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Question. Let $f:\Bbb{R} \to \Bbb{R}$ be a continuous function and $A\subset \Bbb{R}$ be defined by $$A=\{y \in \Bbb{R}: y=\lim_{n\to \infty}f(x_n), \text{for some sequence}~ x_n \to +\infty\}$$ Then the set $A$ is necessarily

A. A connected set

B. A compact set

C. A singleton set

D. None of the above

My attempt. I don't have enough progress in this problem.

First, I want to understand the definition of $A$ properly. Since $f$ is given continuous, then for a sequence $(x_n)$ with $x_n \to +\infty$, $f(x_n)$ may converge or diverge to $+\infty$ or may be oscilatory. So, for a $(x_n)$ with $x_n \to +\infty$ if it happens that $f(x_n)$ is convergent then we collect that limit, $\lim f(x_n)$, in the set $A$.

Now if we take $f(x)=\cos\pi x$ and $x_n=2n+1, y_n=2n$ then both $x_n,y_n\to +\infty$ but $\lim f(x_n)=-1$ and $\lim f(y_n)=1$. So, $\{1,-1\} \subset A$. Therefore Option C is False.

Now I try to prove Option A:

To prove $A$ to be connected I have to show that any continuos function $g: A\to \{\pm 1\}$ is constant. But I cannot proceed with this further.

So I try to prove: $A=cl(B)$ for some connected set $B \subset \Bbb{R}$. Now looking on the description of the set $A$ I try to find out $B$. It seems $B$ should contains all $f(x_n)$'s for some $x_n \to +\infty$. Since, every such $(x_n)$ can be made $>1$ after all but finitely many terms, so $B=\{f(x):x>1\}=f((1,+\infty))$. Since $f$ is continuous so $B$ is connected being continuous image of a connected set and so is $A=cl(B)$. Therefore, Option A is True.

But I cannot conclude Option B. Please help me in B. Is this solution correct? Please tell if there is any mistake in my solution.

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    $\begingroup$ Try $e^x \sin(x)$. $\endgroup$ – JonathanZ Oct 23 '18 at 14:25
  • $\begingroup$ @JonathanZ ...But how to compute $A$ for $e^x \sin(x)$? OR, How to prove it to be non compact? $\endgroup$ – Indrajit Ghosh Oct 23 '18 at 14:29
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    $\begingroup$ If you try to picture the function you will see that for $x \rightarrow + \infty$ it forms periodic waves of increasing amplitude. Thus for any $y \in \mathbb{R}$ there will be a wave that touches it, moreover all the following waves will reach that value. so it is easy to see that you can find an increasing sequence whose limit of the image is that $y$. Therefore $A= \mathbb{R}$. $\endgroup$ – N.B. Oct 23 '18 at 14:34
  • $\begingroup$ @IndrajitGhosh (+1) Very interesting problem. If you don't mind can I ask you what is the source of this problem. I mean from where you got this. $\endgroup$ – StammeringMathematician Oct 23 '18 at 15:11
  • $\begingroup$ @StammeringMathematician...It was given in the TIFR PhD entrance examination in 2015. $\endgroup$ – Indrajit Ghosh Oct 23 '18 at 15:12
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I think the understanding of the question is hampering your progress here.

The meaning of $x_n \to +\infty$ is that for all $r > 0$ there exists $N$ such that $n > N$ implies $x_n > r$. In other words, $x_n$ is eventually larger than every real number.

Now, we are asking if $f(x_n)$ has a limit for any such sequence $x_n \to + \infty$. $A$ is the set of all such limits.


To see the intuition for $A$, try to think of points that would obviously belong to $A$. For example, if there is a sequence $x_n \to +\infty$ such that $f(x_n)$ are all equal. Then, this number will be in $A$.

This allows us to construct $\sin x$ as a counterexample to $A$ being a singleton : it is periodic, so $A$ is the range of this function, which is not a singleton.


For connectedness, you seem to have made a mistake in interpreting $A$. $A$ does not contain $f(x_n)$ for any sequence $x_n \to +\infty$, but rather any limit point that may arise from a choice of $x_n$ for which this is convergent. Points of a sequence are not its limit points, therefore we must be more careful in this aspect.

To show that $A$ is actually connected, we note that $A$ is a subset of the real numbers, and here we have the description that $A$ is connected if and only if it is an interval. This is what must be used : suppose that $a < c < b$ is there with $a,b \in A$, then show that $c \in A$ as well. This is the definition of an interval.

If $a,b \in A$, then there exist two sequences $x_n , w_n \to \infty$ such that $f(x_n) \to a$ and $f(w_n) \to b$. We want to construct $z_n$ such that $z_n \to \infty$ and $f(z_n)$ goes to $c$.

Let us perform a trimming : Let $y_n$ be a subsequence of $w_n$ defined as follows : $y_n = w_{l_n}$, where $l_n = min\{ k > l_{n-1} : w_{k} > x_n\}$, with $l_0 = 0$. Now, $y_n \to +\infty$ (why?) and $y_n > x_n$ (why?). Furthermore, a subsequence of a convergent sequence is convergent, so $f(y_n) \to b$.

First, denote $\delta = \frac{\min(c-a,b-c)}{2}$. Then, by definition of $f(y_n) \to b$ we have $N$ such that $f(y_n) > b-\delta > c$ for $n > N$. Similarly, we have $f(x_n) < a+\delta < c$ for some $n > M$. Taking the maximum, we have for $L = \max \{N,M\}$ that $n > L$ implies $f(x_n) < c < f(y_n)$.

Now, by the intermediate value theorem (applied infinitely many times) for each $x_n < y_n$, after $n > L$ there exists a $z_n$ in between such that $f(z_n) = c$, since $c$ lies between $f(x_n)$ and $f(y_n)$. Reindex $z_n$(it is now defined only for $n> L$) to start from $1$ by shifting. This $z_n$ goes to $+\infty$ (why?) , however we even get that $f(z_n) \to c$ , because it is the constant sequence $c,c,...$!

Hence, the connectedness follows.


As for compactness, the idea is that a compact set is bounded. A continuous function can possible have very wild oscillaing behaviour (if you have seen $\sin \frac 1x$ near zero, like that) near infinity. This oscillating behaviour, like $\sin x$, leads to the creation of many elements of $A$. If this oscillation can be made unbounded, then $A$ can possibly be made unbounded!

That is the thinking behind the counterexample function $x \sin x$.

If you look at this function, it will have increasing oscillation with increasing $x$. This increasing oscillation means, that if you draw a horizontal line at any height on the grid, it will intersect the graph of this function at infinitely many points, starting from some point onwards, and will have a subsequence increasing to infinity.

To prove this a little more rigorously, we will show that $A = \mathbb R$ in this case, with a few steps skipped.

The easy way to see this is to see that $f(n\pi + \frac \pi 2) = n\pi + \frac \pi 2$, and $f(n \pi - \frac{\pi}{2}) = n \pi - \frac \pi 2$ for every $n \geq 1$, so you have two subsequences going to infinity : now use the intermediate value theorem like we did for connectedness between these points to conclude that every point has in its preimage a sequence going to infinity.

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