# 5.4: Continuous Functions

- Page ID
- 22667

## 5.4.1 Continuity at a Point

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 \(a .\)

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 \(x \in \mathbb{R}\).

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

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

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 \(a .\)

Prove the previous proposition.

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 \(a .\)

Prove the previous proposition.

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

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 \(x \in I \cap D .\)

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 \(a\).

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 \(a .\)

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.

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

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 \(a \in D\).

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

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

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

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

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

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.

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

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

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

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 \(F \subset \mathbb{R} .\)

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

(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.

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 \(a\).

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 \(x \geq 0\).

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

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.

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

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.

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

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

##### 5.4.4 Extreme Value Theorem

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.

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

(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.

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 \([a, b] .\)

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 \(I .\)

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 \(x \in K\).

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.

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

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 \(x, y \in D\),

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

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.

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 \((0,+\infty)\).

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 \(\mathbb{R}\).

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

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.

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 \(f(D) .\)

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

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.