Don't understand the proof: continuous function defined on a closed and bounded interval is uniformly continuous?

Question

I am reading book An Introduction To Analysis By William R.Wade, the theorem is stated as:

Suppose that $I$ is a closed, bounded interval. If $f:I\rightarrow R$ is continuous on $I$. Then $f$ is uniformly continuous on $I$.

The given proof starts from:

Suppose to the contrary that $f$ is continuous but not uniformly continuous on $I$. Then there is an $\epsilon_0>0$ and points $x_n,y_n\in I$ such that $|x_n-y_n|<\frac{1}{n}$ and $$|f(x_n)-f(y_n)|\geq\epsilon_0,n\in N.$$

I have no difficulty to follow this assumption and derive the contradiction (relates to this question), to me the even harder part is to understand why the contrary to

If $f:I\rightarrow R$ is continuous on $I$. Then $f$ is uniformly continuous on $I$

Can is formed into

There is an $\epsilon_0>0$ and points $x_n,y_n\in I$ such that $|x_n-y_n|<\frac{1}{n}$ and $|f(x_n)-f(y_n)|\geq\epsilon_0,n\in N.$

Shouldn't it be easier to say

There is an $\epsilon_0>0$ and points $x,y\in I$ such that $|x-y|<\delta$ for arbitrary small $\delta$ and $|f(x)-f(y)|\geq\epsilon_0,n\in N?$

Because then by theorem $$|x|<\epsilon,\forall\epsilon>0\Longleftrightarrow x=0,$$

we have $x=y,$ therefore $|f(x)-f(y)|=0$ contrary to $|f(x)-f(y)|>\epsilon_0>0$. Why would it be necessary to constract sequence and use Bolzano-Weierstrass and Comparison Theorem to complete the proof?

Answer 1

$\textbf{Not uniform continuity:---}$ There is $\epsilon_0$ such that, for each $\delta>0$ we have $x_{\delta},y_{\delta}\in I$,

this $x_{\delta},y_{\delta}$ may depend on the choice of $\delta$,

with $|x_{\delta}-y_{\delta}|<\delta$ but $|f(x_{\delta})-f(y_{\delta})|\geq \epsilon_0$.

$\textbf{Proof:---}$Hence there is an $\epsilon_0>0$ and points $x_n,y_n\in I$ such that $|x_n-y_n|<\frac{1}{n}$ and $|f(x_n)-f(y_n)|\geq\epsilon_0,n\in \Bbb N.$ Bolzano-Weierstrass theorem says, $\{x_n\}$ has a convergent subsequnce say, $\{x_{n_k}\}$ converges to $x\in I$. Then, $| x_{n_k}-y_{n_k}|<\frac{1}{n_k}$ implies $\{y_{n_k}\}$ also converges to $x$. Continuity of $f$ implies both $\{f(x_{n_k})\}$ and $\{f(y_{n_k})\}$ converges to $f(y)$. Therefore, $$\lim_{k\to \infty}|f(x_{n_k})-f(y_{n_k})|=0.$$

Answer 2

The negation of

If $f\colon I\longrightarrow\mathbb R$ is continuous, then $f$ is uniformly continuous.

is

There is a function $f\colon I\longrightarrow\mathbb R$ which is continuous but not uniformly continuous.

And asserting that $f$ is not uniformly continuous means that$$(\exists\varepsilon>0)(\forall\delta>0)(\exists x,y\in I):\lvert x-y\rvert<\delta\wedge\bigl\lvert f(x)-f(y)\bigr\rvert\geqslant\varepsilon.$$

Now, suppose that this takes place. Take such a $\varepsilon>0$. Then, if $n\in\mathbb N$, $\frac1n>0$. Therefore, there are numbers $x_n,y_n\in I$ such that $\lvert x_n-y_n\rvert<\frac1n$ and that $\bigl\lvert f(x_n)-f(y_n)\bigr\rvert\geqslant\varepsilon$.

Answer 3

Negation of uniform continuity: \begin{align*} &\neg(\forall\epsilon\exists\delta\forall x,y(|x-y|<\delta\rightarrow|f(x)-f(y)|<\epsilon))\\ &=\exists\epsilon\neg(\exists\delta\forall x,y(|x-y|<\delta\rightarrow|f(x)-f(y)|<\epsilon))\\ &=\exists\epsilon\forall\delta\neg(\forall x,y(|x-y|<\delta\rightarrow|f(x)-f(y)|<\epsilon))\\ &=\exists\epsilon\forall\delta\exists x,y\neg(|x-y|<\delta\rightarrow|f(x)-f(y)|<\epsilon)\\ &=\exists\epsilon\forall\delta\exists x,y\neg((\neg(|x-y|<\delta))\vee(|f(x)-f(y)|<\epsilon))\\ &=\exists\epsilon\forall\delta\exists x,y((|x-y|<\delta)\wedge(\neg(|f(x)-f(y)|<\epsilon))).\\ &=\exists\epsilon\forall\delta\exists x,y(|x-y|<\delta)\wedge(|f(x)-f(y)|\geq\epsilon) \end{align*}

Answer 4

The existing answers seem to me to be giving insufficient attention to the most important point here, which is exactly why your simplified version is wrong.

Suppose $f$ is not uniformly continuous. Then there exists $\epsilon_0>0$ such that

True Fact(TF). For every $\delta>0$ there exist $x,y$ with $|f(x)-f(y)|\ge\epsilon_0$ and $|x-y|<\delta$.

Here $x$ and $y$ depend on $\delta$, in particular we do not know

False Fact(FF). There exist $x,y$ such that $|f(x)-f(y)|\ge\epsilon_0$ and $|x-y|<\delta$ for all $\delta>0$.

Your simplified proof of the theorem would be right if the FF were true, but it's not so it's not.