Anyone with a reasonably decent proof-based calculus or linear algebra course (or the equivalent in self-study) would be completely prepared for Rudin. To see that the inclusion can be proper, let $X = (0,1)$ seen as a subspace of $\mathbb R$, and $f$ the inclusion into $\mathbb R$. See for example how he shows that the set of rationals whose square is less than 2 does not have a least upper bound in Q. Let $g(x)=f(x)$ for $x\in E$ and $g(x)=g_i(x)$ for $x\in(a_i,b_i)$. Get ready to fill in lots of details in comments, lemmas and theorems throughout the text. This makes $f$ continuous. Especially in Papa Rudin, there's a ton of arguments that are absolutely beautiful and clever, but entirely non-obvious. You just gave me an idea for an illustrated analysis book, perhaps in Dr. Seuss style. Otherwise, on $[a_i,b_i]$, let $g_i$ be the linear function $$g_i(x)=\frac{f(b_i)-f(a_i)}{b_i-a_i}x+\frac{f(a_i)b_i-f(b_i)a_i}{b_i-a_i}.$$ Note that $g_i(a_i)=f(a_i)$, $g_i(b_i)=f(b_i)$, and the values of $g_i$ lie between $f(a_i)$ and $f(b_i)$ on $(a_i,b_i)$. Unless your class is using this book, don't go through it on your own. Since $E$ is compact, so is $\Gamma(E)$, which is the graph of $F$. Out of curiosity, since your advice does stand out a lot, how would you estimate your 'level of doing maths' or how serious you take mathematics? Hence $f$ is uniformly continuous. It's not a passive text.. which I believe is a good thing if you're willing to put in the time into it. Hence $\mathop{\rm diam}f(E)<\varepsilon$. In some cases (e.g. By the construction of the $g_i$, we must have $\big|g(x)-g(y)\big|<\varepsilon$ for $y\in(x-\delta_1,x+\delta_2)$. My (4) was “fall asleep in topology every day and learn it minimally – become a theoretical computer scientist” but the rest of my list was the same. Maybe "You Can Count On Monsters"? I'm not criticizing your guide at all, I'm just saying the need for all of these supplementary resources (including an entire other textbook on real analysis) is proof that Rudin makes for a poor first textbook to use on the subject. Make sure you really absorb definitions. Download George Bergman's notes about Rudin's exercises. (4) was close to home. It may take you hours, and often you'll have to sleep on it before an idea comes to mind. Hence $\bar f$ is bounded on $\bar E$, so $f$ is bounded on $E$. My tip is to get another book. While there are books that are more intuitive insofar as learning Analysis, going through a book like Baby Rudin is a great experience for forcing you to learn how to think for yourself when reading a math book. It is clear that $g$ is continuous at any $x\in(a_i,b_i)$, so suppose $x\in E$ and let $\varepsilon>0$. This also gives you an understanding of what your professor wants you to learn, and you can develop a sense of what's coming next. I found this helped me absorb it better. Try to get a feel for common threads between theorems. Extend each of the $f_n$ to a continuous function $g_n$ defined on all of $\mathbf R$. Get Rudin and some other book with pictures/drawings (I liked Abbott's and Pugh's). Its more comparable to rudins real and complex analysis book. Then try to prove it before reading the proof given. If $\Gamma(E)$ is compact, let $V$ be a closed subset of $Y$. I disagree. Baby Rudin is one of the clearest introductory level analysis textbooks out there. Whenever the author says something is obvious, you should be able to explicitly prove it. As an example, take the function \\[ … Please only read these solutions after thinking about the problems carefully. (By analambanomenos) Suppose $f$ is a uniformly continuous function from the metric space $X$ to the metric space $Y$. For $y\neq 0$, $g(ky^3,y)=k/\bigl((k^3+1)y\bigr)\rightarrow\infty$ as $y\rightarrow 0$, so $g$ is unbounded in every neighborhood of $(0,0)$. Right. \[ f(\bar E) = f((0,1)) = (0,1) \subsetneq \overline{(0,1)} = [0,1] \]. Exercises are easier to do if you know how difficult they are upfront, and if you do the easier ones first. Similarly, if $x+\delta\notin E$, we can replace $x+\delta$ with some $x+\delta_2=a_j\in E$. Hence, if