I am reading Abbott's Understanding Analysis wherein the sequential criterion for the absence of uniform continuity is given as follows:
A function $f : A \to\mathbb R$ fails to be uniformly continuous on $A$ if and only if there exists a particular $\epsilon_0 > 0$ and two sequences ($x_n$) and ($y_n$) in A satisfying $|x_n − y_n| \to 0$ but $|f(x_n) − f(y_n)| \ge \epsilon_0$.
I am not able to understand this completely. When I look at it as a negation of the definition of uniform continuity, it seems to be okay to my understanding but when I consider $f$ to be only continuous, I don't understand how to comprehend it.
So if $f$ is continuous on $A$ then for every $\epsilon$ nbd of $f(x)$, there exists a $\delta$ nbd of $x$ such that whenever $|x - y| < \delta$, $|f(x) - f(y)| < \epsilon$. Since $f$ is only continuous, the criterion above is valid here which says that there does exist an $\epsilon_0$ for which no $\delta$ will work.
Doesn't this contradict the definition of continuity which say that for every possible $\epsilon$, we can find a $\delta$?
Also as an exercise, I have to prove that if $f$ is uniformly continuous and $A$ is bounded then $f(A)$ is bounded.
I think I will have to use the criterion mentioned above to prove this. By considering a convergent subsequence in $A$, since $A$ is bounded and then by assuming that $f(A)$ is unbounded I may arrive at a contradiction that $f$ is not uniformly continuous.
I am hesitant to use the criterion in the proof because clearly, I don't understand it completely or maybe, not at all.
Please help me understand the criterion. Thanks in advance :)
The idea behind uniform convergence is that there is a single $\varepsilon$ which serves to control the rate of convergence. In other words, if you have a sequence $\{x_n\}_{n\in {\mathbb N}}$ and a function $f$ such that $f(x_n)$ is uniformly convergent then there is some $\delta(\varepsilon)$ which depends on $\varepsilon$ so that $|f(x_n)-f(x_m)|<\varepsilon$ whenever $|x_n-x_m|<\delta$. In particular, once you choose your $\varepsilon$ that specifies the values you can choose for $\delta$ and whenever you take points of the sequence that are that close together, the corresponding values of $f$ are no further apart than $\varepsilon$.
Now, take two (infinite) subsequences of $\{x_n\}_{n\in {\mathbb N}}$, say $\{y_{n_k}\}_{k\in {\mathbb N}}$ and $\{z_{n_k}\}_{k\in {\mathbb N}}$. Since they are subsequences of a convergent sequence they must converge to the same point, i.e.
$$y_{n_k} \rightarrow x \quad \mbox{ and } \quad z_{n_k} \rightarrow x $$ which means that $|y_{n_k} - z_{n_k}| \rightarrow 0$. Since $f$ is uniformly continuous, this means that there exists some $\delta(\varepsilon)$ which depends on $\varepsilon$ so that $|f(y_{n_k})-f(z_{n_k})|<\varepsilon$ whenever $|y_{n_k}-z_{n_k}|<\delta$.
So, if $f$ is not uniformly convergent, there must be some $\varepsilon_0$ such that $|f(y_{n_k})-f(z_{n_k})|>\varepsilon_0$ even though $|y_{n_k}-z_{n_k}|<\delta$, i.e. $|f(y_{n_k})-f(z_{n_k})|>\varepsilon_0$ as $|y_{n_k}-z_{n_k}| \rightarrow 0$.
We can generalise this to your definition by noting that if there are any two sequences with this property then $f$ cannot be uniformly continuous (because we can combine those two sequences into one, of which they are subsequences, and then the reasoning above holds).
I think the comment above should cover the second part for you quite adequately.