Consider the metrics 𝜌₁ and 𝜌₂ on ℝ, defined by

𝜌₁ (𝑥, 𝑦) = |𝑥 − 𝑦| and 𝜌₂ (𝑥, 𝑦) = \(\rm \left\{\begin{matrix}0,& if\ x=y\\\ 1,&if\ x\ne y\end{matrix}\right.\)

Let 𝑋 = {𝑛 ∈ ℕ ∶ 𝑛 ≥ 3} and 𝑌 = { 𝑛 + \(\frac{1}{n}\) ∶ 𝑛 ∈ ℕ}.

Define 𝑓: 𝑋 ∪ 𝑌 → ℝ by 𝑓(𝑥) = \(\rm \left\{\begin{matrix}2,& if\ x∈ X\\\ 3,&if\ x∈ Y\end{matrix}\right.\)

Consider the following statements:

𝑃: The function 𝑓: (𝑋 ∪ 𝑌, 𝜌₁ ) → (ℝ, 𝜌₁) is uniformly continuous.

𝑄: The function 𝑓: (𝑋 ∪ 𝑌, 𝜌₂ ) → (ℝ, 𝜌₁) is uniformly continuous.

Then 

Concept:

(i) Uniformly continuous: A function f: X → Y is said to be uniformly continuous if, for every ϵ > 0, there exists a δ > 0 such that for every x, y ∈ X, |x - y| < δ ⇒ |f(x) - f(y)| < ϵ 

(ii) Any function from a discrete metric space to any other metric space is uniformly continuous.

Explanation:

𝑋 = {𝑛 ∈ ℕ ∶ 𝑛 ≥ 3} and 𝑌 = { 𝑛 + \(\frac{1}{n}\) ∶ 𝑛 ∈ ℕ}

So, X = {3, 4, 5, 6, ...} and Y = {\(2, 2\frac{1}{2}, 3\frac{1}{3}, 4\frac{1}{4}, ...\)}

X ∪ Y contain all the natural numbers and  𝑛 + \(\frac{1}{n}\), 𝑛 ∈ ℕ 

𝑓: 𝑋 ∪ 𝑌 → ℝ by 𝑓(𝑥) = \(\rm \left\{\begin{matrix}2,& if\ x∈ X\\\ 3,&if\ x∈ Y\end{matrix}\right.\)

𝜌1 (𝑥, 𝑦) = |𝑥 − 𝑦| and 𝜌2 (𝑥, 𝑦) = \(\rm \left\{\begin{matrix}0,& if\ x=y\\\ 1,&if\ x\ne y\end{matrix}\right.\)

So, 𝜌1 is the usual metric and 𝜌2 is a discrete metric

Hence by concept (ii),  𝑓: (𝑋 ∪ 𝑌, 𝜌2 ) → (ℝ, 𝜌1) is uniformly continuous.

Q is true

For P, 

𝑓: 𝑋 ∪ 𝑌 → ℝ by 𝑓(𝑥) = \(\rm \left\{\begin{matrix}2,& if\ x∈ X\\\ 3,&if\ x∈ Y\end{matrix}\right.\)

Let x ∈ X and y ∈ Y then 

|f(x) - f(y)| = |2 - 3| = 1 \(\nless\) ϵ for |x - y| < δ 

Hence 𝑓: (𝑋 ∪ 𝑌, 𝜌1 ) → (ℝ, 𝜌1) is not uniformly continuous.

P is false

∴ 𝑃 is FALSE and 𝑄 is TRUE

(2) is correct