Suppose $ f: \mathbb{R} \rightarrow \mathbb{R} $ is a function satisfying: $\displaystyle \lim_{x\rightarrow +\infty} f(x) = \lim_{x\rightarrow -\infty} f(x) = -\infty $

part a) Show that if $(x_n)$ is any sequence of real numbers such that the sequence $(f(x_n))$ either converges to a real number or diverges to $+\infty$, then $(x_n)$ is bounded.

I've done part a)

Part b) Prove that if f is continuous everywhere on $\mathbb{R}$, then f is bounded above. (hint: Show first that if f is not founded above, then there exists a sequence $(x_n)$ with $f(x_n)\rightarrow + \infty$. Does $(x_n)$ have a convergent subsequence?)

For part b), I have constructed a proof without using the hint:

Firstly, consider $f:[R,R] \rightarrow \mathbb{R}$ $R\in \mathbb{R}$ now since $f$ is continuous then as all continuous functions over a closed interval are bounded, so we have $|f(x)| \leq M$ for some real $M$.

We also know that$\displaystyle \lim_{x\rightarrow +\infty} f(x) = \lim_{x\rightarrow -\infty} f(x) = -\infty $ meaning for $x>B>R$ we eventually get $f(x) < 1$ and similarly $x<-B<-R$ we get $f(x) < 1$ for $B>0$ therefore $f$ is bounded above.

I'm I missing anything from the proof above? (1)

Ideally, I'd like to prove it using the hint, here's my attempt using the hint:

I believe the hint is trying to get us to prove the contrapositive, so that was my goal:

if $f$ is not bounded above then take a sequence $(x_n)$ where $x_n>n$ $\forall n \in \mathbb{N}$ then we get $f(x_n) \rightarrow + \infty $as $n\rightarrow \infty$ (is this step correct?).

edit: new attempt: since $\displaystyle \lim_{x\rightarrow +\infty} f(x) = \lim_{x\rightarrow -\infty} f(x) = -\infty $ eventually the function $f$ will begin to decrease, let the point where f begins to decrease be point R: we know that f is bounded in the region $[-R,R]$ by the same argument before, and now we can say that $f(x) < 1$ for some point $B>0$ where $x<-B<-R$ and $x>B>R$

From a), as $(x_n)$ is bounded we know that $|x_n| \leq K $ for some real K. By the Bolzano-Weierstrass theorem we know that there exists a subsequence $(x_{jn})$ s.t. $x_{jn} \rightarrow l$ for some real $l$. Now since f is continuous we have $f(x_{jn}) \rightarrow f(l)$

but I don't really know how to conclude that f is not continuous from here? (2)

Any help answering (1) and/or (2) please, thanks.


Almost correct :)

In your second attempt I think the person grading your homework would appreciate if you would be more explicit at the point where you apply the result of (a). After the subsequence you should be almost done. You conclude that $f(x_{j_n})\to l$, is that a typo? Anyway, you had constructed the sequence $(x_n)$ to such that $f(x_n)$ has a certain property. Now, is there a contradiction somewhere?

Regarding your first attempt: This idea is perfectly fine. Indeed here you use that a continuous function has a maximum on a closed and bounded interval, while in the second attempt you do something which resembles the proof of that fact, so the two are related. However, your execution of the idea is not completely correct. You seem to start with a certain interval $[-R,R]$. Where does that $R$ come from? How is it useful? You later say something about $f(x)$ for $|x|>B>R$. What about $f(x)$ for $R<|x|<B$?

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  • $\begingroup$ On my second attempt: Yes that was a typo, edited now, thanks. Okay, so we constructed a sequence $(x_n)$ so that $f(x_n) \rightarrow +\infty$ but $(x_{jn}) \rightarrow l$ and $ f(x_{jn}) \rightarrow l$ so I'm assuming that as both are going to different limits there is a contradiction as all subsequences have the same limit as the main sequence, so f is bounded above? I'm not very comfortable using this argument as I'm dealing with a function, so if some function approaches infinity for some sequence $x_n$, then if another subsequences $x_{jn}$ causes the function to converge then does that $\endgroup$ – DH. Dec 3 '13 at 20:04
  • $\begingroup$ mean it's a contradiction? (sorry wrote too much): addressing my first attempt: I understand what you're saying. I'm trying to construct f to be bounded between an interval before f starts to decrease, then I wouldn't have to deal with $R<|x|<B$. How would I construct this? $\endgroup$ – DH. Dec 3 '13 at 20:05
  • $\begingroup$ I'm also uncomfortable with saying x_n > n, if $x_n > n$ then how is $x_n$ bounded? (from a) $\endgroup$ – DH. Dec 3 '13 at 20:56

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