Question

Consider the metrics \(\rho_1\) and \(\rho_2\) on \(\mathbb{R}\), defined by
\[ \rho_1(x, y) = |x-y| \quad \text{and} \quad \rho_2(x, y) = \begin{cases} 0, & \text{if } x = y \\ 1, & \text{if } x \neq y \end{cases} \]
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

• A. P is TRUE and Q is FALSE
• B. P is FALSE and Q is TRUE
• C. both P and Q are FALSE
  (D) both P and Q are TRUE
• D.

Hint:

A standard way to disprove uniform continuity is to find two sequences \(x_n, y_n\) such that \(d(x_n, y_n) \to 0\) but \(d(f(x_n), f(y_n))\) does not converge to 0. Conversely, any function on a space with the discrete metric is always uniformly continuous.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question tests the definition of uniform continuity. A function \(f\) is uniformly continuous if for any \(\epsilon>0\), there is a \(\delta>0\) (depending only on \(\epsilon\)) such that for any two points \(x, y\) in the domain, if the distance between them is less than \(\delta\), the distance between their images is less than \(\epsilon\). We will examine this property with two different metrics on the domain: the usual metric (\(\rho_1\)) and the discrete metric (\(\rho_2\)).
Step 3: Detailed Explanation:
Analysis of Statement P:
The domain space is \(D = X \cup Y\) with the usual metric \(\rho_1(x,y)=|x-y|\).
To show \(f\) is NOT uniformly continuous, we need to find an \(\epsilon_0>0\) such that for every \(\delta>0\), there exist points \(x, y \in D\) with \(|x-y|<\delta\) but \(|f(x)-f(y)| \ge \epsilon_0\).
Let's choose \(\epsilon_0 = 1/2\). Now consider the difference in function values. If we pick one point from \(X\) and one from \(Y\), the difference is \(|f(x)-f(y)| = |2-3| = 1\). This is greater than our \(\epsilon_0\).
Let's see if we can make points from \(X\) and \(Y\) arbitrarily close. Consider the sequence of points \(x_n = n \in X\) (for \(n \ge 3\)) and \(y_n = n + 1/n \in Y\).
The distance between these points is:
\[ |x_n - y_n| = \left| n - \left(n + \frac{1}{n}\right) \right| = \left| -\frac{1}{n} \right| = \frac{1}{n} \] As \(n \to \infty\), the distance \(|x_n - y_n| \to 0\). This means for any \(\delta>0\), we can find a large enough \(n\) such that \(|x_n - y_n|<\delta\).
However, the distance between their function values is: \[ |f(x_n) - f(y_n)| = |f(n) - f(n+1/n)| = |2 - 3| = 1 \] This distance is always 1, which is greater than our chosen \(\epsilon_0 = 1/2\). Since we can find pairs of points that are arbitrarily close but whose images remain a fixed distance apart, the function is not uniformly continuous.
Thus, P is FALSE.
Analysis of Statement Q:
The domain space is \(D = X \cup Y\) with the discrete metric \(\rho_2\). A metric space with the discrete metric is called a discrete space.
A function whose domain is a discrete space is always uniformly continuous. Let's prove it. Let \(\epsilon>0\) be given. We need to find a \(\delta>0\) that satisfies the definition.
Choose \(\delta = 1/2\).
Now, let \(x, y\) be any two points in the domain \(D\) such that \(\rho_2(x,y)<\delta\). Since \(\delta = 1/2\) and the only values \(\rho_2\) can take are 0 and 1, the condition \(\rho_2(x,y)<1/2\) implies that \(\rho_2(x,y)\) must be 0. By definition of \(\rho_2\), \(\rho_2(x,y)=0\) means that \(x=y\). If \(x=y\), then \(f(x)=f(y)\), and the distance in the codomain is \(|f(x)-f(y)| = 0\). Certainly, \(0<\epsilon\) for any positive \(\epsilon\). Therefore, the condition for uniform continuity is satisfied.
Thus, Q is TRUE.
Step 4: Final Answer:
P is FALSE and Q is TRUE.
Step 5: Why This is Correct:
P is false because we can find pairs of points, one from X and one from Y, that get arbitrarily close as \(n \to \infty\), while their function values remain a constant distance of 1 apart. Q is true because any function defined on a discrete metric space is uniformly continuous; choosing \(\delta<1\) forces any two points within that distance to be identical.