Continuity

Suppose $f$ is a real function defined on $\mathbb {R}^{1}$ which satisfies

for every $x \in \mathbb {R}^{1}$. Does this imply that $f$ is continuous?

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that

for every set $E \subset X$. ($\overline{E}$ denotes the closure of $E$). Show, by an example, that $f(E)$ can be a proper subset of $\overline{f(E)}$.

Let $f$ be a continuous real function on a metric space $X$. Let $Z(f)$ (the zero set of $f$) be the set of all $p \in X$ at which $f(p) = 0$. Prove that $Z(f)$ is closed.

Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. If $g(p) = f(p)$ for all $p \in E$, prove that $g(p) = f(p)$ for all $p \in X$. (In other words, a continuous mapping is determined by its values on a dense subset of its domain).

If $f$ is a real continuous function defined on a closed set $E \subset \mathbb {R}^{1}$, prove that there exist continuous real functions $g$ on $\mathbb {R}^{1}$ such that $g(x) = f(x)$ for all $x \in E$. (Such functions $g$ are called continuous extensions of $f$ from $E$ to $\mathbb {R}^{1}$). Show that the result becomes false if the word "closed" is omitted. Extend the result to vectorvalued functions. Hint: Let the graph of $g$ be a straight line on each of the segments which constitute the complement of $E$ (compare Exercise 29, Chap. 2). The result remains true if $\mathbb {R}^{1}$ is replaced by any metric space, but the proof is not so simple.

If $f$ is defined on $E$, the graph of $f$ is the set of points $(x, f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane.

Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.

If $E \subset X$ and if $f$ is a function defined on $X$, the restriction of $f$ to $E$ is the function $g$ whose domain of definition is $E$, such that $g(p) = f(p)$ for $p \in E$. Define $f$ and $g$ on $\mathbb {R}^{2}$ by: $f(0,0) = g(0,0) = 0, f(x, y) = x y^{2} /\left(x^{2}+y^{4}\right), g(x, y) = x y^{2} /\left(x^{2}+y^{6}\right)$ if $(x, y) \neq (0,0)$. Prove that $f$ is bounded on $\mathbb {R}^{2}$, that $g$ is unbounded in every neighborhood of $(0,0)$, and that $f$ is not continuous at $(0,0)$; nevertheless, the restrictions of both $f$ and $g$ to every straight line in $\mathbb {R}^{2}$ are continuous!

Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb {R}^{1}$. Prove that $f$ is bounded on $E$.

Show that the conclusion is false if boundedness of $E$ is omitted from the hypothesis.

Show that the requirement in the definition of uniform continuity can be rephrased as follows, in terms of diameters of sets: To every $\varepsilon > 0$ there exists a $\delta > 0$ such that $\operatorname {diam} f(E) < \varepsilon$ for all $E \subset X$ with $\operatorname {diam} E < \delta$.

Complete the details of the following alternative proof of Theorem 4.19: If $f$ is not uniformly continuous, then for some $\varepsilon > 0$ there are sequences $\left\{ p_{n}\right\} ,\left\{ q_{n}\right\}$ in $X$ such that $d_{\mathrm{x}}\left(p_{n}, q_{n}\right) \rightarrow 0$ but $d_{\mathrm{y}}\left(f\left(p_{n}\right), f\left(q_{n}\right)\right) > \varepsilon$. Use Theorem 2.37 to obtain a contradiction.

Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\left\{ f\left(x_{n}\right)\right\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\left\{ x_{n}\right\}$ in $X$. Use this result to give an alternative proof of the theorem stated in Exercise 13.

A uniformly continuous function of a uniformly continuous function is uniformly continuous.

State this more precisely and prove it.

Let $E$ be a dense subset of a metric space $X$, and let $f$ be a uniformly continuous real function defined on $E$. Prove that $f$ has a continuous extension from $E$ to $X$ (see Exercise 5 for terminology). (Uniqueness follows from Exercise 4). Hint: For each $p \in X$ and each positive integer $n$, let $V_{n}(p)$ be the set of all $q \in E$ with $d(p, q) < 1 / n$. Use Exercise 9 to show that the intersection of the closures of the sets $f\left(V_{1}(p)\right), f\left(V_{2}(p)\right), \ldots$, consists of a single point, say $g(p)$, of $\mathbb {R}^{1}$. Prove that the function $g$ so defined on $X$ is the desired extension of $f$.

Could the range space $\mathbb {R}^{1}$ be replaced by $\mathbb {R}^{k}$? By any compact metric space? By any complete metric space? By any metric space?

