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--- math104-s21:s:ryotainagaki:problems [2021/05/12 20:28]
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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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