[problem] Let $(X, d)$ be a metric space and let $U \subseteq X$ be open. Prove that for every $x \in U$, the open ball $B(x, \varepsilon)$ with $\varepsilon = \sup\{r > 0 : B(x, r) \subseteq U\}$ (the maximal radius) satisfies $B(x, \varepsilon) \subseteq U$, and that if $\varepsilon < \infty$, then $B(x, \varepsilon)$ is the largest open ball centred at $x$ contained in $U$. [/problem]

[solution] **Step 1 (Define the maximal radius).** Since $U$ is open and $x \in U$, there exists $r_0 > 0$ with $B(x, r_0) \subseteq U$. Define \begin{align*} \varepsilon := \sup\{r > 0 : B(x, r) \subseteq U\}. \end{align*} This supremum exists in $(0, \infty]$ because the set $\{r > 0 : B(x, r) \subseteq U\}$ is nonempty (it contains $r_0$). **Step 2 (Prove $B(x, \varepsilon) \subseteq U$).** Let $y \in B(x, \varepsilon)$, so $d(x, y) < \varepsilon$. Choose $r$ with $d(x, y) < r < \varepsilon$. By the definition of supremum, $r \in \{s > 0 : B(x, s) \subseteq U\}$ (since $r < \varepsilon = \sup$, there exists $s > r$ with $B(x, s) \subseteq U$, and $B(x, r) \subseteq B(x, s) \subseteq U$). Therefore $y \in B(x, r) \subseteq U$. **Step 3 (Maximality).** If $B(x, R) \subseteq U$ for some $R > 0$, then $R \in \{r > 0 : B(x, r) \subseteq U\}$, so $R \le \varepsilon$ by definition of supremum. Therefore $B(x, R) \subseteq B(x, \varepsilon)$, and $B(x, \varepsilon)$ is the largest open ball centred at $x$ contained in $U$. **Step 4 (The ball at radius $\varepsilon$ is not necessarily in $U$ when $\varepsilon < \infty$).** The supremum need not be attained: consider $U = (0, 2) \subset \mathbb{R}$ and $x = 1$. Then $\varepsilon = 1$ and $B(1, 1) = (0, 2) = U \subseteq U$. But for $U = (0, 2)$ and $x = 1/2$, we get $\varepsilon = 1/2$ and $B(1/2, 1/2) = (0, 1) \subseteq U$. Here $B(x, \varepsilon) \subsetneq U$, and the closed ball $\overline{B}(x, \varepsilon) = [0, 1] \not\subseteq U$. $\blacksquare$ [/solution]

[problem] Prove that a subset $U \subseteq \mathbb{R}^n$ (with the standard topology) is open if and only if for every $x \in U$, there exists an open cube $Q = (a_1, b_1) \times \cdots \times (a_n, b_n)$ with $x \in Q \subseteq U$. [/problem]

[solution] **Step 1 (Open implies cube neighbourhood).** Let $U$ be open and $x = (x_1, \ldots, x_n) \in U$. By the definition of openness in the metric topology (using the Euclidean metric $d_2$), there exists $\varepsilon > 0$ with $B_{d_2}(x, \varepsilon) \subseteq U$. Define the open cube \begin{align*} Q := \left(x_1 - \frac{\varepsilon}{\sqrt{n}}, x_1 + \frac{\varepsilon}{\sqrt{n}}\right) \times \cdots \times \left(x_n - \frac{\varepsilon}{\sqrt{n}}, x_n + \frac{\varepsilon}{\sqrt{n}}\right). \end{align*} For any $y \in Q$, each coordinate satisfies $|y_k - x_k| < \varepsilon/\sqrt{n}$, so \begin{align*} d_2(x, y) = \left(\sum_{k=1}^n |y_k - x_k|^2\right)^{1/2} < \left(n \cdot \frac{\varepsilon^2}{n}\right)^{1/2} = \varepsilon. \end{align*} Therefore $Q \subseteq B_{d_2}(x, \varepsilon) \subseteq U$, and $x \in Q$ by construction. **Step 2 (Cube neighbourhood implies open).** Suppose every $x \in U$ has an open cube $Q_x$ with $x \in Q_x \subseteq U$. The open cube $Q_x = (a_1, b_1) \times \cdots \times (a_n, b_n)$ contains the Euclidean ball $B_{d_2}(x, \delta)$ where $\delta = \min_{k}(\min(x_k - a_k, b_k - x_k)) > 0$ (since $x$ is in the interior of each interval factor). Therefore $B_{d_2}(x, \delta) \subseteq Q_x \subseteq U$, so $x$ is an interior point of $U$. **Step 3 (Conclusion).** Every point of $U$ is an interior point, so $U$ is open. The equivalence holds because open cubes and open Euclidean balls generate the same topology on $\mathbb{R}^n$ — they are both bases for the standard topology. $\blacksquare$ [/solution]

[problem] Let $f: X \to Y$ be a continuous bijection between topological spaces. Prove that $f$ is a homeomorphism if and only if $f$ is an open map. [/problem]

[solution] **Step 1 (Forward: homeomorphism implies open).** Suppose $f$ is a homeomorphism, so $f^{-1}: Y \to X$ is continuous. Let $U \subseteq X$ be open. We must show $f(U)$ is open in $Y$. Since $f$ is a bijection, $f(U) = (f^{-1})^{-1}(U)$ — the preimage of $U$ under $f^{-1}$. Since $f^{-1}$ is continuous and $U$ is open, $(f^{-1})^{-1}(U)$ is open in $Y$. Therefore $f(U)$ is open. **Step 2 (Reverse: continuous open bijection implies homeomorphism).** Suppose $f$ is continuous, bijective, and open. We must show $f^{-1}: Y \to X$ is continuous, i.e., for every open $U \subseteq X$, the preimage $(f^{-1})^{-1}(U)$ is open in $Y$. Since $f$ is a bijection, $(f^{-1})^{-1}(U) = f(U)$. Since $f$ is an open map and $U$ is open, $f(U)$ is open in $Y$. Therefore $f^{-1}$ is continuous, and $f$ is a homeomorphism. **Step 3 (Conclusion).** A continuous bijection is a homeomorphism if and only if it is an open map. The key identity is $(f^{-1})^{-1}(U) = f(U)$, which converts the continuity of $f^{-1}$ into the openness of $f$. $\blacksquare$ [/solution]