Proving or disproving uniform continuity of various real functions 
Okay before anyone does anything, I don't want any of you guys to just write out proofs for all of these, that's asking a bit much :P
Maybe just do one, and make it detailed because I really need to see how a problem like this is done. I need a lot of exposure and practice to how to write a proof like this..
also, if anyone can clarify this for me, i'd really appreciate it: I fail to see how $\epsilon$ and $\delta$ relate in the definition of uniform continuity. How does the $\delta$ (unique if the function actually is uniform continuous) actually impose anything on $\epsilon$? this is actually even less clear in "regular" continuity.
 A: We say that $a_n$ and $b_n$ are equivalent if $a_n-b_n \to 0$. If $f$ is uniformly continuous then maps equivalent sequences to equivalent sequences.
1) Consider the sequences $a_n =(1/n)$  and $b_n=(1/2n)$ both are equivalent in $(0,1]$. But $f(a_n)$, $f(b_n)$ are not equivalent. So $f$ is not uniformly continuous in $(0,1]$. 
[Alternatively you can show that if $f$ is  uniformly continuous then maps Cauchy sequences to Cauchy sequences, and in particular for this case, $(1/n)$ is a Cauchy sequence  in $(0,1]$, but $f(1/n)$ is not a Cauchy sequence].
2) You can show that uniformly continuous using $\varepsilon$-$\delta$ definition is logically equivalent to say that $f$ maps equivalent sequences to equivalent sequences, i.e., $f(a_n)-f(b_n)\to 0$ whenever $a_n -b_n \to 0$. Using this second definition: 
Given $(a_n), (b_n) \subset [1, \infty)$ and $a_n-b_n \to 0$. We shall show that $f(a_n)-f(b_n) \to 0$. Let $\varepsilon>0$ be arbitrary and choose $n_0>0$ such that $|a_n-b_n|<\varepsilon$ for all $n\ge n_0$. Thus
$$|f(a_n)-f(b_n)|= \bigg|\frac{1}{a_n}-\frac{1}{b_n} \bigg|=\frac{|a_n-b_n|}{a_n b_n}\le |a_n-b_n|< \varepsilon$$
This is possible since $1\le a_n, b_n$. Hence $f(a_n)-f(b_n) \to 0$.
3) Consider the sequences $a_n = n$ and $b_n = n+1/n$ both are equivalent sequences in $[0,\infty)$. But $f(b_n)=n^2+2+1/n^2=f(a_n)+2+1/n^2$, so $f(b_n)-f(a_n)\ge2$. Thus we can conclude that is not uniformly continuous. 
4) Since $f$ is continuous on $[0,1]$ then must be uniformly continuous (why?). Let $(0,1) \hookrightarrow [0,1]$. We claim that is uniformly continuous, let $x_n-y_n\to 0$ and $(x_n),(y_n) \subset (0,1)$, so $i(x_n)-i(y_n) = x_n -y_n \to 0$, since the sequences are arbitrary it holds for all equivalent sequences and so $i$ is uniformly continuous.  
We claim that $f\circ i$ is uniformly continuous. We shall show that maps equivalent sequences to equivalent sequences. Since $i(x_n)-i(y_n)\to 0$ and $f$ is uniformly continuous then $ f(i(x_n))-f(i(y_n))=(f\circ i)(x_n)-(f \circ i)(y_n) \to 0$. 
Hence $f\circ i$ is uniformly continuous. But $f\circ i=f \restriction _{(0,1)} $  which is just the square root function define on $(0,1)$.
5) Suppose that  $(a_n),(b_n) \subset [1,\infty)$ and $a_n-b_n \to 0$. We shall show that $a_n^{1/2}-b_n^{1/2} \to 0$. Given $\varepsilon>0$, choose $n_0$ such that $|a_n-b_n|< \varepsilon$ for all $n\ge n_0$. Thus 
$$|f(a_n)-f(b_n)|=|\sqrt{a_n}-\sqrt{b_n}|= \bigg|\frac{a_n-b_n}{\sqrt{a_n}+\sqrt{b_n}} \bigg|=\frac{|a_n-b_n|}{\sqrt{a_n}+\sqrt{b_n}}\le \frac{|a_n-b_n|}{2}< \varepsilon$$
Since $1\le a_n, b_n$.
A: As you're lacking a positive answer, I'll do the second. We have $$|f(x)-f(y)|=|\frac{1}{x}-\frac{1}{y}|=\frac{|x-y|}{xy}\le|x-y|$$
as you have $x\ge1$ and $y\ge1$.
A: My answer is:
i) No.
ii) Yes.
iii) No.
iv) Yes.
v) Yes.