any of the open intervals $(a_i,b_i)$ intersect $(x-\delta_1,x+\delta_2)$, then both $a_i$ and $b_i$ must be in $(x-\delta_1,x+\delta_2)$. Some results in Rudin are proven by contradiction, I think it is productive to find (yourself, or on the internet) more direct or constructive ways to prove them. Do not just copy these solutions. Actively write down the definitions, your thoughts on them, sketches of proofs, and so on. Let $\varepsilon>0$ and $\delta>0$ such that $$\mathop{\rm diam}f(E)<\varepsilon$$ for all $E\subset X$ with $\mathop{\rm diam}E<\delta$. take a point set topology class and now undergrad analysis makes perfect sense, It's definitely helpful to see why a lot of the things about metric spaces that seemed "easy" to prove were easy because they hold more generally (e.g. Rudin is not for learning, it's for reorganizing what you should already know to be more rigorous, for personal training. Let $\varepsilon>0$. To take this to the next level, you can develop exercises that are probably of the right difficulty, which gives you more opportunities to practice. If you still have more free time, do the difficulty 5 ones. (By analambanomenos) Assume $f : X \to Y$ is continuous. This is the bread and butter. Know your definitions in-and-out.. you should be able to write a definition out and its negation if need be. Don't be surprised if you cannot solve a particular exercise right away. If $p,q\in E$, then $d_X(p,q)\le\mathop{\rm diam}E<\delta$, so $d_Y\bigl(f(p),f(q)\bigr)<\varepsilon$. They won't be easy. Letting $E=\{p,q\}$ we have $\mathop{\rm diam}E<\delta$, so $\mathop{\rm diam}f(E)=d_Y\bigl(f(p),f(q)\bigr)<\varepsilon$. Otherwise I liked using office hours. Don't be upset if Rudin gives an argument that's really really clever and it seems like you can't see where he got it from. Read every theorem statement carefully. Authors (and professors) have tendencies in favorite techniques and theorems, and if you can ID those, you can develop a base package of tricks. That's the primary one we use at my university. The projection $ \pi : X \times Y \to X$ is continuous, so $ f^{-1}(V) = \pi(V^\prime) $ is compact, hence closed (since $X$ is a metric space and therefore Hausdorff). Solution to Principles of Mathematical Analysis Chapter 4 Part A, Solution to Principles of Mathematical Analysis Chapter 3 Part C, Solution to Principles of Mathematical Analysis Chapter 4 Part B, Solution to Principles of Mathematical Analysis Chapter 10, Solution to Principles of Mathematical Analysis Chapter 9 Part C, Solution to Principles of Mathematical Analysis Chapter 9 Part B, Solution to Principles of Mathematical Analysis Chapter 9 Part A, Solution to Principles of Mathematical Analysis Chapter 8 Part C, Solution to Principles of Mathematical Analysis Chapter 8 Part B, Solution to Principles of Mathematical Analysis Chapter 8 Part A, Solution to Principles of Mathematical Analysis Chapter 7 Part C, Solution to Principles of Mathematical Analysis Chapter 7 Part B, Solution to Principles of Mathematical Analysis Chapter 7 Part A, Solution to Linear Algebra Hoffman & Kunze Chapter 4.3. Then when going through a hard exercise go down your technique list and how it can be used in combination to the definitions that appear in the exercise. I honestly don’t think I could have counted the number of times I had just learned something in analysis and then had it mentioned in topology as a trivial consequence of such and such theorem of Urysohn or Alexandroff or whatever. Maybe skip some of Chapter 1. (By analambanomenos) Note that both $f$ and $g$ are equal to 0 on the $x$ and $y$ axes. Let $p,q\in X$ such that $d_X(p,q)<\delta$. They were a thing of beauty. Try to do as many exercises as you can. If $b_i=\infty$ define $g_i$ on $[a_i,\infty)$ to take the constant value $f(a_i)$. Previous Post Solution to Principles of Mathematical Analysis Chapter 1 Part B. when they were assigned to be collected) my colleagues and I googled them and reverse engineered the solutions we found until we understood how to arrive at them.

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