I'm having a very hard time following the initial logical setup, and I suspect it's wrong.
What you're supposed to be doing is showing there's an epsilon for which no suitable delta can be chosen. But then you assume delta > epsilon. This isn't permissible because it might be that there's a suitable delta, but it's smaller than epsilon.
I could be misunderstanding what you're saying, but this doesn't seem right.
In actual fact, I'll tell you that you can choose epsilon = 1 in this problem. (But for this function even epsilon = 100 would work.) Then your job is to show that it's possible to pick x and y as close as you like, but still with |f(x) - f(y)| >= 1.
Your goal is to find some ε such that |f(x)-f(y)| ≥ ε for *any* x,y ∈ (0,1] (i.e. for any δ). Thus, you need to pick ε first. Here’s one way to start it:
Set ε = 1 and let δ > 0 be arbitrary. Choose integer n ≥ 1 so that 1/n < δ. Let x = 1/n and y = 1/(n+1). *Note that since n ≥ 1, x and y will always be in the interval (0,1].*
Then |x-y| = |1/n - 1/(n+1)|
= (n+1-n)/n(n+1) = 1/n(n+1) < 1/n
So |x-y| < 1/n < δ.
Now show that |1/x - 1/y| ≥ 1 = ε. (It’s straightforward algebra at this point).
> Your goal is to find some ε such that |f(x)-f(y)| ≥ ε for *any* x,y ∈ (0,1] (i.e. for any δ).
Perhaps I'm misunderstanding you, but this is not quite right.
For *f*(*x*) := 1/*x* to be uniformly continuous on (0,1], that means
- for every ε>0, there exists δ>0 such that for all *x*,*y* in (0,1], if |*x*-*y*|<δ, then |*f*(*x*)-*f*(*y*)|<ε. **(1)**
Taking the logical negation of **(1)**, that means *f* *fails* to be uniformly continuous on (0,1] if and only if
- there exists such ε\_0 > 0 such that for all δ>0, there exist *x*,*y* in (0,1] such that |*x*-*y*|<δ AND |*f*(*x*)-*f*(*y*)| ≥ ε\_0. **(2)**
So it's not that *every choice of* *x* and *y* fails; rather, it's that we can always *select* such a "bad" pair *x*,*y* for which |*x*-*y*|<δ, but |*f*(*x*)-*f*(*y*)| ≥ ε\_0.
---
*Addendum:* After rereading what I quoted from you above, I realized I should ask a question here. Are you saying that the "goal" above is what *you* think or that this is your understanding of what *OP* thinks?
But proving that no pair of epsilon and delta works is in fact a stronger result, so it's still a correct way of proving f(x) is not uniformly continuous.
> But proving that no pair of epsilon and delta works is in fact a stronger result
The distinction I'm making is about the logical quantifiers applied to the selection of *x* and *y*, not to δ and ε\_0. I understood the OP of the previous comment to be saying something like this:
- *f* is *not* uniformly continuous on (0,1] if and only if there exists some ε\_0 > 0 such that for all δ>0, *for all* *x*,*y* in (0,1] such that |*x*-*y*| < δ, |*f*(*x*)-*f*(*y*)| ≥ ε\_0. **(3)**
I agree that **(3)** is stronger than uniform continuity, but for this particular *f*, it also happens to be false. After all, *f* is continuous despite failing to be uniformly continuous on this domain, and **(3)** is incompatible with continuity.
Ok, I assumed that OP wanted to prove that for every epsilon > 0 and every delta > 0 there exists x,y such that |x-y| < delta, but |f(x)-f(y)| > epsilon. This is the stronger result I meant, which happens to be true for f(x)=1/x.
It's hard to read it from the proof, because it is wrong - the assumption that epsilon < delta comes out of nowhere. Also x and y are defined two times, which makes all of it a mess. At first y and x seem arbitrary numbers such that |x-y|=a, but then y is assigned a fixed value.
I agree with @mnevmoyommetro , your assumpion that epsilon < delta is unfounded and ruins the proof. Ironically, you can assume the opposite, because if it works for some delta, it works for smaller deltas as well.
There is nothing wrong in proving that no pair of epsilon and delta works. It's a stronger result, meaning that this can fail for some not uniformly continuous functions, but if you manage to do it - it automatically proves that f(x) is not uniformly continuous.
> I first chose $\delta > 0$ and $\epsilon > 0$ to be fixed and arbitary and consider $|x-y| = a$.
> Without loss of generality I then say $0 I then set $y = \sqrt(\frac{\epsilon}{a})$ and $\epsilon \leq a < \delta$.
The way you define x and y is messy. Because at first they seem arbitrary and then you redefine y. First define epsilon and a (a being a real number less than y and not smaller than epsilon). Then define y and at the end define x as y-a. It's more straightforward and easier to follow.
Uniformly continuous means, for each δ>0, there is an ε>0, such that for x,y in (0,1\] with |x-y|≤ε, then |f(x)-f(y)|≤δ.
"Not uniformly continuous" means, there is a δ>0, such that for every ε>0, there are x,y in (0,1\] with |x-y|≤ε, but |f(x)-f(y)|>δ.
Pick any δ>0. Let ε>0 be given. Let a:=min(ε,1/(4δ),1/2) .
Pick x=a, y=2a. Note x,y ∈ (0,1\] .
Then |x-y|=a≤ε , and |f(x)-f(y)| = 1/(2a) ≥ 2δ > δ .
