1.9: Continuity

The angle measure is also assumed to be continuous. Namely, the following property of angle measure will become a part of the axioms:

The function

$$\measuredangle: (A, O, B) \mapsto \measuredangle AOB$$

is continuous at any triple of poitns $(A, O, B)$ such that $O \ne A$ and $O \ne B$ and $\measuredangle AOB \ne \pi$.

To explain this property, we need to extend the notion of continuity to the functions between metric spaces. The definition is a straightforward generalization of the standard definition for the real-to-real functions.

Further, let $\mathcal{X}$ and $\mathcal{Y}$ be two metric spaces, and $d_{\mathcal{X}}, d_{\mathcal{Y}}$ be their metrics.

A map $f: \mathcal{X} \to \mathcal{Y}$ is called continuous at point $A \in \mathcal{X}$ if for any $\varepsilon > 0$ there is $\delta > 0$, such that

$d_{\mathcal{X}} (A, A') < \delta \Rightarrow d_{\mathcal{Y}}(f(A), f(A')) < \varepsilon.$

(Informally it means that sufficiently small changes of $A$ result in arbitrarily small changes of $f(A)$.)

A map $f: \mathcal{X} \to \mathcal{Y}$ is called continuous if it is continuous at every point $A \in \mathcal{X}$.

One may define a continuous map of several variables the same way. Assume $f(A, B, C)$ is a function which returns a point in the space $\mathcal{Y}$ for a triple of points $(A, B, C)$ in the space $\mathcal{X}$. The map $f$ might be defined only for some triples in $\mathcal{X}$.

Assume $f(A, B, C)$ is defined. Then, we say that $f$ is continuous at the triple $(A, B, C)$ if for any $\varepsilon > 0$ there is $\delta > 0$ such that

$d_{\mathcal{Y}} (f(A, B, C), f(A',B',C')) < \varepsilon.$

if $d_{\mathcal{X}} (A, A') < \delta, d_{\mathcal{X}} (B, B') < \delta$, and $d_{\mathcal{X}} (C, C') < \delta$.

Exercise $\PageIndex{1}$

Let $\mathcal{X}$ be a metric space.

(a) Let $A \in \mathcal{X}$ be a fixed point. Show that the function
$$f(B):= d_{\mathcal{X}} (A, B)$$
is continuous at any point $B$.

(b) Show that $d_{\mathcal{X}} (A, B)$ is continuous at any pair $A, B \in \mathcal{X}$.

Hint

(a). By the triangle inequality, $|f(A') - f(A)| \le d(A', A)$. Therefore, we can take $\delta = \varepsilon$.

(b). By the triangle inequality,

$$\begin{array} {rcl} {|f(A',B') - f(A, B)|} & \le & {|f(A',B') - F(A, B')| + |F(A, B') - F(A, B)|} \\ {} & \le & {d(A',A) + d(B',B).} \end{array}$$

Therefore, we can take $\delta = \dfrac{\varepsilon}{2}$.

Exercise $\PageIndex{2}$

Let $\mathcal{X}, \mathcal{Y}$, and $\mathcal{Z}$ be metric spaces. Assume that the functions $f: \mathcal{X} \to \mathcal{Y}$ and $g: \mathcal{Y} \to \mathcal{Z}$ are continuous at any point, and $h = g \circ f$ is their composition; that is, $h(A) = g(f(A))$ for any $A \in \mathcal{X}$. Show that $h: \mathcal{X} \to \mathcal{Z}$ is continuous at any point.

Hint

Fix $A \in \mathcal{X}$ and $B \in \mathcal{Y}$ such that $f(A) = B$.

Fix $\varepsilon > 0$. Since $g$ is continuous at $B$, there is a positive value $\delta_1$ such that

$d_{\mathcal{Z}} (g(B'), g(B)) < \varepsilon$ if $d_{\mathcal{Y}} (B', B) < \delta_1$.

Since $f$ is continuous at $A$, there is $\delta_2 > 0$ such that

$d_{\mathcal{Y}} (f(A'), f(A)) < \delta_1$ if $d_{\mathcal{X}} (A', A) < \delta_2$.

Since $f(A) = B$, we get that

$d_{\mathcal{Z}} (h(A'), h(A)) < \varepsilon$ if $d_{\mathcal{X}} (A',A) < \delta_2$.

Hence the result.

Exercise $\PageIndex{3}$

Show that any distance-preserving map is continuous at any point.