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)?

  • $\begingroup$ The answer is NO. In 3) $\delta$ can depend on $c$. If it happens to be independent of $c$ then you get uniform continuity. $\endgroup$ Aug 22, 2022 at 7:19
  • $\begingroup$ 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...". $\endgroup$
    – GEdgar
    Aug 22, 2022 at 7:22
  • $\begingroup$ Let's play with the example $f:(0, \infty) \to\Bbb{R}$ defined by $$f(x)= \frac{1}{x}$$ $\endgroup$ Aug 22, 2022 at 7:47

2 Answers 2


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!


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