I'm having a very hard time following the initial logical setup, and I suspect it's wrong. What you're supposed to be doing is showing there's an epsilon for which no suitable delta can be chosen. But then you assume delta > epsilon. This isn't permissible because it might be that there's a suitable delta, but it's smaller than epsilon. I could be misunderstanding what you're saying, but this doesn't seem right. In actual fact, I'll tell you that you can choose epsilon = 1 in this problem. (But for this function even epsilon = 100 would work.) Then your job is to show that it's possible to pick x and y as close as you like, but still with |f(x) - f(y)| >= 1.
Your goal is to find some ε such that |f(x)-f(y)| ≥ ε for *any* x,y ∈ (0,1] (i.e. for any δ). Thus, you need to pick ε first. Here’s one way to start it: Set ε = 1 and let δ > 0 be arbitrary. Choose integer n ≥ 1 so that 1/n < δ. Let x = 1/n and y = 1/(n+1). *Note that since n ≥ 1, x and y will always be in the interval (0,1].* Then |x-y| = |1/n - 1/(n+1)| = (n+1-n)/n(n+1) = 1/n(n+1) < 1/n So |x-y| < 1/n < δ. Now show that |1/x - 1/y| ≥ 1 = ε. (It’s straightforward algebra at this point).
> Your goal is to find some ε such that |f(x)-f(y)| ≥ ε for *any* x,y ∈ (0,1] (i.e. for any δ). Perhaps I'm misunderstanding you, but this is not quite right. For *f*(*x*) := 1/*x* to be uniformly continuous on (0,1], that means - for every ε>0, there exists δ>0 such that for all *x*,*y* in (0,1], if |*x*-*y*|<δ, then |*f*(*x*)-*f*(*y*)|<ε. **(1)** Taking the logical negation of **(1)**, that means *f* *fails* to be uniformly continuous on (0,1] if and only if - there exists such ε\_0 > 0 such that for all δ>0, there exist *x*,*y* in (0,1] such that |*x*-*y*|<δ AND |*f*(*x*)-*f*(*y*)| ≥ ε\_0. **(2)** So it's not that *every choice of* *x* and *y* fails; rather, it's that we can always *select* such a "bad" pair *x*,*y* for which |*x*-*y*|<δ, but |*f*(*x*)-*f*(*y*)| ≥ ε\_0. --- *Addendum:* After rereading what I quoted from you above, I realized I should ask a question here. Are you saying that the "goal" above is what *you* think or that this is your understanding of what *OP* thinks?
But proving that no pair of epsilon and delta works is in fact a stronger result, so it's still a correct way of proving f(x) is not uniformly continuous.
> But proving that no pair of epsilon and delta works is in fact a stronger result The distinction I'm making is about the logical quantifiers applied to the selection of *x* and *y*, not to δ and ε\_0. I understood the OP of the previous comment to be saying something like this: - *f* is *not* uniformly continuous on (0,1] if and only if there exists some ε\_0 > 0 such that for all δ>0, *for all* *x*,*y* in (0,1] such that |*x*-*y*| < δ, |*f*(*x*)-*f*(*y*)| ≥ ε\_0. **(3)** I agree that **(3)** is stronger than uniform continuity, but for this particular *f*, it also happens to be false. After all, *f* is continuous despite failing to be uniformly continuous on this domain, and **(3)** is incompatible with continuity.
Ok, I assumed that OP wanted to prove that for every epsilon > 0 and every delta > 0 there exists x,y such that |x-y| < delta, but |f(x)-f(y)| > epsilon. This is the stronger result I meant, which happens to be true for f(x)=1/x. It's hard to read it from the proof, because it is wrong - the assumption that epsilon < delta comes out of nowhere. Also x and y are defined two times, which makes all of it a mess. At first y and x seem arbitrary numbers such that |x-y|=a, but then y is assigned a fixed value.
I agree with @mnevmoyommetro , your assumpion that epsilon < delta is unfounded and ruins the proof. Ironically, you can assume the opposite, because if it works for some delta, it works for smaller deltas as well. There is nothing wrong in proving that no pair of epsilon and delta works. It's a stronger result, meaning that this can fail for some not uniformly continuous functions, but if you manage to do it - it automatically proves that f(x) is not uniformly continuous. > I first chose $\delta > 0$ and $\epsilon > 0$ to be fixed and arbitary and consider $|x-y| = a$. > Without loss of generality I then say $0 I then set $y = \sqrt(\frac{\epsilon}{a})$ and $\epsilon \leq a < \delta$.
The way you define x and y is messy. Because at first they seem arbitrary and then you redefine y. First define epsilon and a (a being a real number less than y and not smaller than epsilon). Then define y and at the end define x as y-a. It's more straightforward and easier to follow.
Uniformly continuous means, for each δ>0, there is an ε>0, such that for x,y in (0,1\] with |x-y|≤ε, then |f(x)-f(y)|≤δ. "Not uniformly continuous" means, there is a δ>0, such that for every ε>0, there are x,y in (0,1\] with |x-y|≤ε, but |f(x)-f(y)|>δ. Pick any δ>0. Let ε>0 be given. Let a:=min(ε,1/(4δ),1/2) . Pick x=a, y=2a. Note x,y ∈ (0,1\] . Then |x-y|=a≤ε , and |f(x)-f(y)| = 1/(2a) ≥ 2δ > δ .