5.4: Continuous Functions

5.4.1 Continuity at a Point

Definition

Suppose $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $a \in D .$ We say $f$ is continuous at $a$ if either $a$ is an isolated point of $D$ or $\lim _{x \rightarrow a} f(x)=f(a) .$ If $f$ is not continuous at $a,$ we say $f$ is discontinuous at $a,$ or that $f$ has a discontinuity at $a .$

Example $\PageIndex{1}$

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$$f(x)=\left\{\begin{array}{ll}{1,} & {\text { if } x \text { is rational, }} \\ {0,} & {\text { if } x \text { is irrational. }}\end{array}\right.$$

Then, by Example $5.1 .5, f$ is discontinuous at every $x \in \mathbb{R}$.

Example $\PageIndex{2}$

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$$f(x)=\left\{\begin{array}{ll}{x,} & {\text { if } x \text { is rational, }} \\ {0,} & {\text { if } x \text { is irrational. }}\end{array}\right.$$

Then, by Example 5.1 .6 and Exercise $5.1 .10, f$ is continuous at $0,$ but discontinuous at every $x \neq 0 .$

If $D \subset \mathbb{R}, \alpha \in \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $g: D \rightarrow \mathbb{R},$ then we define $\alpha f: D \rightarrow \mathbb{R}$ by

$$(\alpha f)(x)=\alpha f(x),$$

$f+g: D \rightarrow \mathbb{R} \mathrm{by}$

$$(f+g)(x)=f(x)+g(x),$$

and $f g: D \rightarrow \mathbb{R}$ by

$$(f g)(x)=f(x) g(x).$$

Moreover, if $g(x) \neq 0$ for all $x \in D,$ we define $\frac{f}{g}: D \rightarrow \mathbb{R}$ by

$$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}.$$

Proposition $\PageIndex{1}$

Suppose $D \subset \mathbb{R}, \alpha \in \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $g: D \rightarrow \mathbb{R} .$ If $f$ and $g$ are continuous at $a,$ then $\alpha f, f+g,$ and $f g$ are all continuous at $a .$ Moreover, if $g(x) \neq 0$ for all $x \in D,$ then $\frac{f}{g}$ is continuous at $a .$

Exercise $\PageIndex{1}$

Prove the previous proposition.

Proposition $\PageIndex{2}$

Suppose $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R}, f(x) \geq 0$ for all $x \in D,$ and $f$ is continuous at $a \in D .$ If $g: D \rightarrow \mathbb{R}$ is defined by $g(x)=\sqrt{f(x)},$ then $g$ is continuous at $a .$

Exercise $\PageIndex{2}$

Prove the previous proposition.

Proposition $\PageIndex{3}$

Suppose $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $a \in D .$ Then $f$ is continuous at $a$ if and only if for every $\epsilon>0$ there exists $\delta>0$ such that

$$|f(x)-f(a)|<\epsilon \text { whenever } x \in(a-\delta, a+\delta) \cap D.$$

Proof

Suppose $f$ is continuous at $a$. If $a$ is an isolated point of $D,$ then there exists a $\delta>0$ such that

$$(a-\delta, a+\delta) \cap D=\{a\}.$$

Then for any $\epsilon>0,$ if $x \in(a-\delta, a+\delta) \cap D,$ then $x=a,$ and so

$$|f(x)-f(a)|=|f(a)-f(a)|=0<\epsilon.$$

If $a$ is a limit point of $D,$ then $\lim _{x \rightarrow a} f(x)=f(a)$ implies that for any $\epsilon>0$ there exists $\delta>0$ such that

$$|f(x)-f(a)|<\epsilon \text { whenever } x \in(a-\delta, a+\delta) \cap D.$$

Now suppose that for every $\epsilon>0$ there exists $\delta>0$ such that

$$|f(x)-f(a)|<\epsilon \text { whenever } x \in(a-\delta, a+\delta) \cap D.$$

If $a$ is an isolated point, then $f$ is continuous at $a$. If $a$ is a limit point, then this condition implies $\lim _{x \rightarrow a} f(x)=f(a),$ and so $f$ is continuous at $a .$ $\quad$ Q.E.D.

