intro to analysis - continuity & sequences

Watch
zoxe
Badges: 7
Rep:
?
#1
Report Thread starter 6 years ago
#1
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

(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.

DONE a)

(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?)

attempt of proof not using the hint:

Since\displaystyle \lim_{x\rightarrow +\infty} f(x) = \lim_{x\rightarrow -\infty} f(x) = -\infty
we know that at some point, say  M the function will be always below the point M, i.e. for some  M >0  f(x) < M always. similarly for a point  -M now consider  f: [-M,M] \rightarrow \mathbb{R} since, f is continuous on the reals we know that for any closed interval f will be bounded (we proved this theorem in lecture). Hence, since  f is bounded in the interval  [-M,M] and it's bounded above everywhere outside that region then f is always bounded above.

Is there anything wrong with the above proof?

Now, I'm trying to prove part (b) using the hint:

so, suppose f is not bounded above, i.e.  f(x) > R for some real R. Now, if we construct a sequence  x_n > n for all natural numbers n, then  f(x_n) \rightarrow +\infty as  n \rightarrow \infty From part (a), we know that  x_n is bounded, i.e.  |x_n |\leq H for some real H. Now by the bolzano-weierstrass theorem we know that  (x_n) has a convergent subsequence. Denote this subsequence by  (x_{jn}) . We now know that  x_{jn} \rightarrow l so  f(x_{jn}) \rightarrow f(l) .

Now i'm not sure where to proceed. I'm trying to prove the contrapositive, i.e. if f is not bounded, then it is not continuous. I think I see a contradiction, i.e. we have  f(x_n) \rightarrow +\infty but  f(x_{jn}) \rightarrow l and since  (x_{jn}) is a subsequence of  (x_n) should  \lim_{n\rightarrow \infty} f(x_n) = \lim_{n\rightarrow \infty} f(x_{jn}) = +\infty ? If so we haven't proved that in lecture so I'm not sure that's the answer needed. Furthermore, I'm uncomfortable with construction  (x_n) , firstly I said  x_n > n and  f(x_n) \rightarrow + \infty but is this true only if  x_n > n always? As that's not the case because I later on stated (from the proof of part a) that  (x_n) is bounded!

Any help please, thank you.

(Original post by Farhan.Hanif93)
.
Helped me with my last thread, any ideas?
0
reply
BlueSam3
Badges: 17
Rep:
?
#2
Report 6 years ago
#2
(Original post by zoxe)
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

(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.

DONE a)

(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?)

attempt of proof not using the hint:

Since\displaystyle \lim_{x\rightarrow +\infty} f(x) = \lim_{x\rightarrow -\infty} f(x) = -\infty
we know that at some point, say  M the function will be always below the point M, i.e. for some  M >0  f(x) < M always.
This is assuming the result.

similarly for a point  -M now consider  f: [-M,M] \rightarrow \mathbb{R} since, f is continuous on the reals we know that for any closed interval f will be bounded (we proved this theorem in lecture). Hence, since  f is bounded in the interval  [-M,M] and it's bounded above everywhere outside that region then f is always bounded above.

Is there anything wrong with the above proof?
Yes - you assume your result. You are also using M to mean at least two completely different things.

Now, I'm trying to prove part (b) using the hint:

so, suppose f is not bounded above, i.e.  f(x) > R for some real R. Now, if we construct a sequence  x_n > n for all natural numbers n, then  f(x_n) \rightarrow +\infty as  n \rightarrow \infty From part (a), we know that  x_n is bounded, i.e.  |x_n |\leq H for some real H. Now by the bolzano-weierstrass theorem we know that  (x_n) has a convergent subsequence. Denote this subsequence by  (x_{jn}) . We now know that  x_{jn} \rightarrow l so  f(x_{jn}) \rightarrow f(l) .
At this point, note that since f is continuous, f(x_n) \to f(\lim_{n\to\infty} x_n) as n \to \infty, and use the uniqueness of limits.

Furthermore, I'm uncomfortable with construction  (x_n) , firstly I said  x_n > n and  f(x_n) \rightarrow + \infty but is this true only if  x_n > n always? As that's not the case because I later on stated (from the proof of part a) that  (x_n) is bounded![/quote]

I have no idea why you insisted on x_n > n - it's entirely unnecessary. All you need to do is note that if f is unbounded above, then for any natural n, there is a point, call it x_n, such that f(x_n) > n (by the definition of "unbounded above".
0
reply
zoxe
Badges: 7
Rep:
?
#3
Report Thread starter 6 years ago
#3
(Original post by BlueSam3;45408060At this point, note that since  is continuous, [tex]f(x_n) \to f(\lim_{n\to\infty} x_n) as n \to \infty, and use the uniqueness of limits.
But what is  f(\lim_{n\to\infty} x_n) ? could it not be  f(l)? How do I know it's different from f(l) (where I can use the uniqueness of limits?)

Also, for the first proof, I don't think I'm assuming the result? I'm just assuming what is given to me.
0
reply
BlueSam3
Badges: 17
Rep:
?
#4
Report 6 years ago
#4
(Original post by zoxe)
But what is  f(\lim_{n\to\infty} x_n) ? could it not be  f(l)? How do I know it's different from f(l) (where I can use the uniqueness of limits?)


You know that said limit is precisely f(l), since you have that f is continuous. Thus, f(l) \in \mathbb{R}. However, look at how you defined your x_n

Also, for the first proof, I don't think I'm assuming the result? I'm just assuming what is given to me.
You assumed that there is some M > 0 such that for all x \in \mathbb{R}, f(x) < M. This is exactly and precisely the definition of a function being bounded above.
0
reply
zoxe
Badges: 7
Rep:
?
#5
Report Thread starter 6 years ago
#5
(Original post by BlueSam3)
You know that said limit is precisely f(l), since you have that f is continuous. Thus, f(l) \in \mathbb{R}. However, look at how you defined your x_n
double post
0
reply
zoxe
Badges: 7
Rep:
?
#6
Report Thread starter 6 years ago
#6
(Original post by BlueSam3)
You know that said limit is precisely f(l), since you have that f is continuous. Thus, f(l) \in \mathbb{R}. However, look at how you defined your x_n
So we assumed it's continuous, and we have f(x_{jn}) \to f(l), but from the uniquness of limits f(x_jn) should also diverge as  f(x_n) \rightarrow +\infty so we've proved the contrapositive (as it's therefore not continuous), and we're done?
0
reply
BlueSam3
Badges: 17
Rep:
?
#7
Report 6 years ago
#7
(Original post by zoxe)
So we assumed it's continuous, and we have f(x_{jn}) \to f(l), but from the uniquness of limits f(x_jn) should also diverge as  f(x_n) \rightarrow +\infty so we've proved the contrapositive (as it's therefore not continuous), and we're done?
Yes.
0
reply
zoxe
Badges: 7
Rep:
?
#8
Report Thread starter 6 years ago
#8
(Original post by BlueSam3)
Yes.
Thank you for your help
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (706)
33.93%
Yes, I like the idea of applying to uni after I received my grades (PQA) (889)
42.72%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (393)
18.89%
I think there is a better option than the ones suggested (let us know in the thread!) (93)
4.47%

Watched Threads

View All
Latest
My Feed