# Pick out the correct choices -TIFR 2015

Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function and $A \subset \mathbb R$ be defined by $A=\{y \in \mathbb R:y=\lim _{n\rightarrow \infty}f(x_n),$for some sequence $x_n\rightarrow \infty\}$

Then the set $A$ is necessarily

A.a connected set

B.compact set

C. a singleton set

D.None of above

Now since $f$ is a continuous function and $x_n$ diverges so $f(x_n)$ will also diverge.Hence the set can't be bounded and hence not compact

Also it may not be singleton as $f(x_n)$ may be either $\infty$ or -$\infty$

Not sure with A.Can someone please check my solution and suggest required edits

Continuous functions take convergent sequences to convergent sequences, but you can't say the same for divergent sequences. For instance, consider a constant function. For a better example, which will also give a hint for how to solve your problem, take $f(x) = \sin(x)$. For a still better example, think of how you could modify this $f$ to change the set $A$...
• May be $x_n=n\pi$ divergent but $f(x_n)$ is not.But I did not get what you are asking about modification – Learnmore Jan 15 '15 at 14:47
• @learnmore: if $f(x)=\sin(x)$, then $A=[-1,1]$, which is both compact and connected. But if you modify $f$ slightly, you can make $A$ non-compact. Think about that for an hour or two before coming straight back with "I don't get it". – TonyK Jan 15 '15 at 14:56
• should I modify $f$ such that its image becomes $[-1,1)$ or $(-1,1]$ which is not closed and hence not compact @TonyK – Learnmore Jan 16 '15 at 2:34