1. 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, $\forall \epsilon$ $\exists q, q \neq x$ 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.

2. Question: Is the set $(0,2) \cap \mathbb{Q}$ connected on $\mathbb{Q}$?

Answer:

3. 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:

4. 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.