, 2^{-k} \epsilon+1 / n\right)$ is within $1 / n$ of $B\left(q_k, 2^{-k} \epsilon\right)$. But for any $x \in[0,1]$, there is a rational within $1 / n$ of $x$, so $$ [0,1] \subset \bigcup{k=1}^{\infty} B\left(q_k, 2^{-k} \epsilon+1 / n\right)
$$
Thus $m\left(\mathcal{O}_n\right) \geq 1$ for all $n$. Hence $\lim _n m\left(\mathcal{O}_n\right) \geq 1$. So for any $\epsilon<1 / 2$, we have $m(E) \leq 1 / 2$ but $\lim _n m\left(\mathcal{O}_n\right) \geq 1$.