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# Intermediate Value Theorem. Proving that f(c) = 0 exists watch

1. Suppose f: [1,2] ----> R is continuous function such that for every natural number n, there exists x on [1,2] with |f(x)|<1/n. Show that there exists a number c on [1,2] such that f(c) = 0

I did the following:
|f(x)|<1/n implies -1/n < f(x) <1/n. By sandwich theorem as n tends to infinity, f(x) converges to 0.
Then I thought of applying the Intermediate Value Theorem to prove that f(c) = 0 exists. But for that I need to have f(1)<=f(0)<=f(2) and I am not sure if convergence of f(x) to 0 proves that f(0) lies on that interval (since it tends to 0 but how do we know that it is ever equal to 0?)

2. (Original post by spacewalker)
Suppose f: [1,2] ----> R is continuous function such that for every natural number n, there exists x on [1,2] with |f(x)|<1/n. Show that there exists a number c on [1,2] such that f(c) = 0

I did the following:
|f(x)|<1/n implies -1/n < f(x) <1/n. By sandwich theorem as n tends to infinity, f(x) converges to 0.
This isn't true. When it says that "for every natural number n we can find x with |f(x)|<1/n", it means (colloquially) "give me an n, and I'll give you an x". In other words, the value of x will depend on n. Your argument relies on x not changing with n.

You could say something like:

For all n, we can find x_n with |f(x_n)| < 1/n

and then make the deduction that |f(x_n)| -> 0.

It's not immediate how this helps you; in particular if x_n keeps changing, you've got no "obvious" value to try for 'c'.

That said, I think this is the right place to start, but you've got some work to do to finish from here.

FWIW, unless I'm missing something I don't see the IVT being useful to you here - the tricky case is when f(x) never changes sign.

Edit: this is kind of a spoiler, but have you encountered the Bolzano-Weierstrauss theorem?

2nd edit: You can actually do this just using theorems you should know with very little actual work. I'm not sure it's what would be expected, but it can be done.
3. (Original post by spacewalker)
Suppose f: [1,2] ----> R is continuous function such that for every natural number n, there exists x on [1,2] with |f(x)|<1/n. Show that there exists a number c on [1,2] such that f(c) = 0

I did the following:
|f(x)|<1/n implies -1/n < f(x) <1/n. By sandwich theorem as n tends to infinity, f(x) converges to 0.
Then I thought of applying the Intermediate Value Theorem to prove that f(c) = 0 exists. But for that I need to have f(1)<=f(0)<=f(2) and I am not sure if convergence of f(x) to 0 proves that f(0) lies on that interval (since it tends to 0 but how do we know that it is ever equal to 0?)

The given data implies that f(x) is infinitesimal for some x (due to the 1/n condition), and the only real infinitesimal is 0, so as the range is the reals rather than the hyperreals, f(x) is 0 for some x. Thus the required result is established. For more on infinitesimals and hyperreals, see https://www.math.washington.edu/~mor...ers/gianni.pdf
4. (Original post by HapaxOromenon3)
The given data implies that f(x) is infinitesimal for some x (due to the 1/n condition), and the only real infinitesimal is 0, so as the range is the reals rather than the hyperreals, f(x) is 0 for some x. Thus the required result is established. For more on infinitesimals and hyperreals, see https://www.math.washington.edu/~mor...ers/gianni.pdf
Aside from the fact that hyperreals are definitely not the intended approach, I'm not entirely convinced you have justified that "f(x) is infinitesimal for some x".
5. (Original post by DFranklin)
Aside from the fact that hyperreals are definitely not the intended approach, I'm not entirely convinced you have justified that "f(x) is infinitesimal for some x".
This follows from letting n be infinite, so that 1/n is infinitesimal, and so |f(x)|<1/n implies f(x) is infinitesimal.
6. (Original post by HapaxOromenon3)
This follows from letting n be infinite, so that 1/n is infinitesimal, and so |f(x)|<1/n implies f(x) is infinitesimal.
As I said, x depends on n. Because of this, I don't see how your conclusion follows. It absolutely requires more justification than you have provided.
7. (Original post by DFranklin)
As I said, x depends on n. Because of this, I don't see how your conclusion follows. It absolutely requires more justification than you have provided.
"for every natural number n, there exists x on [1,2] with |f(x)|<1/n"
so if we let n be infinite (which is part of the standard extension of the natural numbers to the hyperreals), there exists x with 1<=x<=2 such that |f(x)|<1/infinity, i.e. f(x) is infinitesimal.
8. (Original post by HapaxOromenon3)
"for every natural number n, there exists x on [1,2] with |f(x)|<1/n"
so if we let n be infinite (which is part of the standard extension of the natural numbers to the hyperreals), there exists x with 1<=x<=2 such that |f(x)|<1/infinity, i.e. f(x) is infinitesimal.
The given hypotheses apply to the usual natural numbers, which don't include infinity, so I don't see how "letting n be infinity" is an option. Even if it can be done this way, where have you used continuity? This question is not true for discontinuous functions.

