Continuity

This section examines some of the exposition in chapter 4 of Rudin’s book “Principles of Mathematical Analysis.”

definition: let $X,Y$ be metric spaces $p$ a limit point of $E$ where $E$ is a subset of $X$ . Then $\lim_{x \rightarrow p} f(x) = q$ iff $q \in Y$ such that for any $\epsilon > 0$ there is a $\delta > 0$ such that $0 < d(x,p) < \delta$ implies $d(fx,q) < \epsilon$ .

theorem: $\lim_{x \rightarrow p} f(x) = q$ iff $\lim_{n \rightarrow \infty} f(x_n) = q$ for any sequence $\left<x_n\right>$ such that $\lim_{n \rightarrow \infty} x_n = p$ and $x_n \neq p$ .

proof: Suppose $\lim_{x \rightarrow p} f(x) = q$ . Assume there is a sequence $\left<x_n\right>$ such that $\lim_{n \rightarrow \infty} x_n = p$ , $x_n \neq p$ and $\lim_{n \rightarrow \infty} f(x_n) \neq q$ . Then there is an $\epsilon > 0$ such that for all $N$ there is an $n \ge N$ such that $d(fx_n,q) \ge \epsilon$ . But from the initial supposition, there is a $\delta > 0$ such that for all $x$ such that $0 < d(x,p) < \delta$ it is true that $d(fx,q) < \epsilon$ . Since there is an $M$ such that for all $m \ge M$ $d(x_n,p) < \delta$ it follows that $d(fx_n,q) < \epsilon$ . Hence the assumption must be false.

Conversely, suppose for any sequence $\left<x_n\right>$ such that $\lim_{n \rightarrow \infty} x_n = p$ and $x_n \neq p$ it is true that $\lim_{n \rightarrow \infty} f(x_n) = q$ . Assume that $\lim_{x \rightarrow p} f(x) \neq q$ . Hence there is a $\epsilon > 0$ such that for all $\delta > 0$ there is an $x$ such that $0 < d(x,p) < \delta$ and $d(fx,q) \ge \epsilon$ . Choose $x_i : 0 < d(x_i,p) < 2^{-i}$ and $d(fx,q) \ge \epsilon$ . Since $p$ is a limit point, this procedure is always feasible. Therefore $\left< x_n \right>$ is a sequence convergent to $p$ and so $\left< fx_n \right>$ is convergent to $q$ . But this is false. Hence the assumption is false.

corollary: if $f$ has a limit at $p$ then the limit is unique.

proof: suppose $\lim_{x \rightarrow p} fx = q$ and $\lim_{x \rightarrow p} fx = r$ . Then every sequence $\left< x_n \right>$ convergent and not equal to $p$ is such that $\lim_n fx_n = q$ . But it is also true for this sequence that $\lim_n fx_n = r$ . Since a sequence can only converge to one limit, $q = r$ .

Continuous Functions

definition: suppose $X,Y$ are metric spaces, $E$ is a subset of $X$ , $p \in E$ and $f : E \rightarrow Y$ . $f$ is continuous at $p$ iff for any $\epsilon > 0$ there is a $\delta > 0$ such that $d(fx,fp) < \epsilon$ when $d(x,p) < \delta$ .

definition: $f$ is continuous if for any $x \in E$ , $f$ is continuous at $x$ .

This definition implies that $f$ is continuous at $x$ when $x$ is an isolated point of $E$ .

lemma: $f$ is continuous at $p$ and $g$ is continuous at $fp$ then $g\circ f$ is continuous at $p$ .

proof: Let $\epsilon > 0$ be given. Choose $\delta > 0$ such that $d(g(fp),g(y)) < \epsilon$ when $d(fp,y) < \delta$ . Choose $\delta_2 > 0$ such that $d(fp,fx) < \delta$ when $d(p,x) <\delta_2$ . Then $d(g(fp),g(fx)) < \epsilon$ when $d(p,x) < \delta_2$ . Since this condition can be satisfied for any $\epsilon > 0$ , it follows that $g\circ f$ is continuous at $p$ .

theorem: $f$ is continuous if $f^{-1} V$ is open set whenever $V$ is an open set.

proof: Suppose $f : X \rightarrow Y$ is continuous. Let $V \subseteq Y$ is an open set. Then $V$ is the union of balls $B_{Y,V}$ of $Y$ contained in $V$ . Let $x \in f^{-1} V$ . Then $fx \in V$ and so there is a $b \in B_{Y,V}$ and a ball $c \in B_X$ such that $f c \subseteq b$ . Hence $x \in c \subseteq f^{-1} b \subseteq f^{-1} V$ . Hence $x$ is an interior point of $f^{-1} V$ . Hence every point of $f^{-1} V$ is an interior point and therefore is open.

