$f: \mathbb R \to \mathbb R$ be a continuous function and $A$ is a proper subset of $\mathbb R$ such that $A =\{y \in \mathbb R: y = \lim_{n \to \infty} f(x_{n}),$ for a sequence $x_{n} \to + \infty\}$. Then the set $A$ is necessarily $-$

$a$ compact set

$b$ closed set

$c$ Singleton set

$d$ none of these

My attempt :

If I take the function $f(x) = \sin x$ and $x_{n} = n \pi$ and another sequence $y_{n}$ for same function $\sin x$ that converges to $1$ or $-1$. Then, I'll have at least two element in $A$ so it is not necessarily a connected set. But this set is compact as well as closed.

I don't know how to choose between other three options. Can I find a set $A$ such that it's not bounded or it's not closed?

  • 1
    $\begingroup$ The set for $\sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit. $\endgroup$ – Paul K Jan 4 at 13:32
  • $\begingroup$ It's given in question, "for a sequence" I didn't paid attention to it. Also, for a sequence I could find only one element in A. $\endgroup$ – Mathsaddict Jan 4 at 13:37
  • $\begingroup$ @Paul K I took sin x as function, and $n\pi$ as sequence, then $f(x_{n})$ converges to 0 $\endgroup$ – Mathsaddict Jan 4 at 13:41

a. $A$ may not be a compact set

$f(x)= \begin{cases} 0 & x \le 0\\ x \left(\sin x +1 \right)& x>0 \end{cases}$

You have $A = [0,\infty)$

c. $A$ may not be a singleton set

$f(x) = \sin x$

$A = [-1,1]$.

b. $A$ is a closed set

If $(y_n)$ is a sequence of $A$ converging to $y$, use a diagonal argument to build a sequence $(x_n)$ such that $\lim\limits_{n \to \infty} x_n = \infty$ and $\lim\limits_{n \to \infty} f(x_n) = y$. Proving that $y \in A$ and that $A$ is closed.


  • $\begingroup$ So I can choose a continuous function and sequence, if I take the set of all limit points of $f(x_{n})$ then I will get $A$, since there always exists a subsequence converging to a limit point. Then I have the counterexamples you mentioned. Now, since the set of limit points is a closed set, so d is correct. Am I right? $\endgroup$ – Mathsaddict Jan 5 at 3:09
  • $\begingroup$ No, as b. is correct, d. is not the right answer. $\endgroup$ – mathcounterexamples.net Jan 5 at 7:46
  • $\begingroup$ Yes , I meant 'd'. It was typing error. But what I got from your answer is correct? $\endgroup$ – Mathsaddict Jan 5 at 14:27
  • $\begingroup$ Yes, that is correct. $\endgroup$ – mathcounterexamples.net Jan 5 at 15:23

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