# Dense set and continuous functionsWatch

#1
I've been thinking about this problem for the past week and I've put my attempt below. The question is as follows: if f is continuous and f(x) = 0 for all x in a dense set then f(x) = 0 for all x.

Attempt

is not empty and bounded above since f is continuous. In fact, it attains its bounds at, say, t for the supremum.

Surely since for some (If sup S<0 then we are lead to a contradiction.)

Since is continuous at , for every there exists a such that for all if then Note that since is the sup of f then Therefore

(Dubious bit follows!)
This holds for all x satisfying , but there is an where by density. This means that and therefore that .

The approach for is essentially the same and we conclude that Hence f(x) = 0 for all

And finally my questions:
1. Does the attempted proof only hold for [a,b]? I mean it says it needs to be proved "for all x", but surely I can extend the interval [a,b] as large as I please? Sorry for asking this in a confusing way, but I'm really confused and I'm not even sure what I mean by "as large as I please". I guess given any number x, I can find an interval where x is contained in and the (attempted) proof holds. Hence the "for all x".
2. If it doesn't hold for all x, and just [a,b], is the attempted proof correct?

Thanks for any help!
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9 years ago
#2
Is f: R --> R?
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#3
(Original post by generalebriety)
Is f: R --> R?
Sorry, I don't know. The question is almost exactly as to how the book states it. I presume the "for all x" in the question means the domain is R? The book doesn't really use any notation for an indication of what the domain of f is...
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9 years ago
#4
Alright, I shall assume f: R --> R.

(Original post by AsakuraMinamiFan)
there is an where by density.
Why? (It's fairly intuitively obvious, but you might want to refer to the definition of 'dense'.)

(Original post by AsakuraMinamiFan)
1. Does the attempted proof only hold for [a,b]?
It holds for every [a,b]. That means, if you pick an x, you can always pick a<x<b, show it's true on [a,b], and therefore show f(x) = 0. So f(x) = 0 for all x.
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#5
(Original post by generalebriety)
Why? (It's fairly intuitively obvious, but you might want to refer to the definition of 'dense'.)
Definition: every open interval contains a point of A. Because is an open interval, and thus there is a point of A where f(x) = 0.

(Original post by generalebriety)
It holds for every [a,b]. That means, if you pick an x, you can always pick a<x<b, show it's true on [a,b], and therefore show f(x) = 0. So f(x) = 0 for all x.
Thanks, that sort of makes sense. (I mean, it's what I was getting at, I think?) But it's still sort of confusing; I'll give it a while to ponder over.
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9 years ago
#6
(Original post by AsakuraMinamiFan)
Definition: every open interval contains a point of A.
Ok, cool. Not the definition I'm used to, but if it's yours, then that's fine.
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#7
(Original post by generalebriety)
Ok, cool. Not the definition I'm used to, but if it's yours, then that's fine.
I wasn't aware of another definition. Sorry, I should have probably stated it.

Can I assume you have no objections to the rest of the proof? Sorry to be hasty but this method is quite new to me and I'm surprised I stumbled upon it. (I was stuck at the 'dubious' bit for a while.) I wonder if I can use a similar method to prove the IVT...

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9 years ago
#8
(Original post by AsakuraMinamiFan)
Since is continuous at , for every there exists a such that for all if then
This isn't right. The conclusion should be .

You can "fix" this by using the result that a cts function on a closed interval attaines it's supremum (that is, you can find a t with ), but it's not really the right approach.

Better approach: suppose there is an x with . Then by continuity, we can find so that f(y) > a/2 for all y such that . This contradicts f being zero on a dense subset of [a,b].
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#9
(Original post by DFranklin)
You can "fix" this by using the result that a cts function on a closed interval attaines it's supremum (that is, you can find a t with ), but it's not really the right approach.
Sorry, this is my poor writing in demonstration! I did say in my original post that it attains it at t, but it was way at the top hidden from everything.

(Original post by DFranklin)
Better approach: suppose there is an x with . Then by continuity, we can find so that f(y) > a/2 for all y such that . This contradicts f being zero on a dense subset of [a,b].
Ah, I had completely forgot about this lemma, even though it was used in the proof of the IVT! That makes sense though.

I see you didn't seem to use the completeness of R in your proof? (I think so, at least.) Is that what you mean by saying "it's not really right"? I must admit, I only attempted it this way because the chapter is called "least upper bounds"...
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9 years ago
#10
(Original post by AsakuraMinamiFan)
Sorry, this is my poor writing in demonstration! I did say in my original post that it attains it at t, but it was way at the top hidden from everything.
Yes you did, sorry. But in that case you need to explictly state why it attains it's bounds. That is, something like "f is cts on a closed interval, therefore it is bounded and there exists t with f(t) = sup f".

I see you didn't seem to use the completeness of R in your proof? (I think so, at least.) Is that what you mean by saying "it's not really right"? I must admit, I only attempted it this way because the chapter is called "least upper bounds"...
Yes. You're bringing in lots of extra machinery when it isn't needed. The actual result is a straightforward consequence of the definitions of continuity and density.
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9 years ago
#11
If not, then there's a point where f is non-zero. But then there's an open set where f is non-zero, by continuity. So the set where f is zero is not dense in the reals.
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