From the preceding, it should be clear that a function $f: D \rightarrow \mathbb{R}$ is continuous at a point $a$ of $D$ if and only if for every sequence $\left\{x_{n}\right\}_{n \in I}$ with $x_{n} \in D$ for every $n \in I$ and $\lim _{n \rightarrow \infty} x_{n}=a, \lim _{n \rightarrow \infty} f\left(x_{n}\right)=f(a)$.

Exercise $\PageIndex{1}$

Show that if $f: D \rightarrow \mathbb{R}$ is continuous at $a \in D$ and $f(a)>0$, then there exists an open interval $I$ such that $a \in I$ and $f(x)>0$ for every $x \in I \cap D .$

Proposition $\PageIndex{4}$

Suppose $D \subset \mathbb{R}, E \subset \mathbb{R}, g: D \rightarrow \mathbb{R}, f: E \rightarrow \mathbb{R}, g(D) \subset E$ and $a \in D .$ If $g$ is continuous at $a$ and $f$ is continuous at $g(a),$ then $f \circ g$ is continuous at $a .$

Proof

Let $\left\{x_{n}\right\}_{n \in I}$ be a sequence with $x_{n} \in D$ for every $n \in I$ and $\lim _{n \rightarrow \infty} x_{n}=a$. Then, since $g$ is continuous at $a,\left\{g\left(x_{n}\right)\right\}_{n \in I}$ is a sequence with $g\left(x_{n}\right) \in E$ for every $n \in I$ and $\lim _{n \rightarrow \infty} g\left(x_{n}\right)=g(a) .$ Hence, since $f$ is continuous at $g(a),$ $\lim _{n \rightarrow \infty} f\left(g\left(x_{n}\right)\right)=f(g(a))$. That is,

$$\lim _{n \rightarrow \infty}(f \circ g)\left(x_{n}\right)=(f \circ g)(a).$$

Hence $f \circ g$ is continuous at $a$.

Definition

Let $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $a \in D .$ If $f$ is not continuous at $a$ but both $f(a-)$ and $f(a+)$ exist, then we say $f$ has a simple discontinuity at $a .$

Proposition $\PageIndex{5}$

Suppose $f$ is monotonic on the interval $(a, b) .$ Then every discontinuity of $f$ in $(a, b)$ is a simple discontinuity. Moreover, if $E$ is the set of points in $(a, b)$ at which $f$ is discontinuous, then either $E=\emptyset, E$ is finite, or $E$ is countable.

Proof

The first statement follows immediately from Proposition 5.2.1. For the second statement, suppose $f$ is nondecreasing and suppose $E$ is nonempty. From Exercise 2.1 .26 and the the proof of Proposition $5.2 .1,$ it follows that for every $x \in(a, b)$,

$$f(x-) \leq f(x) \leq f(x+).$$

Hence $x \in E$ if and only if $f(x-)<f(x+) .$ Hence for every $x \in E,$ we may choose a rational number $r_{x}$ such that $f(x-)<r_{x}<f(x+) .$ Now if $x, y \in E$ with $x<y,$ then, by Proposition $5.2 .2,$

$$r_{x}<f(x+) \leq f(y-)<r_{y},$$

so $r_{x} \neq r_{y}$. Thus we have a one-to-one correspondence between $E$ and a subset of $\mathbb{Q},$ and so $E$ is either finite or countable. A similar argument holds if $f$ is nonincreasing. $\quad$ Q.E.D.

Exercise $\PageIndex{4}$

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$$f(x)=\left\{\begin{array}{ll}{\frac{1}{q},} & {\text { if } x \text { is rational and } x=\frac{p}{q},} \\ {0,} & {\text { if } x \text { is irrational. }}\end{array}\right.$$

where $p$ and $q$ are taken to be relatively prime integers with $q>0,$ and we take $q=1$ when $x=0 .$ Show that $f$ is continuous at every irrational number and has a simple discontinuity at every rational number.

5.4.2 Continuity on a Set

Definition

Suppose $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R} .$ We say $f$ is continuous on $D$ if $f$ is continuous at every point $a \in D$.

Proposition $\PageIndex{6}$

If $f$ is a polynomial, then $f$ is continuous on $\mathbb{R}$.

