[problem] Let $(X, d)$ be a metric space and let $A \subseteq X$ be nonempty. Prove that the **distance function** $d_A: X \to [0, \infty)$ defined by $d_A(x) := \inf_{a \in A} d(x, a)$ is continuous (in fact, Lipschitz with constant $1$), and that $\overline{A} = \{x \in X : d_A(x) = 0\}$. [/problem]

[solution] **Step 1 (Lipschitz continuity).** Let $x, y \in X$. For any $a \in A$, the triangle inequality gives $d(x, a) \le d(x, y) + d(y, a)$. Taking the infimum over $a \in A$: \begin{align*} d_A(x) = \inf_{a \in A} d(x, a) \le d(x, y) + \inf_{a \in A} d(y, a) = d(x, y) + d_A(y). \end{align*} Rearranging: $d_A(x) - d_A(y) \le d(x, y)$. Symmetrically, $d_A(y) - d_A(x) \le d(x, y)$. Therefore $|d_A(x) - d_A(y)| \le d(x, y)$, so $d_A$ is Lipschitz with constant $1$, hence continuous. **Step 2 (Characterise $d_A(x) = 0$).** We show $d_A(x) = 0$ if and only if $x \in \overline{A}$. *Forward:* Suppose $d_A(x) = 0$. Then $\inf_{a \in A} d(x, a) = 0$, so for every $n \in \mathbb{N}$ there exists $a_n \in A$ with $d(x, a_n) < 1/n$. Every open ball $B(x, \varepsilon)$ contains $a_n$ for $n > 1/\varepsilon$, so $B(x, \varepsilon) \cap A \neq \varnothing$. Therefore $x \in \overline{A}$. *Reverse:* Suppose $x \in \overline{A}$. For every $n \in \mathbb{N}$, the ball $B(x, 1/n)$ meets $A$, so there exists $a_n \in A$ with $d(x, a_n) < 1/n$. Therefore $0 \le d_A(x) \le d(x, a_n) < 1/n$ for all $n$, giving $d_A(x) = 0$. **Step 3 (Conclusion).** The distance function $d_A$ is $1$-Lipschitz, and $\overline{A} = d_A^{-1}(\{0\})$. Since $d_A$ is continuous and $\{0\}$ is closed in $\mathbb{R}$, this gives an alternative proof that $\overline{A}$ is closed. $\blacksquare$ [/solution]

[problem] Let $f, g: X \to \mathbb{R}$ be continuous functions on a Hausdorff topological space $X$. Prove that the **agreement set** $\{x \in X : f(x) = g(x)\}$ is closed. [/problem]

[solution] **Step 1 (Express as a level set).** Define $h: X \to \mathbb{R}$ by $h(x) := f(x) - g(x)$. Since $f$ and $g$ are continuous and subtraction $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is continuous, the composition $h$ is continuous. The agreement set is \begin{align*} \{x \in X : f(x) = g(x)\} = \{x \in X : h(x) = 0\} = h^{-1}(\{0\}). \end{align*} **Step 2 (Apply the closed-set criterion).** The singleton $\{0\}$ is closed in $\mathbb{R}$ (since $\mathbb{R}$ is Hausdorff, all singletons are closed). Since $h$ is continuous, the preimage $h^{-1}(\{0\})$ is closed in $X$. **Step 3 (Consequence).** If two continuous functions $f, g: X \to \mathbb{R}$ agree on a dense subset $D \subseteq X$, then they agree everywhere: the agreement set $\{f = g\}$ is closed and contains $D$, hence contains $\overline{D} = X$. This is the **identity principle** for continuous functions, and it follows purely from the closedness of level sets. $\blacksquare$ [/solution]

[problem] Let $(X, d)$ be a complete metric space and let $F_1 \supseteq F_2 \supseteq F_3 \supseteq \cdots$ be a decreasing sequence of nonempty closed subsets with $\operatorname{diam}(F_n) \to 0$. Prove directly (without citing the Cantor Intersection Theorem) that $\bigcap_{n=1}^\infty F_n$ consists of exactly one point. [/problem]

[solution] **Step 1 (Construct a Cauchy sequence).** For each $n \in \mathbb{N}$, choose $x_n \in F_n$. Since the sets are nested, $x_m \in F_n$ for all $m \ge n$. For $m, m' \ge n$, both $x_m$ and $x_{m'}$ belong to $F_n$, so \begin{align*} d(x_m, x_{m'}) \le \operatorname{diam}(F_n). \end{align*} Since $\operatorname{diam}(F_n) \to 0$, for any $\varepsilon > 0$ there exists $N$ with $\operatorname{diam}(F_N) < \varepsilon$, and then $d(x_m, x_{m'}) < \varepsilon$ for all $m, m' \ge N$. The sequence $\{x_n\}$ is Cauchy. **Step 2 (The limit lies in every $F_n$).** Since $(X, d)$ is complete, there exists $x \in X$ with $x_n \to x$. Fix $n \in \mathbb{N}$. For all $m \ge n$, $x_m \in F_n$. Since $F_n$ is closed and $\{x_m\}_{m \ge n}$ is a sequence in $F_n$ converging to $x$, the sequential characterisation of closed sets gives $x \in F_n$. Since $n$ was arbitrary, $x \in \bigcap_{n=1}^\infty F_n$. **Step 3 (Uniqueness).** Suppose $y \in \bigcap_{n=1}^\infty F_n$. Then $x, y \in F_n$ for all $n$, so $d(x, y) \le \operatorname{diam}(F_n)$ for all $n$. Since $\operatorname{diam}(F_n) \to 0$, we conclude $d(x, y) = 0$, hence $x = y$. **Step 4 (Conclusion).** The intersection $\bigcap_{n=1}^\infty F_n = \{x\}$ is a singleton. $\blacksquare$ [/solution]