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This should be a very easy question but it might just be that I'm confusing myself. So we have the definition of a function $f$ on $S$ being continuous at $x_0$:

For any $\epsilon$>0, there exists $\delta>0$ such that: whenever $|x-x_0|<\delta$, we have $|f(x)-f(x_0)|<\epsilon$

And I assume the negation is

There exists $\epsilon$>0 such that for all $\delta>0$, $|x-x_0|<\delta$ yet $|f(x)-f(x_0)|\ge \epsilon$.

Now I want to show that the function $f(x)=\sin(\frac{1}{x})$ together with $f(0)=0$ cannot be made into a continuous function at $x=0$. So I need to show that there exists $\epsilon>0$ such that for all $\delta>0$, $|x|<\delta$ yet $|f(x)|\ge\epsilon$.

Let $\epsilon = \frac{1}{2}$. Then no matter what $\delta$ we choose, let $|x|<\frac{1}{2}$. It is certainly possible that $|f(x)|\ge \frac{1}{2}$, because, well, $\frac{1}{x}$ can really take on arbitrarily large value as $x$ is small.

Now, what confuses me is that, as $x$ gets small, $f(x)$ can certainly be greater than $\frac{1}{2}$ for infinitely many times, but it will be less than that infinitely many times, too. But I suppose it doesn't really matter. So I think there's something wrong with my negation but I couldn't figure out where.

Update: The correct version can be found here. Watch for Lemma 4.6

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  • $\begingroup$ @DavidMitra I knew it... So I guess there should be "for all $x\in S$" in the original definition in front of "whenever", and should be "there exists $x' \in S$ in the negation in from of "$|x-x_0| < \delta$"? $\endgroup$ – 3x89g2 Jul 13 '16 at 8:53
  • $\begingroup$ Oh, I was off... You're right: in the negation, write: ... for each $\delta>0$, there exists $x'$ such that $|x'-x_0|<\delta$ yet ... $\endgroup$ – David Mitra Jul 13 '16 at 9:02
  • $\begingroup$ But is the definition (quote 2 in post) not incomplete? excerpt: ... $\delta>0$, $|x-x_0| < \delta$... it could be $\forall x_0$ or it could be $\exists x_0$ which makes a significant change. How is this implicit? $\endgroup$ – IceFire Jul 13 '16 at 9:30
  • $\begingroup$ It is always better to write the definition of continuity using quantifiers because it will help you to formulate the negation more easily. The definition of continuity can be written as, $$(∀ε>0)[∃δ(ε,a)>0[(∀x∈S)[|x−a|<δ⟹|f(x)−f(a)|<ε]]]$$ Its negation is, $$(∃ε>0)[∀δ(ε,a)>0[(∃x∈S)[|x−a|<δ⟹|f(x)−f(a)|≥ε]]]$$ Compare the statement to the original one to see how I have formulated the negation. $\endgroup$ – user170039 Jul 14 '16 at 9:30
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The negation is:

There exists $ϵ>0$ such that for all $δ>0$, there is an $x_\delta$ such that $ |x_\delta−x_0|<δ$ yet $|f(x_\delta)−f(x_0)|≥ϵ$

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Your negation is correct; you should specify though that what you're defining is continuity at the point $x_0$, which is distinct from continuity in the whole domain $S$.

Your choice of $\epsilon=1/2$ is fine. However you need to do some more work to show that $f$ can't be continuous. Suppose we try to make $f$ into a continuous function by assigning $f(0)=y_0$. Take any $\delta>0$.

Case 1: Suppose $y_0<0$. Let $x=1/(\pi/2+2\pi N)$ where $N$ is chosen large enough so $|x|<\delta$. Then $|f(x)-f(x_0)|=|1-y_0|\geq1>\epsilon$ which proves discontinuity.

Case 2: Suppose $y_0\geq0$. Let $x=1/(-\pi/2+2\pi N)$ where $N$ is chosen large enough so $|x|<\delta$. Then $|f(x)-f(x_0)|=|-1-y_0|\geq1>\epsilon$ which again proves discontinuity.

Thus we conclude there's no choice of $y_0=f(0)$ which makes $f$ continuous at zero.

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The negation is:

there exists $\epsilon >0$ such that for any $\delta>0$ we can find an $x$ such that $|x-x_0|<\delta$ and $|f(x)-f(x_0)| > \epsilon$.

And you have just proved this.

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  • $\begingroup$ That's exactly what I wanted to confirm. Thanks. $\endgroup$ – 3x89g2 Jul 13 '16 at 9:40
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    $\begingroup$ Surely this is not the negation. It has the quantifiers for $\delta$ and $\epsilon$ the wrong way around: it begins $\forall\delta\exists\epsilon$ whereas the original proposition is $\forall\epsilon\exists\delta$. $\endgroup$ – Gareth McCaughan Jul 13 '16 at 13:10
  • $\begingroup$ But $\epsilon$ can get smaller and smaller when choosing smaller $\delta$s I think, so that doesn't necessarily mean that $f$ isn't continuous on $x_0$. @GarethMcCaughan I don't know much about maths, so can you please explain how do you know what the correct order of the quantifiers is? $\endgroup$ – zagoku Jul 13 '16 at 16:16
  • $\begingroup$ It's possible that the two turn out to be equivalent or something; I haven't thought about it. But the proposition in this answer is certainly not simply the negation of the original one, even if they turn out to be equivalent for reasons of topology or real analysis. $\endgroup$ – Gareth McCaughan Jul 13 '16 at 16:51
  • $\begingroup$ The order of quantifiers doesn't change when you negate things. The relevant equivalences are (1) "not (for all x, P(x))" = "for some x, not P(x)" and (2) "not (for some x, P(x))" = "for all x, not P(x)". So the negation of "for all epsilon, for some delta, P" is "for some epsilon, for all delta, not P". $\endgroup$ – Gareth McCaughan Jul 13 '16 at 16:54

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