In any case, this clearly isn't the intended approach for the given question and is likely to just to confuse the OP. The question is a lot more about the Bolzano-Weierstrass Theorem and how continuous functions treat limits.
9. (Original post by RichE)
The given hypotheses apply to the usual natural numbers, which don't include infinity, so I don't see how "letting n be infinity" is an option. Even if it can be done this way, where have you used continuity? This question is not true for discontinuous functions.
I'm thinking continuity comes in via the "non-standard analysis treatment of infinitesimals". However, there's something *else* he hasn't used that is also pretty crucial (the result is false without it). [Not trying to be coy to you, but I don't want to spell it out for Hapax].

The problem I have with non-standard analysis is that it seems very easy to convince yourself that "I don't need to worry about XXX, because ... infinitesimal" when this isn't actually the case. Pretty sure that's going on here.

In any case, this clearly isn't the intended approach for the given question and is likely to just to confuse the OP. The question is a lot more about the Bolzano-Weierstrass Theorem and how continuous functions treat limits.
Agreed on it not being the intended approach. However, there's a "just quote a theorem" proof of the result that I'm wondering might have been what was actually intended:

Spoiler:
Show
|f| attains its bounds, but the bound must be 0 from the condition given.

What do you think?
10. (Original post by DFranklin)

What do you think?
Quite possibly - I guess it depends whether that theorem has appeared in the course yet.
11. (Original post by RichE)
Quite possibly - I guess it depends whether that theorem has appeared in the course yet.
Was just curious since B-W seemed the "natural" way to prove it, and then I although realised the other method was easier, it still seems a lot "sneakier" and harder to spot, somehow. It's a nice compact argument though.
12. (Original post by DFranklin)
This isn't true. When it says that "for every natural number n we can find x with |f(x)|<1/n", it means (colloquially) "give me an n, and I'll give you an x". In other words, the value of x will depend on n. Your argument relies on x not changing with n.

You could say something like:

For all n, we can find x_n with |f(x_n)| < 1/n

and then make the deduction that |f(x_n)| -> 0.

It's not immediate how this helps you; in particular if x_n keeps changing, you've got no "obvious" value to try for 'c'.

That said, I think this is the right place to start, but you've got some work to do to finish from here.

FWIW, unless I'm missing something I don't see the IVT being useful to you here - the tricky case is when f(x) never changes sign.

Edit: this is kind of a spoiler, but have you encountered the Bolzano-Weierstrauss theorem?