Proposition $\PageIndex{7}$

If $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$ is a rational function, then $f$ is continuous on $D .$

Exercise $\PageIndex{5}$

Explain why the function $f(x)=\sqrt{1-x^{2}}$ is continuous on $[-1,1]$.

Exercise $\PageIndex{6}$

Discuss the continuity of the function

$$f(x)=\left\{\begin{array}{ll}{x+1,} & {\text { if } x<0,} \\ {4,} & {\text { if } x=0,} \\ {x^{2},} & {\text { if } x>0.}\end{array}\right.$$

If $D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$ and $E \subset \mathbb{R},$ we let

$$f^{-1}(E)=\{x: f(x) \in E\}.$$

Proposition $\PageIndex{8}$

Suppose $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R} .$ Then $f$ is continuous on $D$ if and only if for every open set $V \subset \mathbb{R}, f^{-1}(V)=U \cap D$ for some open set $U \subset \mathbb{R} .$

Proof

Suppose $f$ is continuous on $D$ and $V \subset \mathbb{R}$ is an open set. If $V \cap f(D)=\emptyset$, then $f^{-1}(V)=\emptyset,$ which is open. So suppose $V \cap f(D) \neq \emptyset$ and let $a \in f^{-1}(V)$. Since $V$ is open and $f(a) \in V,$ there exists $\epsilon_{a}>0$ such that

$$\left(f(a)-\epsilon_{a}, f(a)+\epsilon_{a}\right) \subset V.$$

Since $f$ is continuous, there exists $\delta_{a}>0$ such that

$$f\left(\left(a-\delta_{a}, a+\delta_{a}\right) \cap D\right) \subset\left(f(a)-\epsilon_{a}, f(a)+\epsilon_{a}\right) \subset V.$$

That is, $\left(a-\delta_{a}, a+\delta_{a}\right) \cap D \subset f^{-1}(V) .$ Let

$$U=\bigcup_{a \in f^{-1}(V)}\left(a-\delta_{a}, a+\delta_{a}\right).$$

Then $U$ is open and $f^{-1}(V)=U \cap D$.

Now suppose that for every open set $V \subset \mathbb{R}, f^{-1}(V)=U \cap D$ for some open set $U \subset \mathbb{R} .$ Let $a \in D$ and let $\epsilon>0$ be given. Since $(f(a)-\epsilon, f(a)+\epsilon)$ is open, there exists an open set $U$ such that

$$U \cap D=f^{-1}((f(a)-\epsilon, f(a)+\epsilon)).$$

Since $U$ is open and $a \in U,$ there exists $\delta>0$ such that $(a-\delta, a+\delta) \subset U .$ But then

$$f((a-\delta, a+\delta) \cap D) \subset(f(a)-\epsilon, f(a)+\epsilon).$$

That is, if $x \in(a-\delta, a+\delta) \cap D,$ then $|f(x)-f(a)|<\epsilon .$ Hence $f$ is continuous at $a .$ $\quad$ Q.E.D.

Exercise $\PageIndex{7}$

Let $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R} .$ For any $E \subset \mathbb{R},$ show that $f^{-1}(\mathbb{R} \backslash E)=\left(\mathbb{R} \backslash f^{-1}(E)\right) \cap D$.

Exercise $\PageIndex{8}$

Let $A$ be a set and, for each $\alpha \in A,$ let $U_{\alpha} \subset \mathbb{R} .$ Given $D \subset \mathbb{R}$ and a function $f: D \rightarrow \mathbb{R},$ show that

$$\bigcup_{\alpha \in A} f^{-1}\left(U_{\alpha}\right)=f^{-1}\left(\bigcup_{\alpha \in A} U_{\alpha}\right)$$

and

$$\bigcap_{\alpha \in A} f^{-1}\left(U_{\alpha}\right)=f^{-1}\left(\bigcap_{\alpha \in A} U_{\alpha}\right).$$

Exercise $\PageIndex{9}$

Suppose $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R} .$ Show that $f$ is continuous on $D$ if and only if for every closed set $C \subset \mathbb{R}, f^{-1}(C)=F \cap D$ for some closed set $F \subset \mathbb{R} .$

Exercise $\PageIndex{10}$

Let $D \subset \mathbb{R} .$ We say a function $f: D \rightarrow \mathbb{R}$ is Lipschitz if there exists $\alpha \in \mathbb{R}, \alpha>0,$ such that $|f(x)-f(y)| \leq \alpha|x-y|$ for all $x, y \in D .$ Show that if $f$ is Lipschitz, then $f$ is continuous.

