Discussion
Consider the metrics \(\rho_{1}\) and \(\rho_{2}\) on \(\mathbb{R}\), defined by
\[ \rho_{1}(x,y) = |x-y| \text{ and } \rho_{2}(x,y) = \begin{cases} 0, & \text{if } x = y \\ 1, & \text{if } x \ne y \end{cases} \]
Let \(X = \{n \in \mathbb{N} : n \ge 3\}\) and \(Y = \{n + \frac{1}{n} : n \in \mathbb{N}\}\). Define \(f : X \cup Y \to \mathbb{R}\) by
\[ f(x) = \begin{cases} 2, & \text{if } x \in X \\ 3, & \text{if } x \in Y \end{cases} \]
Consider the following statements:
P: The function \(f : (X \cup Y, \rho_{1}) \to (\mathbb{R}, \rho_{1})\) is uniformly continuous.
Q: The function \(f : (X \cup Y, \rho_{2}) \to (\mathbb{R}, \rho_{1})\) is uniformly continuous.
Then
Answer: B
P is FALSE and Q is TRUE.
P: with the usual metric, the points n in X and n + 1/n in Y (for n at least 3) are only 1/n apart, yet f takes the values 2 and 3 there. So no single \(\delta\) makes \(|f(x)-f(y)| \lt \tfrac{1}{2}\) whenever \(|x-y| \lt \delta\): f is not uniformly continuous.
Q: with the discrete metric \(\rho_{2}\), taking \(\delta = \tfrac{1}{2}\) means \(\rho_{2}(x,y) \lt \delta\) only when x = y, so \(|f(x)-f(y)| = 0\). Every function on a discrete metric space is uniformly continuous.
Note that f is continuous in P's setting too, since every point of X ∪ Y is isolated; it is the uniformity that fails.
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