Let \((X,d)\) be an unbounded and connected metric space. Prove that for each \(x_0 \in X\), the set \(\{x \in X \, \colon \, d(x,x_0) = r\}\) is nonempty.
Solution
1. Fix \(x_0 \in X\) and \(r > 0\); the map \(d_{x_0} : X \to [0, \infty)\), \(x \mapsto d(x, x_0)\), is continuous.
the triangle inequality gives \(|d(x, x_0) - d(y, x_0)| \le d(x, y)\).
2. \(d_{x_0}\) is unbounded: for every \(M\) there is \(x \in X\) with \(d(x, x_0) > M\).
\(X\) is unbounded, so no ball \(B(x_0, M)\) contains all of \(X\).
3. \(d_{x_0}(X)\) is an interval containing \(0\): if \(u, v \in d_{x_0}(X)\) with \(u \le w \le v\), then \(w \in d_{x_0}(X)\).
\(d_{x_0}(X)\) is the continuous image of the connected space \(X\), hence connected; the connected subsets of \(\RR\) are exactly the intervals; \(0 = d(x_0, x_0) \in d_{x_0}(X)\).
4. \(r \in d_{x_0}(X)\) for every \(r \ge 0\).
\(0 \in d_{x_0}(X)\) (<1>3) and \(d_{x_0}\) is unbounded (<1>2), so the interval \(d_{x_0}(X) \supseteq [0, \infty)\); in particular \(r\) lies in the image.
5. Q.E.D.: \(\{x \in X : d(x, x_0) = r\} = \inverseof{d_{x_0}}(r) \neq \varnothing\) for every \(r \ge 0\).
4 says \(r\) has a preimage under \(d_{x_0}\), which is exactly a point at distance \(r\) from \(x_0\).