Let $I = [0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x) = x$ for at least one $x \in I$.

Call a mapping of $X$ into $Y$ open if $f(V)$ is an open set in $Y$ whenever $V$ is an open set in $X$.

Prove that every continuous open mapping of $\mathbb {R}^{1}$ into $\mathbb {R}^{1}$ is monotonic.

Let $[x]$ denote the largest integer contained in $x$, that is, $[x]$ is the integer such that $x-1 < [x] \leq x$; and let $(x) = x-[x]$ denote the fractional part of $x$. What discontinuities do the functions $[x]$ and $(x)$ have?

Let $f$ be a real function defined on $(a, b)$. Prove that the set of points at which $f$ has a simple discontinuity is at most countable. Hint: Let $E$ be the set on which $f(x-) < f(x+)$. With each point $x$ of $E$, associate a triple $(p, q, r)$ of rational numbers such that

Every rational $x$ can be written in the form $x = m / n$, where $n > 0$, and $m$ and $n$ are integers without any common divisors. When $x = 0$, we take $n = 1$. Consider the function $f$ defined on $\mathbb {R}^{1}$ by

Prove that $f$ is continuous at every irrational point, and that $f$ has a simple discontinuity at every rational point.

Suppose $f$ is a real function with domain $\mathbb {R}^{1}$ which has the intermediate value property: If $f(a) < c < f(b)$, then $f(x) = c$ for some $x$ between $a$ and $b$.

Suppose also, for every rational $r$, that the set of all $x$ with $f(x) = r$ is closed. Prove that $f$ is continuous. Hint: If $x_{n} \rightarrow x_{0}$ but $f\left(x_{n}\right) > r > f\left(x_{0}\right)$ for some $r$ and all $n$, then $f\left(t_{n}\right) = r$ for some $t_{n}$ between $x_{0}$ and $x_{n}$; thus $t_{n} \rightarrow x_{0}$. Find a contradiction. (N. J. Fine, Amer. Math. Monthly, vol. 73, 1966, p. 782).

If $E$ is a nonempty subset of a metric space $X$, define the distance from $x \in X$ to $E$ by

Suppose $K$ and $F$ are disjoint sets in a metric space $X, K$ is compact, $F$ is closed. Prove that there exists $\delta > 0$ such that $d(p, q) > \delta$ if $p \in K, q \in F$. Hint: $\rho_{F}$ is a continuous positive function on $K$.

Show that the conclusion may fail for two disjoint closed sets if neither is compact.

Let $A$ and $B$ be disjoint nonempty closed sets in a metric space $X$, and define

Show that $f$ is a continuous function on $X$ whose range lies in $[0,1]$, that $f(p) = 0$ precisely on $A$ and $f(p) = 1$ precisely on $B$. This establishes a converse of Exercise 3: Every closed set $A \subset X$ is $Z(f)$ for some continuous real $f$ on $X$. Setting

show that $V$ and $W$ are open and disjoint, and that $A \subset V, B \subset W$. (Thus pairs of dis joint closed sets in a metric space can be covered by pairs of disjoint open sets. This property of metric spaces is called normality).

A real-valued function $f$ defined in $(a, b)$ is said to be convex if

whenever $a < x < b, a < y < b, 0 < \lambda < 1$. Prove that every convex function is continuous. Prove that every increasing convex function of a convex function is convex. (For example, if $f$ is convex, so is $e^{f}$).

If $f$ is convex in $(a, b)$ and if $a < s < t < u < b$, show that

Assume that $f$ is a continuous real function defined in $(a, b)$ such that

for all $x, y \in (a, b)$. Prove that $f$ is convex.

If $A \subset \mathbb {R}^{k}$ and $B \subset \mathbb {R}^{k}$, define $A+B$ to be the set of all sums $\mathbf{x}+\mathbf{y}$ with $\mathbf{x} \in A$, $\mathbf{y} \in B$.

Suppose $X, Y, Z$ are metric spaces, and $Y$ is compact. Let $f$ map $X$ into $Y$, let $g$ be a continuous one-to-one mapping of $Y$ into $Z$, and put $h(x) = g(f(x))$ for $x \in X$.

Prove that $f$ is uniformly continuous if $h$ is uniformly continuous. Hint: $g^{-1}$ has compact domain $g(Y)$, and $f(x) = g^{-1}(h(x))$. Prove also that $f$ is continuous if $h$ is continuous. Show (by modifying Example 4.21, or by finding a diffierent example) that the compactness of $Y$ cannot be omitted from the hypotheses, even when $X$ and $Z$ are compact.