Let 𝐶[0, 1] = { 𝑓 ∶ [0, 1] → ℝ ∶ 𝑓 is continuous}.

Consider the metric space (𝐶[0,1], 𝑑_∞), where

𝑑_∞(𝑓, 𝑔) = sup{ |𝑓(𝑥) − 𝑔(𝑥)| ∶ 𝑥 ∈ [0, 1] } for 𝑓, 𝑔 ∈ 𝐶[0,1].

Let 𝑓₀ (𝑥) = 0 for all 𝑥 ∈ [0,1] and

𝑋 = {𝑓 ∈ (𝐶[0, 1], 𝑑_∞) ∶ 𝑑_∞(𝑓₀, 𝑓) ≥ \(\frac{1}{2}\)}.

Let 𝑓₁, 𝑓₂ ∈ 𝐶[0, 1] be defined by 𝑓₁(𝑥) = 𝑥 and 𝑓₂ (𝑥) = 1 − 𝑥 for all 𝑥 ∈ [0,1].

Consider the following statements:

𝑃: 𝑓₁ is in the interior of 𝑋.

𝑄: 𝑓₂ is in the interior of 𝑋.

Which of the following statements is correct? 

Concept:

Given a subset A of a metric space M, its interior A^◦ is defined as the set of all points x ∈ A such that some open ball around x is a subset of A.

Explanation:

𝐶[0, 1] = { 𝑓 ∶ [0, 1] → ℝ ∶ 𝑓 is continuous}

𝑑∞(𝑓, 𝑔) = sup{ |𝑓(𝑥) − 𝑔(𝑥)| ∶ 𝑥 ∈ [0, 1] } for 𝑓, 𝑔 ∈ 𝐶[0,1]

 𝑓0 (𝑥) = 0 for all 𝑥 ∈ [0,1] and

𝑋 = {𝑓 ∈ (𝐶[0, 1], 𝑑∞) ∶ 𝑑∞(𝑓0, 𝑓) ≥ \(\frac{1}{2}\)}.

𝑓1(𝑥) = 𝑥 

Now, 𝑑∞(𝑓₁, 𝑓₀) = sup{ |x − 0| ∶ 𝑥 ∈ [0, 1] }  = sup{ |x| ∶ 𝑥 ∈ [0, 1] } = 1 > \(\frac{1}{2}\)

So, 𝑓1 ∈ X^◦

i.e., 𝑓1 is in the interior of 𝑋.

Also, 𝑓2 (𝑥) = 1 − 𝑥  

𝑑∞(𝑓2, 𝑓0) = sup{ |1- x − 0| ∶ 𝑥 ∈ [0, 1] } = sup{ |1 - x| ∶ 𝑥 ∈ [0, 1] } = 1 > \(\frac{1}{2}\)

So, 𝑓2 ∈ X◦

i.e., 𝑓2 is in the interior of 𝑋.

∴ Both 𝑃 and 𝑄 are TRUE

(4) is correct