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--- math104-s21:s:victoriatuck [2021/05/11 06:43]
73.158.208.111 [Questions]
+++ math104-s21:s:victoriatuck [2026/02/21 14:41] (current)
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 . Question: $(-\sqrt{2}, \sqrt{2}) \cap \mathbb{Q}$ = $[-\sqrt{2}, \sqrt{2}] \cap \mathbb{Q}$ is both closed and open on $\mathbb{Q}$. However, for a given space E, Rudin Theorem 2.27 states that if $X$ is a metric space and $E \subset X$ then $E = \bar{E}$ if and only if E is closed. [$\bar{E} = E \cup E'$, $E'$ denotes the set of all limit points of E in X]. How does this work with the idea that a sequence of rationals can converge to a number outside of the rationals?
-Answer: Let $X = \mathbb{Q}$. $E = (-\sqrt{2}, \sqrt{2}) \cap \mathbb{Q} \subset \mathbb{Q}$. The set of all limit points E' must be in X, so it can only contain rationals. We can prove that the limit points of E must be inside the set E by contradiction. Assume there is a point $x \in \mathbb{Q}: x \notin E$. This may be a limit point of E because it is a rational. Given an $\epsilon > 0$, in order for x to be a limit point, there must exist a sequence $x_n: \forall \epsilon$ $\exists N$ st $n > N => |x - x_n| < \epsilon$. Let $\epsilon = \frac{min(x - \sqrt{2}, -x - \sqrt{2})}{2}$. Given this value, we can not have a sequence of numbers $x_{n'} \in E$ where $|x_{n'} - x| < \epsilon$. Therefore, no rational number outside of E can be a limit point of E, and E contains all of its limit points.
+Answer: Let $X = \mathbb{Q}$. $E = (-\sqrt{2}, \sqrt{2}) \cap \mathbb{Q} \subset \mathbb{Q}$. The set of all limit points E' must be in X, so it can only contain rationals. We can prove that the limit points of E must be inside the set E by contradiction. Assume there is a point $x \in \mathbb{Q}: x \notin E$. This may be a limit point of E because it is a rational. Given an $\epsilon > 0$, in order for x to be a limit point, $\forall \epsilon$ $\exists q, q \neq x$ st $q \in E$. Let $\epsilon = \frac{min(x - \sqrt{2}, -x - \sqrt{2})}{2}$. Given this value, we can not have a point $q \in E$ where $|q - x| < \epsilon$. Therefore, no rational number outside of E can be a limit point of E, and E contains all of its limit points.
 The confusion with this question for me lay with the use of the limit point argument for situations such as $E = (0, 1) \cap \mathbb{Q}$, which is not closed. However, the difference here is that 1 and 0 are rational numbers.
-. Question: Let $f: X -> Y$ be a continuous mapping with $A \subset X$ and $B \subset Y$. I know that if B is open/closed then $f^{-1}(B)$ is open/closed, respectively, but why is this true?
+. Question: Is the set $E = (0,2) \cap \mathbb{Q}$ connected on $\mathbb{Q}$? [With explanation]
-Answer:
+Answer: No, this set is disconnected. $\mathbb{Q} \subset \mathbb{R}$. Therefore, in general, we can pick an element outside of our set E but inside the metric space $\mathbb{R}$ i.e., $x \in \mathbb{R} \setminus \mathbb{Q}$ to "split" our set E into two open subsets $U_1 = (0, x) \cap \mathbb{Q}$, $U_2 = (x, 2) \cap \mathbb{Q}$ where $(E \cap U_1) \cap (E \cap U_2) = \varnothing$ and $(E \cap U_1) \cup (E \cap U_2) = E$ where $E \cap U_1 \neq \varnothing$ and $E \cap U_2 \neq \varnothing$. Thus, we can use the definition of a disconnected set to show that E is a disconnected set.
-. Question: Exam 2, Problem 5. Let $f: \mathbb{Q} -> \mathbb{R}$ be a continuous map. Is it true that one can always find a continuous map $g: \mathbb{R} -> \mathbb{R}$ extending $f$, namely, $g(x) = f(x)$ for any $x \in \mathbb{Q}$? Prove or find a counterexample.
+. Question: Let $f: X -> Y$ be a continuous mapping with $A \subset X$ and $B \subset Y$. I know that if B is open/closed then $f^{-1}(B)$ is open/closed, respectively, but why is this true?
+Answer: For an open set B, this is by definition of a continuous mapping, i.e., a mapping $f: X -> Y$ is continuous iff $\forall B \subset Y$, $f^{-1}(B)$ is open in X.
+. Question: Exam 2, Problem 5. Let $f: \mathbb{Q} -> \mathbb{R}$ be a continuous map. Is it true that one can always find a continuous map $g: \mathbb{R} -> \mathbb{R}$ extending $f$, namely, $g(x) = f(x)$ for any $x \in \mathbb{Q}$? Prove or find a counterexample.
+Answer:
+. Assume f is a continuous map $f: X -> Y$ with $A \subset X$ and $B \subset Y$
+a) Question: What is a counterexample to "If $A$ is closed, then $f(A)$ is closed".
+Answer: Let $f := arctan(x)$. Then $(-\infty, \infty) -> (-\frac{\pi}{2}, \frac{\pi}{2})$.
+b) Question: What is a counterexample to "If $B$ is connected, then $f^{-1}(B)$ is connected."?
+Answer: Let $f := x^2$. Then $[1, \infty) -> (-\infty, -1] \cup [1, \infty)$, which is disconnected.
+. Question: Given an open cover $\mathcal{U}$ of $f(E)$ where f is a continuous function and E is a compact set, why is $\{f^{-1}(U) : U \in \mathcal{U}\}$ a cover of E? [Question from Ross Theorem 21.4]
+Answer: https://math.stackexchange.com/questions/605195/detail-on-a-theorem-of-continuity-and-compactness
+. Question: For non 1-to-1 functions (like $f = x^2$), how do we define the inverse $f^{-1}$ of a set in the range of that function? E.g., for $x^2$, what would $f^{-1}([1, inf))$ be?
+Answer: In Ross 168, we define the inverse $f^{-1}(U)$ to include any value $s$ st $f(s) \in U$. So $f^{-1}([1, inf)) = (-inf, -1] \cup [1, inf)$.
+. Question:

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