5.4.3 Intermediate Value Theorem

Theorem $\PageIndex{9}$

(Intermediate Value Theorem).

Suppose $a, b \in \mathbb{R}, a<b,$ and $f:[a, b] \rightarrow \mathbb{R} .$ If $f$ is continuous and $s \in \mathbb{R}$ is such that either $f(a) \leq s \leq f(b)$ or $f(b) \leq s \leq f(a),$ then there exists $c \in[a, b]$ such that $f(c)=s$.

Proof

Suppose $f(a)<f(b)$ and $f(a)<s<f(b) .$ Let

$$c=\sup \{x: x \in[a, b], f(x) \leq s\}.$$

Suppose $f(c)<s .$ Then $c<b$ and, since $f$ is continuous at $c,$ there exists a $\delta>0$ such that $f(x)<s$ for all $x \in(c, c+\delta) .$ But then $f\left(c+\frac{\delta}{2}\right)<s$, contradicting the definition of $c .$ Similarly, if $f(c)>s,$ then $c>a$ and there exists $\delta>0$ such that $f(x)>s$ for all $x \in(c-\delta, c),$ again contradicting the definition of $c .$ Hence we must have $f(c)=s .$ $\quad$ Q.E.D.

Example $\PageIndex{3}$

Suppose $a \in \mathbb{R}, a>0,$ and consider $f(x)=x^{n}-a$ where $n \in \mathbb{Z}, n>1 .$ Then $f(0)=-a<0$ and

$$\begin{aligned} f(1+a) &=(1+a)^{n}-a \\ &=1+n a+\sum_{i=2}^{n}\left(\begin{array}{c}{n} \\ {i}\end{array}\right) a^{i}-a \\ &=1+(n-1) a+\sum_{i=2}^{n}\left(\begin{array}{c}{n} \\ {i}\end{array}\right) a^{i}>0, \end{aligned}$$

where $\left(\begin{array}{l}{n} \\ {i}\end{array}\right)$ is the binomial coefficient

$$\left(\begin{array}{l}{n} \\ {i}\end{array}\right)=\frac{n !}{i !(n-i) !}.$$

Hence, by the Intermediate Value Theorem, there exists a real number $\gamma>0$ such that $\gamma^{n}=a .$ Moreover, there is only one such $\gamma$ since $f$ is increasing on $(0,+\infty) .$

We call $\gamma$ the $n$ th root of $a,$ and write

$$\gamma=\sqrt[n]{a}$$

or

$$\gamma=a^{\frac{1}{n}}.$$

Moreover, if $a \in \mathbb{R}, a<0, n \in Z^{+}$ is odd, and $\gamma$ is the nth root of $-a,$ then

$$(-\gamma)^{n}=(-1)^{n}(\gamma)^{n}=(-1)(-a)=a.$$

That is, $-\gamma$ is the $n$ th root of $a$.

Definition

If $n=\frac{p}{q} \in \mathbb{Q}$ with $q \in \mathbb{Z}^{+},$ then we define

$$x^{n}=(\sqrt[q]{x})^{p}$$

for all real $x \geq 0$.

Exercise $\PageIndex{11}$

Explain why the equation $x^{5}+4 x^{2}-16=0$ has a solution in the interval $(0,2)$.

Exercise $\PageIndex{12}$

Give an example of a closed interval $[a, b] \subset \mathbb{R}$ and a function $f:[a, b] \rightarrow \mathbb{R}$ which do not satisfy the conclusion of the Intermediate Value Theorem.

Exercise $\PageIndex{13}$

Show that if $I \subset \mathbb{R}$ is an interval and $f: I \rightarrow \mathbb{R}$ is continuous, then $f(I)$ is an interval.

Exercise $\PageIndex{14}$

Suppose $f:(a, b) \rightarrow \mathbb{R}$ is continuous and strictly monotonic. Let $(c, d)=f((a, b)) .$ Show that $f^{-1}:(c, d) \rightarrow(a, b)$ is strictly monotonic and continuous.

