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-** 32. Give an example of a function that is continuous on $\mathbb{Q}$ but not on $\mathbb{R}$ and RIGOROUSLY prove that your functions satisfies such property. **
+** 32. (Credits to Midterm 2) In less than 3 sentences and in less than 3 minutes, formally prove or disprove: given $A \subseteq B$ is bounded and f is a continuous map $f: B \to Y$, then $f(A)$ is bounded. **
-**Answer: ** Consider the function $f(x)$ defined as $1$ if $x \in \mathbb{Q}$ and $0$ otherwise. We can say that on $\mathbb{Q}$ the function is rational since on $\mathbb{Q}$ $f(x) = 1$. However on $\mathbb{R}$ the function is not continuous since for any point x_0 in $\mathbb{R}$, $\lim_{x \to x_0}f(x)$ does not exist. Why? Using a sequence $x_n$ of irrational numbers, I get that $f(x_n) = 0 $ and $ f(x_n) \to 0$. Using a sequence $x_n$ of rational numbers, I get that $f(x_n) = 1$ and $ f(x_n) \to 1$.
+**Answer: On the spot and on a timed fast exam, this may seem like a hard problem and deceptive. ** We know that the given statement is false. To show that consider the metric space $(S, d(x, y )=|x- y|)$, $U = (0, 1)$, and $f(x) = \ln(x)$. We know that $U$ is bounded but $f(U) = (-\infty, 0)$ by the property of natural log and is NOT bounded. Thus, just because $A$ is bounded doesn't mean that $f(A)$ is bounded.
 ** 33. Suppose that (a_n)_n is a sequence in a metric space (M, d), which converges to a limit
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 However, we cannot say that $f$ is Riemann integrable over $[-1, 1]$. This is due to the behaviour of $f$. And when we consider interval $[-1, 0]$ we start to understand that $f$ is not integrable because of the upper integral for that interval is 1 whereas the lower integral for that interval would be 0.
-** 51. (From MIT Opencourseweare Math 18.100B Course) True or False? If $f_n : [a, b] \to R$is a sequence of almost everywhere continuous functions, and $f_n \to f$ converges uniformly, then the limit f is almost everywhere continuous.**
+** 51. (From MIT Opencourseweare Math 18.100B Course) True or False? If $f_n : [a, b] \to R$is a sequence of almost everywhere continuous functions, and $f_n \to f$ converges uniformly, then the limit f is almost everywhere continuous. (I wanted an explanation with the specific theorem numbers from Rudin)**
-**Answer: ** I kind of do not like how the term "almost everywhere continuous" is used in this problem; it just sounds imprecise. It turns out that the problem meant f has only a finite number of discontinuities.  In order to address this problem we may want to first note that since $f_n \to f$ continuous on $[a, b]$ except for a finite number of points on $[a, b]$ we know that $f_n$ is Riemann integrable on $[a, b]$. Because of that and by theorem 7.16 in Rudin, we can supposedly say that $f$ is integrable as well. And since f is integrable, f MUST have only finitely many discontinuities.(Bonus: Try to prove that if $f$ has infinitely many discontinuities, then it must not be integrable.)
+**Answer: ** I kind of do not like how the term "almost everywhere continuous" is used in this problem; it just sounds imprecise. It turns out that the problem meant f has only a finite number of discontinuities.  In order to address this problem we may want to first note that since $f_n \to f$ continuous on $[a, b]$ except for a finite number of points on $[a, b]$ we know that $f_n$ is Riemann integrable ($\alpha(x) = x$ and that is continuous for all $x \in \mathbb{R}$) on $[a, b]$ by theorem 6.10. Because of that and by ** theorem 7.16 ** in Rudin, we can supposedly say that $f$ is integrable as well. And since f is integrable, f MUST have only finitely many discontinuities.(This one does not seem to have a direct theorem from Rudin I can use.)
 One thing that I am still rightfully confused about is whether it is true that if $f$ has only finite number of discontinuities then and only then it is riemann integrable. I may be able to puncture this argument by considering $f(x) = \frac{1}{x^2}$. We may all know that $f(x)$ is continuous at any point but at $x = 0$, where $\lim_{x \to 0} f(x) = \infty$. Now, we know that f has only a finite number (1) of discontinuities on $[-a, b]$ (let a, b be any positive number of your choosing), but we cannot integrate $\int_{x= -a}^{x = b}f(x)dx$. It will diverge to infinity (not a real number)!
 Sources for the Questions and Inspiration: Past midterms and/or finals from Dr. Peng Zhou, Ian Charlesworth, Sebastian Eterovic, Yu-Wei Fan, Dr. Charles Rycroft, MIT Opencourseware (MIT's Math 18.100 courses), summer 2018 math 104, spring 2017 math 104, //Principles of Analysis// by Rudin, //A Problem Book in Real Analysis//, and //Principles of Analysis// by Ross.

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