Consider the set :A={x∈Q:0<(2−1)x<2+1}A = \{ x \in $$\mathbb{Q}$$ : 0 < ($$\sqrt{2}$$ - 1)x < $$\sqrt{2}$$ + 1 \}A={x∈Q:0<(2\sqrt {2}2−1)x<$$2\sqrt{2}2$$+1} as a subset of R$$\mathbb{R}$$R. Which of the following statements is true?

set A consists of elements "x" such that 0<($$2\\sqrt{2}2$$ - 1)x< (2+$$1\\sqrt{2}$$ +12+1) ⟹ \\implies⟹ 0<x<2+12−10<x<\\frac{$$\\sqrt{2}$$+1}{$$\\sqrt{2}$$-1}0<x<2−12+1 ⟹ 0<x<2+12−1∗2+12+1\\implies 0<x<\\frac{$$\\sqrt{2}$$+1}{$$\\sqrt{2}$$-1}*\\frac{$$\\sqrt{2}$$+1}{$$\\sqrt{2}$$+1}⟹0<x<2−12+1∗2+12+1 (rationalising) ⟹ \\implies⟹ 0<x<(2+1)2(2)2−120<x<\\frac{($$\\sqrt{2}$$+1)^2}{($$\\sqrt{2}$$)^2 - 1^2}0<x<(2)2−12(2+1)2 ⟹ 0<x<2+1+222−1\\implies 0<x<\\frac{2+1+$$2\\sqrt{2}$$}{2-1}⟹0<x<2−12+1+22 ⟹ 0<x<3+22\\implies 0<x<3+$$2\\sqrt{2}$$⟹0<x<3+22 hence , A = (0 , 3+$$22\\sqrt{2}2$$) ⟹ infA=0,supA=3+22\\implies inf A = 0 , sup A = 3 + $$2\\sqrt{2}$$ ⟹infA=0,supA=3+22