Exercise $\PageIndex{15}$

Let $n \in \mathbb{Z}^{+} .$ Show that the function $f(x)=\sqrt[n]{x}$ is continuous on $(0,+\infty)$.

Exercise $\PageIndex{16}$

Use the method of bisection to give another proof of the Intermediate Value Theorem.

5.4.4 Extreme Value Theorem

Theorem $\PageIndex{10}$

Suppose $D \subset \mathbb{R}$ is compact and $f: D \rightarrow \mathbb{R}$ is continuous. Then $f(D)$ is compact.

Proof

Given a sequence $\left\{y_{n}\right\}_{n \in I}$ in $f(D),$ choose a sequence $\left\{x_{n}\right\}_{n \in I}$ such that $f\left(x_{n}\right)=y_{n} .$ Since $D$ is compact, $\left\{x_{n}\right\}_{n \in I}$ has a convergent subsequence $\left\{x_{n_{k}}\right\}_{k=1}^{\infty}$ with

$$\lim _{k \rightarrow \infty} x_{n_{k}}=x \in D.$$

Let $y=f(x) .$ Then $y \in f(D)$ and, since $f$ is continuous,

$$y=\lim _{k \rightarrow \infty} f\left(x_{n_{k}}\right)=\lim _{k \rightarrow \infty} y_{n_{k}}.$$

Hence $f(D)$ is compact.

Exercise $\PageIndex{17}$

Prove the previous theorem using the open cover definition of a compact set.

Theorem $\PageIndex{11}$

(Extreme Value Theorem).

Suppose $D \subset \mathbb{R}$ is compact and $f: D \rightarrow \mathbb{R}$ is continuous. Then there exists $a \in D$ such that $f(a) \geq f(x)$ for all $x \in D$ and there exists $b \in D$ such that $f(b) \leq f(x)$ for all $x \in D .$ $\quad$ Q.E.D.

As a consequence of the Extreme Value Theorem, a continuous function on a closed bounded interval attains both a maximum and a minimum value.

Exercise $\PageIndex{18}$

Find an example of a closed bounded interval $[a, b]$ and a function $f:[a, b] \rightarrow \mathbb{R}$ such that $f$ attains neither a maximum nor a minimum value on $[a, b] .$

Exercise $\PageIndex{19}$

Find an example of a bounded interval $I$ and a function $f: I \rightarrow \mathbb{R}$ which is continuous on $I$ such that $f$ attains neither a maximum nor a minimum value on $I .$

Exercise $\PageIndex{20}$

Suppose $K \subset \mathbb{R}$ is compact and $a \notin K .$ Show that there exists $b \in K$ such that $|b-a| \leq|x-a|$ for all $x \in K$.

Proposition $\PageIndex{12}$

Suppose $D \subset \mathbb{R}$ is compact, $f: D \rightarrow \mathbb{R}$ is continuous and one-to-one, and $E=f(D) .$ Then $f^{-1}: E \rightarrow D$ is continuous.

Proof

Let $V \subset \mathbb{R}$ be an open set. We need to show that $f(V \cap D)=U \cap E$ for some open set $U \subset \mathbb{R}$. Let $C=D \cap(\mathbb{R} \backslash V) .$ Then $C$ is a closed subset of $D,$ and so is compact. Hence $f(C)$ is a compact subset of $E .$ Thus $f(C)$ is closed, and so $U=\mathbb{R} \backslash f(C)$ is open. Moreover, $U \cap E=E \backslash f(C)=f(V \cap D) .$ Thus $f^{-1}$ is continuous.

Exercise $\PageIndex{21}$

Suppose $f:[0,1] \cup(2,3] \rightarrow[0,2]$ by

$$f(x)=\left\{\begin{array}{ll}{x,} & {\text { if } 0 \leq x \leq 1,} \\ {x-1,} & {\text { if } 2<x \leq 3.}\end{array}\right.$$

Show that $f$ is continuous, one-to-one, and onto, but that $f^{-1}$ is not continuous.

