ISI2019-MMA-24

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $\lim _{n\rightarrow \infty} f^n(x)$ exists for every $x \in \mathbb{R}$, where $f^n(x) = f  \circ f^{n-1}(x)$ for $n \geq 2$. Define 

$$S=\left\{\lim _{n \rightarrow \infty} f^n(x): x \in \mathbb{R}\right\}  \text{ and } T=\left\{x  \in \mathbb{R}:f(x)=x\right\}$$

Then which of the following is necessarily true?

• A. $S \subset T$
• B. $T \subset S$
• C. $S = T$
• D. None of the above

Answer 1

Let

$h(x)  = \lim_{n \rightarrow \infty} f^n(x) = f(\lim_{n \rightarrow \infty}f^{n-1}(x))$

We can perform the last step as the function is continuous for every x. Continuing on,

$h(x) = f(h(x))$ $ \forall x \in R$

because $n\rightarrow \infty\implies n-1 \rightarrow \infty$

It means $S = \{f(x) = x : \forall x \in Range(h(x))\}$

The function defined by two sets in same but $S$ is defined over some elements of real numbers whereas $T$ is defined for all real numbers.

$\implies S \subset T$

$A$ is correct answer.

Comments

How do u know $T$ mapped on all real numbers. It also can be mapped on some real numbers

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Given in question. f is defined on all real numbers because domain is R . and most importantly in the definition of T,  $ x\in R$ which means no restriction. And this relation holds necessarily. But it could be the case that both S and T are equal if $Range(h(x)) = R$ , not otherwise.

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Your answer is Partially correct but wrong as per the official key. The correct answer is option $C$.

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you are right. C is the correct answer. A function by definition is defined over all the elements of the domain. Otherwise it is not a function.

Answer 2

$
f^{n}(x)=f \circ f^{n-1}(x)
$

$taking$  n = 2 ,
$
f^{2}(x)=f(f(x))
$

$
accordingly,f^{3}(x)=f(f(f(x)))
$
putting the value of $f(x)=x$

S={f(x), where  $x \in \mathbb{R}$}

So T is a subset of S

*Correct me if I am wrong

Comments

Your answer is definitely wrong. It should be Option $C$.