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

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(A) Given the function $f:[0,1] ightarrow [0,1]$ defined as $f(x) = \begin{cases} x & 	ext{if } x \in Q \ 1-x & 	ext{if } x 
otin Q \end{cases}$

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$[0,1]$

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(A) Given the function $f:[0,1] ightarrow [0,1]$ defined as $f(x) = \begin{cases} x & 	ext{if } x \in Q \ 1-x & 	ext{if } x 
otin Q \end{cases}$.We need to find the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$.Case $1$: If $x \in Q$, then $f(x) = x$. Since $x \in [0,1]$, $x$ is rational, so $f(x) \in Q$. Thus, $(f \circ f)(x) = f(f(x)) = f(x) = x$. This holds for all $x \in Q \cap [0,1]$.Case $2$: If $x 
otin Q$, then $f(x) = 1-x$. Since $x$ is irrational and $x \in [0,1]$, $1-x$ is also irrational and $1-x \in [0,1]$. Thus, $(f \circ f)(x) = f(f(x)) = f(1-x) = 1-(1-x) = x$. This holds for all $x \in [0,1] \setminus Q$.Since $(f \circ f)(x) = x$ for all $x \in [0,1]$, the set $S$ is the entire interval $[0,1]$.

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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