# $f: \mathbb R \to \mathbb R$ be a continuous function, $A =\{y \in \mathbb R: y = \lim_{n \to \infty} f(x_{n}),$ for a sequence $x_{n} \to + \infty$.

$$f: \mathbb R \to \mathbb R$$ 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?

• The set for $\sin(x)$ is connected. Every element in $[-1,1]$ occurs as a limit. – Paul K Jan 4 at 13:32
• 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. – Mathsaddict Jan 4 at 13:37
• @Paul K I took sin x as function, and $n\pi$ as sequence, then $f(x_{n})$ converges to 0 – 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.

FINALLY, THE RIGHT ANSWER IS b.

• 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? – Mathsaddict Jan 5 at 3:09
• No, as b. is correct, d. is not the right answer. – mathcounterexamples.net Jan 5 at 7:46
• Yes , I meant 'd'. It was typing error. But what I got from your answer is correct? – Mathsaddict Jan 5 at 14:27
• Yes, that is correct. – mathcounterexamples.net Jan 5 at 15:23