5.4.5 Uniform Continuity

Definition

Suppose $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R} .$ We say $f$ is uniformly continuous on $D$ if for every $\epsilon>0$ there exists $\delta>0$ such that for any $x, y \in D$,

$$|f(x)-f(y)|<\epsilon \text { whenever }|x-y|<\delta .$$

Exercise $\PageIndex{22}$

Suppose $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$ is Lipschitz (see Exercise $5.4 .10) .$ Show that $f$ is uniformly continuous on $D .$

Clearly, if $f$ is uniformly continuous on $D$ then $f$ is continuous on $D .$ However, a continuous function need not be uniformly continuous.

Example $\PageIndex{4}$

Define $f:(0,+\infty)$ by $f(x)=\frac{1}{x},$ Given any $\delta>0,$ choose $n \in \mathbb{Z}^{+}$ such that $\frac{1}{n(n+1)}<\delta .$ Let $x=\frac{1}{n}$ and $y=\frac{1}{n+1} .$ Then

$$|x-y|=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}<\delta .$$

However,

$$|f(x)-f(y)|=|n-(n+1)|=1.$$

Hence, for example, there does not exist a $\delta>0$ such that

$$|f(x)-f(y)|<\frac{1}{2}$$

whenever $|x-y|<\delta .$ Thus $f$ is not uniformly continuous on $(0,+\infty),$ although $f$ is continuous on $(0,+\infty)$.

Example $\PageIndex{5}$

Define $f: \mathbb{R} \rightarrow \mathbb{R}$ by $f(x)=2 x .$ Let $\epsilon>0$ be given. If $\delta=\frac{\varepsilon}{2}$, then

$$|f(x)-f(y)|=2|x-y|<\epsilon$$

whenever $|x-y|<\delta .$ Hence $f$ is uniformly continuous on $\mathbb{R}$.

Exercise $\PageIndex{23}$

Let $f(x)=x^{2} .$ Show that $f$ is not uniformly continuous on $(-\infty,+\infty)$.

Proposition $\PageIndex{13}$

Suppose $D \subset \mathbb{R}$ is compact and $f: D \rightarrow \mathbb{R}$ is continuous. Then $f$ is uniformly continuous on $D .$

Proof

Let $\epsilon>0$ be given. For every $x \in D,$ choose $\delta_{x}$ such that

$$|f(x)-f(y)|<\frac{\epsilon}{2}$$

whenever $y \in D$ and $|x-y|<\delta_{x} .$ Let

$$J_{x}=\left(x-\frac{\delta_{x}}{2}, x+\frac{\delta_{x}}{2}\right).$$

Then $\left\{J_{x}: x \in D\right\}$ is an open cover of $D$. Since $D$ is compact, there must exist $x_{1}, x_{2}, \ldots, x_{n}, n \in Z^{+},$ such that $J_{x_{1}}, J_{x_{2}}, \ldots, J_{x_{n}}$ is an open cover of $D .$ Let $\delta$ be the smallest of

$$\frac{\delta_{x_{1}}}{2}, \frac{\delta_{x_{2}}}{2}, \ldots, \frac{\delta_{x_{n}}}{2}.$$

Now let $x, y \in D$ with $|x-y|<\delta .$ Then for some integer $k, 1 \leq k \leq n, x \in J_{x_{k}},$ that is,

$$\left|x-x_{k}\right|<\frac{\delta_{x_{k}}}{2}.$$

Moreover,

$$\left|y-x_{k}\right| \leq|y-x|+\left|x-x_{k}\right|<\delta+\frac{\delta_{x_{k}}}{2} \leq \delta_{x_{k}}.$$

Hence

$$|f(x)-f(y)| \leq\left|f(x)-f\left(x_{k}\right)\right|+\left|f\left(x_{k}\right)-f(y)\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .$$

Q.E.D.

Exercise $\PageIndex{24}$

Suppose $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$ is uniformly continuous. Show that if $\left\{x_{n}\right\}_{n \in I}$ is a Cauchy sequence in $D,$ then $\left\{f\left(x_{n}\right)\right\}_{n \in I}$ is a Cauchy sequence in $f(D) .$

Exercise $\PageIndex{25}$

Suppose $f:(0,1) \rightarrow \mathbb{R}$ is uniformly continuous. Show that $f(0+)$ exists.

Exercise $\PageIndex{26}$

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous and $\lim _{x \rightarrow-\infty} f(x)=0$ and $\lim _{x \rightarrow+\infty} f(x)=0 .$ Show that $f$ is uniformly continuous.