数学代写|MATH2400 Mathematical Analysis

Proposition 0.1 (Exercise 1). The middle-thirds Cantor set $\mathcal{C}$ is totally disconnected and perfect.

Proof. First we show that $\mathcal{C}$ is totally disconnected. Let $x, y \in \mathcal{C}$ such that $x \neq y$. Then $|x-y|>0$, so there exists $k \in \mathbb{N}$ such that $(1 / 3)^k<|x-y|$ (because $\left.\lim _k(1 / 3)^k=0\right)$. Since $(1 / 3)^k<|x-y|, x$ and $y$ must be in different subintervals of $C_k$, since the length of each interval in $C_k$ is $(1 / 3)^k$. Thus there is a whole interval of points in $[0,1] \backslash \mathcal{C}$ between $x$ and $y$, so $\mathcal{C}$ is totally disconnected.

Now we show that $\mathcal{C}$ is perfect. Let $x \in \mathcal{C}$. We will show that $x$ is not an isolated point by finding a nearby point in $\mathcal{C}$ in a ball of any radius centered at $x$. Let $r>0$, then choose $k$ large enough that $(1 / 3)^k \leq r$. Since $\mathcal{C}=\cap_j C_j$, we have $x \in C_k$. Since each subinterval of $C_k$ has length $(1 / 3)^k$ and $x$ is in one such subinterval. Let $y$ be an endpoint of that subinterval so that $x \neq y$ (there are two endpoints, so this is always possible). Then we have $x \neq y$ and $|x-y|<(1 / 3)^k<r$, so $y \in B_r(x) \cap \mathcal{C}$. Thus $x$ is not an isolated point, and hence $\mathcal{C}$ is perfect.

Proposition 0.2 (Exercise 2a). Let $\mathcal{C}$ be the middle-thirds Cantor set. Then for $x \in[0,1]$, $x \in \mathcal{C}$ if and only if $x$ has a ternary expansion
$$
x=\sum_{k=1}^{\infty} a_k 3^{-k}
$$
where $a_k \in{0,2}$ for all $k$.
Proof. First, suppose that $x \in[0,1]$ has such a ternary expansion, where each $a_k \in{0,2}$. Since $a_1 \in{0,2}$, we know that $0 \leq x \leq 1 / 3$ or $2 / 3 \leq x \leq 1$, so $x \in C_1$. More generally, we can see by induction that $a_k \in{0,2} \Longrightarrow x \in C_k$. Thus $x \in C_k$ for all $k$, so $x \in \mathcal{C}$.

Now suppose that $x \in \mathcal{C}$. We know $x$ has some ternary expansion $\sum_{k=1}^{\infty} a_k 3^{-k}$ with $a_k \in$ ${0,1,2}$ for all $k$. Consider the remaining intervals after just the first stage of construction, $[0,1 / 3] \cup[2 / 3,1]$. If $x \in[0,1 / 3)$, then $a_1=0$. If $x=1 / 3$, then $x$ also has the ternary expansion $\sum_{k=2}^{\infty} 2 / 3$. If $x \in[2 / 3,1)$, then $a_1=2$, and if $x=1$, then $x$ has the ternary expansion $\sum_{k=1}^{\infty} 2 / 3$.

By induction on the recursive construction, at the $k$ th stage, if $x$ is in the left half of a “split” interval, then $a_k=0$, except possibly at the right endpoint. But at that endpoint, $x$ has a finite ternary expansion terminating with a $1 / 3^k$ term. But this term can be rewritten as a ternary expansion involving only 0 ‘s and 2 ‘s by the following substitution:
$$
\frac{1}{3^k}=\sum_{j=k+1}^{\infty} \frac{2}{3^j}
$$
And for $x$ in the right half of a “split” interval, we get $a_k=2$, except possibly at the right endpoint. But again, if we get a ternary expansion involving a 1 at such an endpoint, it must be a finite expansion terminating in a $1 / 3^k$ term, which we can expand as above. Thus $x$ has the required ternary expansion.

Proposition 0.3 (Exercise 2b). Let $F$ be the Cantor-Lebesgue function on $\mathcal{C}$. $F$ is well defined and continuous, and $F(0)=0$, and $F(1)=1$.

Proof. First we show that $F$ is well defined. To do this, we need to show that for $x=$ $\sum_{k=1}^{\infty} a_k 3^{-k} \in \mathcal{C}$, the value of $F$ converges. By the previous proposition, we know that $x \in \mathcal{C}$ has a ternary expansion where each $a_k$ is 0 or 2 , so if we define $b_k=a_k / 2$, each $b_k$ is 0 or 1 . Then $b_k / 2^k \leq 1 / 2^k$ for each $k$. We know that the geometric series $\sum_{k=1}^{\infty} 1 / 2^k$ converges, so by the comparison test,
$$
F(x)=\sum_{k=1}^{\infty} \frac{b_k}{2^k}
$$
converges as well. Thus $F$ is well defined.
Now we show that $F$ is continuous on $\mathcal{C}$. Let $x_0 \in \mathcal{C}$. To show that $F$ is continuous at $x_0$, we will show that for $\epsilon>0$, there exists $\delta>0$ such that
$$
\left|x-x_0\right|<\delta \Longrightarrow\left|F(x)-F\left(x_0\right)\right|<\epsilon $$ Fix $\epsilon>0$. Then there exists $n \in \mathbb{N}$ such that $2^{-n-1}<\epsilon<2^{-n}$. Let $\delta=3^{-n}$. Let
$$
x=\sum_{k=1}^{\infty} a_k 3^{-k} \quad x_0=\sum_{k=1}^{\infty} c_k 3^{-k}
$$
For convenience, let $b_k=a_k-c_k$. Then
$$
\left|x-x_0\right|=\left|\sum_{k=1}^{\infty} a_k 3^{-k}-\sum_{k=1}^{\infty} c_k 3^{-k}\right|=\left|\sum_{k=1}^{\infty}\left(a_k-c_k\right) 3^{-k}\right|=\left|\sum_{k=1}^{\infty} b_k 3^{-k}\right|
$$
If $\left|x-x_0\right|<\delta=3^{-n}$, then $b_1, b_2, \ldots b_n$ are all zero. Then the first $n$ terms of $\left|F(x)-F\left(x_0\right)\right|$ are also zero:
$$
\begin{aligned}
\left|F(x)-F\left(x_0\right)\right| & =\left|\sum_{k=1}^{\infty}\left(a_k / 2\right) 2^{-k}-\sum_{k=1}^{\infty}\left(c_k / 2\right) 2^{-k}\right|=\left|\sum_{k=1}^{\infty}\left(a_k-c_k\right) 2^{-k-1}\right| \
& =\left|\sum_{k=1}^{\infty} b_k 2^{-k-1}\right|=\left|\sum_{k=n+1}^{\infty} b_k 2^{-k-1}\right| \leq 2^{-n-1}<\epsilon
\end{aligned}
$$

Thus $F$ is continuous at $x_0$.
Now we show that $F(0)=0$ and $F(1)=1$. In the ternary expansion for zero, each $a_k=0$. In the ternary expansion of 1 , each $a_k=2$. Then
$$
\begin{aligned}
& 0=\sum_{k=1}^{\infty}(0) 3^{-k} \leadsto F(0)=\sum_{k=1}^{\infty}(0) 2^{-k}=\sum_{k=1}^{\infty} 0=0 \
& 1=\sum_{k=1}^{\infty}(2) 3^{-k} \leadsto F(1)=\sum_{k=1}^{\infty}(1) 2^{-k}=\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k=1
\end{aligned}
$$
since both are geometric series.