Let

𝑐00 = {(π‘₯1, π‘₯2, π‘₯3, … ) ∢ π‘₯𝑖 ∈ ℝ, 𝑖 ∈ β„•, π‘₯𝑖 ≠ 0 only for finitely many indices 𝑖}.

For (π‘₯1, π‘₯2, π‘₯3, … ) ∈ 𝑐00, let β€–(π‘₯1, π‘₯2, π‘₯3, … )β€– = sup{|π‘₯𝑖 | ∢ 𝑖 ∈ β„•}.

Define 𝐹, 𝐺 ∢ (𝑐00, β€–⋅β€–) → (𝑐00, β€–⋅β€–) by

𝐹((π‘₯1, π‘₯2, … , π‘₯𝑛, … )) = \(\rm \left((1+x)_{x_1}, (2+\frac{1}{2})x_2,..., (n+\frac{1}{n})x_n,...\right)\)

𝐺((π‘₯1, π‘₯2, … , π‘₯𝑛, … )) = \(\rm \left(\frac{x_1}{1+1},\frac{x_2}{2+\frac{1}{2}},..., \frac{x_n}{n+\frac{1}{n}},..\right)\)

for all (π‘₯1, π‘₯2, … , π‘₯𝑛, … ) ∈ 𝑐00.

Then

  1. 𝐹 is continuous but 𝐺 is NOT continuous
  2. 𝐹 is NOT continuous but 𝐺 is continuous
  3. both 𝐹 and 𝐺 are continuous 
  4. NEITHER 𝐹 NOR 𝐺 is continuous 

Answer (Detailed Solution Below)

Option 2 : 𝐹 is NOT continuous but 𝐺 is continuous

Detailed Solution

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Explanation -

The correct option is (2)

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