Conversely, suppose for every open set $V$ of $Y$ , $f^{-1} V$ is an open set of $X$ . Given $x \in X$ , $\epsilon > 0$ . Consider the ball $B_{fx,\epsilon}$ . Then $f^{-1} B_{fx,\epsilon}$ is open and so $x$ is an interior point of $f^{-1} B_{fx,\epsilon}$ . Hence there is a $\delta > 0$ such that $B_{x,\delta} \subseteq f^{-1} B_{fx,\epsilon}$ . Hence $f B_{x,\delta} \subseteq B_{fx,\epsilon}$ and so $f$ is continuous at $x$ . Since this argument applies to all $x \in X$ it follows that $f$ is continuous.

corollary: $f$ is continuous iff $f^{-1} C$ is closed when $C$ is closed.

proof: Suppose $f$ is continuous. Let $C$ be closed. Then $Y-C$ is open and so $f^{-1}(Y-C) = X - f^{-1} C$ is open. But then $X - (X - f^{-1} C) = f^{-1} C$ is closed. The converse is achieved by a similar argument.

theorem: Let $f(x) = (f_1(x),f_2(x))$ where $f_1,f_2$ are real functions on a metric space. Equip the real plane with the product topology. Then $f$ is continuous iff $f_1,f_2$ are continuous.

proof: Suppose $f_1,f_2$ are continuous. Then for any planar real open set $V$ is the union of sets of the form $O_1 \times O_2$ where $O_1, O_2$ are real open sets. Since $f^{-1} (O_1 \times O_2) = f_1^{-1} O_1 \cap f^{-1} O_2$ is an open set it follows that $f^{-1} V$ is an open set. Hence $f$ is continuous. Conversely, suppose that $f$ is continuous. Then for any open real set $V$ , $V \times \mathbb{R}$ is an open set in the product topology and therefore $f^{-1} (V \times \mathbb{R} = f_1^{-1} V$ is open. Hence $f_1$ is open. Dually $f_2$ is open.

Continuity and Compactness

theorem: For a mapping $f : X \rightarrow Y$ where $X$ is compact, $fX$ is compact.

proof: Let $V$ be an open cover of $fX$ . Then $f^{-1}V$ is an open cover of $X$ and so has a finite subcover $\{f^{-1} V_1,\ldots, f^{-1} V_n\}$ . Since $fX = f \cup_{i=1..n} f^{-1} V_i \subseteq \cup_{i=1..n} V_i$ , it follows that $V$ has a finite subcover and so $fX$ is compact.

corollary: If $f$ is a continuous mapping of a continuous function from a compact metric space to $\mathbb{R}^n$ then $fX$ is closed and bounded.

proof: This follows since Heine-Borel theorem asserts that compact sets of $n$-dimensional real space are identified with closed and bounded sets.

theorem: Suppose $f$ is a continuous map from a compact metric space into a metric space such that $f$ is a set isomorphism. Then the inverse map $f^{-1}$ is continuous.

proof: It is sufficient to show that when $V$ is open, $fV$ is open. Since $X-V$ is closed subset of a compact metric space, it is compact. Hence $f(X-V)$ is compact. Since compact subsets of metric spaces are closed, it follows that $f(X-V)$ is closed. But since $f$ is a set isomorphism, $f(X-V) = Y - fV$ and so $fV$ is open.

definition: A function on metric spaces is uniformly continuous if for any $\epsilon > 0$ there is a $\delta > 0$ such that $d(fx,fy) < \epsilon$ when $d(x,y) < \delta$ .

It is clear that uniformly continuous functions are continuous, but the converse does not generally hold. However,

theorem: If $f : X \rightarrow Y$ is a continuous function on metric spaces and $X$ is compact, then $f$ is uniformly continuous.

proof: Suppose $f$ is continuous. Let $\epsilon > 0$ be given. For each $x \in X$ , let $\phi(x) > 0$ be such that $f B(x,\phi(x)) \subseteq B(fx,\epsilon/2)$ . Then $\{ B(x,\phi(x)/2) : x \in X \}$ is an open cover of $X$ and so there is a finite sequence of points $x_1,\ldots,x_n$ such that $X = \cup_{i=1..n} B(x_i,\phi(x_i)/2)$ . Let $\nu = \min_{i=1..n} \phi(x_i)/2$ . Then when $y \in B(x,\nu)$ there is $i = 1..n$ such that $x \in B(x_i,\phi(x_i)/2)$ and so $y \in B(x_i,\phi((x_i))$ . Hence $fx \in B(fx_i,\epsilon/2)$ and $fy \in B(fx_i, \epsilon/2)$ and so $fx \in B(fy,\epsilon)$ .

Continuity and Connectedness

to be examined.

Discontinuities

Consider real-valued function $f$ defined on a segment $(a,b)$ . The left-hand limit if it exists is defined by $f(x-)$ and the right-hand limit if it exists is $f(x+)$ . If both limits exist and are identical, then $f$ is continuous at $x$ . Otherwise, $f$ is discontinuous. If $f(x-),f(x+)$ both exist but are not equal then $f$ has a discontinuity of type 1. Otherwise it has a discontinuity of type 2.

For example, the indicator function on rationals has discontinuities of type 2 at all points since the left and right limits do not exist. (If they did exist then given $x$ , some interval $(x-\delta,x)$ either maps to $\{0\}$ or it maps to $\{1\}$ . But since the rationals/irrationals are dense in the reals, every non-degenerate interval has both rational and irrational numbers. )

Monotonic Functions

A real-valued function $f$ on an interval $(a,b)$ is non-decreasing if for $x,y \in (a,b)$ such that $x<y$ implies $fx \le fy$ . A dual definition applies for non-increasing functions. A function is monotonic if it is either non-decreasing or non-increasing.

lemma: If $f$ is monotonic then both left and right limits exist.

corollary: monotonic functions have no type 2 discontinuities.

lemma: monotonic functions have at most a countable number of discontinuities.

proof: assume $f$ is non-decreasing. let $D$ be the set of points where $f$ is discontinuous. Then for $x \in D$ , $f(x-) < f(x+)$ . Furthermore, the set $\{(f(x-),f(x+) : x \in D\}$ is disjoint. For each interval, choose rational $r(x) \in (f(x-),f(x+))$ . Since $r$ is an injection from $D$ into the rationals, it follows that $D$ is at most countable.