Let Q be the set of all rational numbers in [ | vedclass

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(A) Given $f(x) = \begin{cases} x & 	ext{for } x \in Q \ 1-x & 	ext{for } x 
otin Q \end{cases}$ for $f:[0,1] ightarrow [0,1]$.Case $1$: If $x \

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$[0,1]$

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(A) Given $f(x) = \begin{cases} x & 	ext{for } x \in Q \ 1-x & 	ext{for } x 
otin Q \end{cases}$ for $f:[0,1] ightarrow [0,1]$.Case $1$: If $x \in Q$,then $f(x) = x$. Since $x \in [0,1]$,$x$ is rational,so $f(x) \in Q$. Thus,$(f \circ f)(x) = f(f(x)) = f(x) = x$.Case $2$: If $x 
otin Q$,then $f(x) = 1-x$. Since $x$ is irrational,$1-x$ is also irrational (because if $1-x$ were rational,$x = 1 - (1-x)$ would be rational,a contradiction). Thus,$f(x) 
otin Q$. Therefore,$(f \circ f)(x) = f(f(x)) = f(1-x) = 1-(1-x) = x$.Since $(f \circ f)(x) = x$ for all $x \in [0,1]$,the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$ is the entire domain $[0,1]$.

Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow [0,1]$ be defined by $f(x) = \begin{cases} x & \text{for } x \in Q \\ 1-x & \text{for } x \notin Q \end{cases}$. Then,the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$ is equal to

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