# intro to analysis - continuity & sequences

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#1
Suppose is a function satisfying:

(a) Show that if is any sequence of real numbers such that the sequence either converges to a real number or diverges to , then is bounded.

DONE a)

(b) Prove that if f is continuous everywhere on , then f is bounded above.
(hint: Show first that if f is not founded above, then there exists a sequence with . Does have a convergent subsequence?)

attempt of proof not using the hint:

Since
we know that at some point, say the function will be always below the point M, i.e. for some always. similarly for a point now consider 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 is bounded in the interval 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. for some real R. Now, if we construct a sequence for all natural numbers n, then as From part (a), we know that is bounded, i.e. for some real H. Now by the bolzano-weierstrass theorem we know that has a convergent subsequence. Denote this subsequence by . We now know that so .

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 but and since is a subsequence of should ? 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 , firstly I said and but is this true only if always? As that's not the case because I later on stated (from the proof of part a) that is bounded!

(Original post by Farhan.Hanif93)
.
Helped me with my last thread, any ideas?
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6 years ago
#2
(Original post by zoxe)
Suppose is a function satisfying:

(a) Show that if is any sequence of real numbers such that the sequence either converges to a real number or diverges to , then is bounded.

DONE a)

(b) Prove that if f is continuous everywhere on , then f is bounded above.
(hint: Show first that if f is not founded above, then there exists a sequence with . Does have a convergent subsequence?)

attempt of proof not using the hint:

Since
we know that at some point, say the function will be always below the point M, i.e. for some always.
This is assuming the result.

similarly for a point now consider 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 is bounded in the interval 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 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. for some real R. Now, if we construct a sequence for all natural numbers n, then as From part (a), we know that is bounded, i.e. for some real H. Now by the bolzano-weierstrass theorem we know that has a convergent subsequence. Denote this subsequence by . We now know that so .
At this point, note that since is continuous, as , and use the uniqueness of limits.

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

I have no idea why you insisted on - it's entirely unnecessary. All you need to do is note that if is unbounded above, then for any natural , there is a point, call it , such that (by the definition of "unbounded above".
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#3
(Original post by BlueSam3;45408060At this point, note that since as , and use the uniqueness of limits.
But what is ? could it not be ? 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.
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6 years ago
#4
(Original post by zoxe)
But what is ? could it not be ? 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 , since you have that is continuous. Thus, . However, look at how you defined your

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 such that for all . This is exactly and precisely the definition of a function being bounded above.
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#5
(Original post by BlueSam3)
You know that said limit is precisely , since you have that is continuous. Thus, . However, look at how you defined your
double post
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#6
(Original post by BlueSam3)
You know that said limit is precisely , since you have that is continuous. Thus, . However, look at how you defined your
So we assumed it's continuous, and we have, but from the uniquness of limits f(x_jn) should also diverge as so we've proved the contrapositive (as it's therefore not continuous), and we're done?
0
6 years ago
#7
(Original post by zoxe)
So we assumed it's continuous, and we have, but from the uniquness of limits f(x_jn) should also diverge as so we've proved the contrapositive (as it's therefore not continuous), and we're done?
Yes.
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#8
(Original post by BlueSam3)
Yes.
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