# intro to analysis - continuity & sequences

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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!

Any help please, thank you.

Helped me with my last thread, any ideas?

(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!

Any help please, thank you.

(Original post by

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**Farhan.Hanif93**).

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#2

(Original post by

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.

**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.

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?

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 .

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 .

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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#4

(Original post by

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?)

**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.

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(Original post by

You know that said limit is precisely , since you have that is continuous. Thus, . However, look at how you defined your

**BlueSam3**)You know that said limit is precisely , since you have that is continuous. Thus, . However, look at how you defined your

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**BlueSam3**)

You know that said limit is precisely , since you have that is continuous. Thus, . However, look at how you defined your

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#7

(Original post by

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?

**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?

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