Is saying "the function $f$ is uniformly continuous on I" the same as "the function is continous at each point of $I$"?

I'm confused by my textbook:

Continuity is first defined using its sequential definition:

(1) A function $$f$$ defined on $$A$$ is continuous at a point $$a ∈ A$$ if for each sequence $$(x_n)$$ in $$A$$ such that $$x_n → a$$, we have $$f(x_n) → f(a)$$.

and then:

(2) $$f$$ is continuous (on $$A$$) if $$f$$ is continuous at each point $$a ∈ A$$.

But way later is the $$ε-δ$$ deﬁnition of continuity introduced:

(3) Let the function $$f$$ have domain $$A$$ and let $$c ∈ A$$. Then $$f$$ is continuous at $$c$$ if for each $$ε > 0$$, there exists $$δ > 0$$ such that: $$|f(x) − f(c)| < ε$$, for all $$x ∈ A$$ with $$|x − c| < δ$$.

and the uniform continuity definition:

(4) A function $$f$$ deﬁned on an interval $$I$$ is uniformly continuous on $$I$$ if for each $$ε > 0$$, there exists $$δ > 0$$ such that $$|f(x) −f(y)| < ε$$, for all $$x, y ∈ I$$ with $$|x − y| < δ$$.

It is said then that (1) and (3) are in fact equivalent.

But are (2) and (4) equivalent also?

That is:

Is saying "the function $$f$$ is uniformly continuous on I"(4) the same as "the function is continuous at each point of $$I$$" (2)?

• The answer is NO. In 3) $\delta$ can depend on $c$. If it happens to be independent of $c$ then you get uniform continuity. Aug 22, 2022 at 7:19
• Here $I$ is a compact interval. In that case, it is true that "continuous on $I$" is equivalent to "uniformly continuous on $I$". But that is something that must be proved: here summarized by "It is said then that...". Aug 22, 2022 at 7:22
• Let's play with the example $f:(0, \infty) \to\Bbb{R}$ defined by $$f(x)= \frac{1}{x}$$ Aug 22, 2022 at 7:47

Uniform continuity is a stronger property, i.e. it implies continuity at each point but the reverse is not true. Using $$\varepsilon-\delta$$, a function $$f:A\subset\mathbb{R}\rightarrow\mathbb{R}$$ is uniformly continuous in $$A$$ if $$\forall\varepsilon>0\quad\exists\delta(\varepsilon)>0: \forall x_1,x_2\in A\quad\lvert x_1-x_2\lvert<\delta\implies\lvert f(x_1)-f(x_2)\lvert<\varepsilon$$

On the other hand, $$f$$ is continuous in $$A$$ if f is continuous at each point of $$A$$, i.e.

$$\forall x_0\in A\forall\varepsilon>0\quad\exists\delta(\varepsilon,x_0)>0:\quad\forall x\in A, \lvert x-x_0\lvert<\delta\implies\lvert f(x)-f(x_0)\lvert<\varepsilon$$

Can you see what's different in the two definitions? In the former given $$\varepsilon$$, the number $$\delta$$ is determined, in other words it is the same $$\delta$$ for all the points of the set: uniform continuity is a global property of the function The latter means $$\forall x_0\in A\quad f\text{ is continuous in }x_0$$. The function is continuous (local property) at each point of $$A$$. For each point we have a different situation in general and given $$\varepsilon$$, $$\delta$$ also depends on the point $$x_0$$ we're dealing with.

On an interval $$[a,b]$$ the two statements are equivalent.

i.e. continuity on $$[a,b]$$ implies uniform continuity.

Assume on the contrary that there is an $$\epsilon >0$$ such that for each $$\delta>0$$

there are $$x,y\in [a,b]$$ such that $$|x-y|<\delta$$ and $$|f(x)-f(y)|\geq\,\epsilon$$.

Taking $$\delta=\dfrac{1}{n}$$ we obtain two sequences $$x_{n},y_{n}$$ such that $$|x_{n}-y_{n}|<\dfrac{1}{n}$$ and $$|f(x_{n})-f(y_{n})|\geq\,\epsilon$$.

The sequence $$x_{n}$$ has a convergent subsequence $$x_{n}{_{k}}\to\,\bar{x}$$.

Taking the sequence $$y_{n}{_{k}}$$ we obtain a further subsequence (we don't change indices) such that $$y_{n}{_{k}}\to\,\bar{y}$$.

It is clear by the condition $$|x_{n}{_{k}}-y_{n}{_{k}}|<\dfrac{1}{n_{k}}$$ that $$\bar{x}=\bar{y}$$.

Then by continuity of $$f$$ we get $$lim|f(x_{n}{_{k}})-f(y_{n}{_{k}}|=0\,\geq\,\epsilon>0$$, contradiction.

Therefore continuity on $$[a,b]$$ implies uniform continuity.!

The fact that uniform continuity implies continuity at each point is obvious!