Continuity and uniform continuity of $1/x$ I was reviewing lecture notes from a term I just finished, and my lecturer leaves an example that illustrates the difference between continuity and uniform continuity but I'm not sure I understand it.
We are told that on closed intervals uniform continuity = continuity and on open intervals uniform continuity is strong than continuity.
i.e. Let $f : (0,1] \to [1,\infty) $. $ f $ is continuous on $(0,1]$ but not uniformly continuous. Here is my proof, there are bits that are not correct and I'd like to clear up the confusion.
So we know that f is continuous if $\exists c \in (0,1]$ s.t $\forall \epsilon > 0 \exists \delta>0$ s.t if $|x-c| < \delta \Rightarrow |f(x) - f(c)| < \epsilon $.
i.e. $|x-y| < \delta  \Rightarrow |1/x - 1/c| = |\frac{x-c}{xc}| < \epsilon$
$|\frac{x-c}{xc}| < |\frac{\delta}{xc}|$ but from here do I set epsilon to $|\frac{\delta}{xc}|$ ? But then epsilon depends on $x$? Anyway if I could clean this up then i would be able to show its continuous, but I cant quite manage the last step. After I do this, how do I show its not uniformly continuous?
 A: The method in proving that $f(x)=\frac{1}{x}$ is not uniformly continuous is similar to the method in proving that $f(x)=\frac{1}{x^2}$. I will address $f(x)=\frac{1}{x^2}$ , and you can infer from it how to prove $f(x)=\frac{1}{x}$.
Assume that $f(x)=\frac{1}{x^2}$ is not uniformly continuous. This means that there exists $\varepsilon>0$ and $\delta>0$  such that $$\vert x-y\vert<\delta\Rightarrow\vert f(x)-f(y)\vert\geq\varepsilon \tag 1$$ (This is the reverse of the definition of uniform continuity)
As Michael Hardy pointed out, to prove that $f(x)$ is not uniformly continuous, we simply need to show that (1) is true for some value of $\varepsilon$. This is because the definition of uniform continuity is for all $\varepsilon$. Hence, we only need to show that uniform continuity fails (or that (1) holds) for one value of $\varepsilon$. Now, let $\varepsilon=1$ and let $y=x+\frac{\delta}{2}$ (the value of $\varepsilon$ and $y$ were chosen arbitrarily...we do this to change the equation to one variable which simplifies things). Thus we have
$$\vert f(x)-f(x+\frac{\delta}{2})\vert\geq 1$$
Thus we have that $$\left| \frac{1}{x}-\frac{1}{x+\frac{1}{\delta}}\right|\geq 1$$
We can rearrange the expression in the absolute values by using the fact that $f(x)-f(y)=\frac{(y-x)(y+x)}{x^2y^2}$ (let $y=x+\frac{\delta}{2}$) 
Thus, we have that $$1\leq\frac{(x+\frac{\delta}{2}-x)(x+\frac{\delta}{2}+x)}{x^2(x+\frac{\delta}{2})^2}$$
Arbitrarily let $x=\delta$, then we have 
$$\frac{5\delta^2}{9\delta^4}\geq 1$$. Note that we can pick a delta that satisfies such that the above inequality holds. For example, we could let $\delta=1/3$. Therefore, there exists $\delta$ such that (1) holds.
A: Or you could use the theorem of uniform continuity regarding Cauchy sequences. A theorem states that if $f$ is uniformly continuous on a set $S$ and $(s_n)$ is a Cauchy sequence in $S$, then $f(s_n)$ is a Cauchy sequence. 
Choose $(s_n)=\frac{1}{n}$ (here $n\in\mathbb{N}$), which is a Cauchy sequence. However, $f(s_n)=n$. Since $f((s_n))=n$ is not Cauchy, $f$ is not uniformly continuous.
Both proofs that I have given are from Kenneth Ross's "Elementary Analysis".
A: Setting $\varepsilon$ to something doesn't make sense.  You need to take $\varepsilon$ to be given, and find a value of $\delta$ that's small enough.
Continuity should not say $\exists c\in(0,1]$ etc., where $c$ is in the role you put it in.  Rather, continuity at the point $c$ should be defined by what comes after that.
Uniform continuity says
$$
\forall\varepsilon>0\ \exists\delta>0\ \forall x\in(0,1]\ \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right).
$$
Lack of uniform continuity is the negation of that:
$$
\text{Not }\forall\varepsilon>0\ \exists\delta>0\ \forall x\in(0,1]\ \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right). \tag 1
$$
The way to negate $\forall\varepsilon>0\ \cdots\cdots$ to by a de-Morganesque law that says $\left(\text{not }\forall\varepsilon>0\ \cdots\cdots\right)$ is the same as $(\exists\varepsilon>0\ \text{not }\cdots\cdots)$, and similarly when "not" moves past $\forall$, then that transforms to $\exists$.  So $(1)$ becomes
$$
\exists\varepsilon>0\text{ not }\exists\delta>0\ \forall x\in(0,1]\ \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 2
$$
and that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\text{ not }\forall x\in(0,1]\ \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 3
$$
and that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\ \exists x\in(0,1]\text{ not } \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 4
$$
and that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\ \exists x\in(0,1]\ \exists y\in(0,1]\text{ not } \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 5
$$
and that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\ \exists x\in(0,1]\ \exists y\in(0,1] \left(|x-y|<\delta\text{ and not }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 6
$$
and finally that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\ \exists x\in(0,1]\ \exists y\in(0,1] \left(|x-y|<\delta\text{ and }\left|\frac1x-\frac1y\right|\ge\varepsilon\right). \tag 7
$$
To show that such a value of $\varepsilon$ exists, it is enough to show that $\varepsilon=1$ will serve.  You need to find $x$ and $y$ closer to each other than $\delta$ but having reciprocals differing by more than $1$.  It is enough to make both $x$ and $y$ smaller than $\delta$ and then exploit the fact that there's a vertical asymptote at $0$ to make $x$ and $y$ far apart, by pushing one of them closer to $0$.
A: Take $\epsilon=1$. Let $\delta>0$. We can always choose $n$ sufficiently large and $x=\frac{1}{n}$ and $y=\frac{1}{n+1}$ so that
$$
x-y=\frac{1}{n\left(n+1\right)}<\delta.
$$
However,
$$
\left|f\left(\frac{1}{n}\right)-f\left(\frac{1}{n+1}\right)\right|=\left|n-\left(n+1\right)\right|=1.
$$
