## Math444 Real Analysis课程简介

That sounds like a challenging and interesting course! $\varepsilon-\delta$ analysis is a crucial foundation for understanding calculus and real analysis, and it’s great that this course will help make the concepts more rigorous.

I’m glad to see that the course covers a range of topics, including the real number system, limits, continuity, derivatives, and integrals. These concepts are fundamental to calculus and real analysis, and it’s important to have a solid understanding of them.

It’s also great to see that the course covers both the Darboux and Riemann integrals. Many calculus courses only cover the Riemann integral, so it’s nice to see that students will be exposed to different approaches to integration.

Overall, it seems like Math 444 will be a challenging and rewarding course for students who are interested in mathematics and want to deepen their understanding of calculus and real analysis.

## PREREQUISITES

Exams: This course has three 90-minute midterm tests and a 3-hour final exam.
Proctorship Information: Exams in this course may be taken online.
Students with a course start date prior to January 1st, 2023 should use proctor information on this page to take their exams.

Students with a course start date on or after January 1st, 2023 should refer to this page for exam taking information.
Once registered, students will find relevant exam taking information and all other course requirements within their Nexus student dashboard.

## Math444 Real Analysis HELP（EXAM HELP， ONLINE TUTOR）

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.

The proof looks good! Here’s a slightly more detailed explanation of each step:

To show that $\mathcal{C}$ is totally disconnected, we take any two distinct points $x$ and $y$ in $\mathcal{C}$. Since $\mathcal{C}$ is the intersection of the nested intervals $C_1 \supseteq C_2 \supseteq C_3 \supseteq \cdots$, there exists some index $k$ such that $x$ and $y$ belong to different subintervals of $C_k$. Each subinterval of $C_k$ has length $(1/3)^k$, so the distance between $x$ and $y$ is greater than $(1/3)^k$. Therefore, there exists an interval in $[0,1] \setminus \mathcal{C}$ that contains both $x$ and $y$, showing that $\mathcal{C}$ is totally disconnected.

To show that $\mathcal{C}$ is perfect, we take any point $x$ in $\mathcal{C}$ and any radius $r>0$. We choose an index $k$ such that $(1/3)^k \leq r$. Since $\mathcal{C}=\cap_j C_j$, we have $x \in C_k$, so $x$ belongs to one of the subintervals of $C_k$. Let $y$ be an endpoint of that subinterval. Then $y$ also belongs to $\mathcal{C}$ because it is a point in $[0,1]$ that is not in any of the removed intervals $I_j$. Moreover, we have $|x-y|<(1/3)^k \leq r$, so $y$ is in the open ball $B_r(x)$. Therefore, $x$ is not an isolated point of $\mathcal{C}$, so $\mathcal{C}$ 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 $x \in \mathcal{C}$.

To prove the converse, suppose that $x \in \mathcal{C}$ and let $x \in C_k$ for some $k$. Then $x$ belongs to one of the $2^k$ subintervals of $C_k$, each of which has length $(1/3)^k$. Let $a_k$ be the binary digit that tells us which subinterval $x$ belongs to: if $x$ is in the leftmost subinterval, let $a_k=0$, and if $x$ is in the rightmost subinterval, let $a_k=2$. By repeating this argument for $C_{k-1}, C_{k-2}, \ldots, C_1$, we obtain a sequence of binary digits $(a_1, a_2, a_3, \ldots)$ that determines $x$ uniquely.

We need to show that this sequence of digits corresponds to the ternary expansion of $x$, i.e., that $x=\sum_{k=1}^\infty a_k 3^{-k}$. To do this, note that $C_k$ consists of $2^k$ disjoint subintervals of length $(1/3)^k$. Moreover, $C_{k+1}$ is obtained from $C_k$ by removing the middle third of each subinterval. Therefore, each subinterval of $C_{k+1}$ is the union of two subintervals of $C_k$, one of which is entirely contained in the left third of the subinterval and the other of which is entirely contained in the right third. The two subintervals have the same length, so we can assign them binary digits based on whether they are in the left or right third. Specifically, if a subinterval of $C_{k+1}$ is obtained by removing the middle third of a subinterval with binary digit $b_k$, then the left and right subintervals have binary digits $2b_k$ and $2b_k+2$, respectively.

Using this observation, we can compute the ternary expansion of $x$ as follows. Since $x \in C_k$, we know that $x$ belongs to a subinterval of $C_k$ with binary digit $a_k$. Let $I_0=[0,1]$ and let $I_1$ be the subinterval of $I_0$ with binary digit $a_1$. Then $I_1$ is a subinterval of length $1/3$ that contains $x$. Let $I_2$ be the subinterval of $I_1$ that contains $x$ and has binary digit $a_2$. Then $I_2$ is a subinterval of length $1/9$ that contains $x$. Continuing in this way, we obtain a sequence of nested closed intervals $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ such that $x \in I_n$ for all $n$. Moreover, the length of $I_n$ is $(1/3)^n$, so $I_n$ converges to the single point $x$. Since the ternary expansion of $x$ is the unique sequence of digits that corresponds to the nested sequence of intervals $(I_n)$, we conclude that $x=\sum_{k=1}^\infty a_k 3^{-k}$.

## Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.