2nd edit: You can actually do this just using theorems you should know with very little actual work. I'm not sure it's what would be expected, but it can be done.
Thank you very much for your reply and all your comments! To be honest I find it very difficult to wrap my mind around such problems but I attempted to correct my approach. Please see below:
For each x on [1,2] we can find n such that |f(x)|<1/n.
So we can create a sequence x_n (which is bounded) so that |f(x_n)|<1/n. By Bolzano-Weierstrass theorem any bounded sequence has a convergent subsequence. So, we have x_t ( which a subsequence of x_n) converging to limit L for example with |f(x_t)|<1/t. Since f is continuous we have lim (x_t) = L (L is on (1,2)) implies lim f(x_t) = f(L) and lim (+/-)1/t = 0. Thus, by sandwich theorem 0< f(L)<0. Therefore, f(L) = 0 and so such c exists and it is equal to L
Thanks
13. (Original post by spacewalker)
For each x on [1,2] we can find n such that |f(x)|<1/n.
I don't think this is quite what you mean.

Also you want weak inequalities surrouding f(L).
14. (Original post by RichE)
I don't think this is quite what you mean.

Also you want weak inequalities surrouding f(L).
Thank you for your reply! Could you please explain on how I can get a weak inequality? It is just the question states |f(x)|<1/n and so the inequality is given to be strict.
15. (Original post by spacewalker)
Thank you for your reply! Could you please explain on how I can get a weak inequality? It is just the question states |f(x)|<1/n and so the inequality is given to be strict.
Firstly you don't want to conclude 0<f(L)<0 as *no* number satisfies this.

Secondly limits only respect weak inequalities. Note

1/n > 0 does not imply 0 = lim (1/n) > 0

but does imply

and such a result holds generally for weak inequalities
16. (Original post by RichE)
Firstly you don't want to conclude 0<f(L)<0 as *no* number satisfies this.

Secondly limits only respect weak inequalities. Note

1/n > 0 does not imply 0 = lim (1/n) > 0

but does imply

and such a result holds generally for weak inequalities
I agree. However, as far as I understand the only way to get this is to say that
|f(x_t)|<= 1/t but I am not sure why we can say that since we are given f(x_n)<1/n
17. (Original post by DFranklin)
I'm thinking continuity comes in via the "non-standard analysis treatment of infinitesimals". However, there's something *else* he hasn't used that is also pretty crucial (the result is false without it). [Not trying to be coy to you, but I don't want to spell it out for Hapax].

The problem I have with non-standard analysis is that it seems very easy to convince yourself that "I don't need to worry about XXX, because ... infinitesimal" when this isn't actually the case. Pretty sure that's going on here.
The idea that I was trying to convey is that as n gets larger and larger, f(x) is forced into smaller and smaller intervals of the form [-k, k], and since f(x) is continuous, these intervals must eventually "converge" (I don't know if that's precisely the right word) to the single point 0.
18. (Original post by spacewalker)
I agree. However, as far as I understand the only way to get this is to say that
|f(x_t)|<= 1/t but I am not sure why we can say that since we are given f(x_n)<1/n
If you know |f(x_t)| < 1/t then it is the case that |f(x_t)|<= 1/t
19. (Original post by HapaxOromenon3)
The idea that I was trying to convey is that as n gets larger and larger, f(x) is forced into smaller and smaller intervals of the form [-k, k], and since f(x) is continuous, these intervals must eventually "converge" (I don't know if that's precisely the right word) to the single point 0.
With all due respect, nothing you have posted about this (in any of your posts) seems to get even close to being a coherent argument that could be extended to a valid proof, let alone standing as a proof in its own right.

I don't think it is helpful for you to continue derailing this thread. If you want to discuss non-standard analysis, start your own thread (but be warned that I doubt many will care to reply to it).
20. (Original post by spacewalker)
I agree. However, as far as I understand the only way to get this is to say that
|f(x_t)|<= 1/t but I am not sure why we can say that since we are given f(x_n)<1/n
RichE's comment about is correct, but note that it's also fine to go from

to .

That is, limits don't respect strong inequalities, but applying a limit to a strong inequality gives a valid weak inequality.

[This is really no different from what RichE is saying, just a slightly different way of thinking about it].

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