Generalising Continuity

The biggest advantage of a blog is how it can become dynamic; where we discuss more and more stuff the more we learn. I didn’t want to discuss metrics and topologies in detail, but it turns out dipping our toes in some of these ideas not only introduces these notions well, but also closes our continuity discussions satisfactorily.

We have previously discussed continuous functions in a technical manner, even proving seemingly obvious theorems like the extreme value theorem and the intermediate value theorem. It turns out that these ideas are not restricted to the real numbers, but generalise to any setting with a reasonable notion of distance, known technically in the mathematics business as a metric.

Definition 1. Let $K$ be any set. A function $d: K^2 \to \mathbb R$ is called a metric on $K$ if it satisfies the following properties:

For any $x,y \in K$ , $d(x,y) \geq 0$ .
For any $x,y \in K$ , $d(x,y) = 0 \Rightarrow x=y$ .
For any $x,y \in K$ , $d(x,y) = d(y,x)$ .
For any $x,y,z \in K$ , $d(x,z) \leq d(x,y) + d(y,z)$ . This is called the triangle inequality.

In this case, we call the pair $(K, d)$ a metric space.

Theorem 1. The map $d : \mathbb R^2 \to \mathbb R$ by $d(x,y) = |x-y|$ is a metric on $\mathbb R$ . Unless stated otherwise, $\mathbb R$ as a metric space will always be defined by $( \mathbb R, | \cdot |)$ .

Proof. We will verify the triangle inequality as follows:

$\begin{aligned}d(x,z) &= |x-z| \\ &= |(x-y) + (y-z)| \\ &\leq |x-y| + |y-z| \\ &= d(x,y) + d(y,z).\end{aligned}$

Definition 2. Let $(K, d_1)$ and $(L, d_2)$ be metric spaces. We say that a function $f : K \to L$ is metrically continuous at $c \in K$ if for any $\epsilon > 0$ , there exists $\delta > 0$ such that for any $x \in K$ ,

$d_1(x,c) < \delta \quad \Rightarrow \quad d_2(f(x),f(c)).$

We say that $f$ is metrically continuous on $K$ if it is metrically continuous at every $c \in K$ .

Theorem 2. A function $f : \mathbb R \to \mathbb R$ is metrically continuous at $c$ if and only if it is continuous at $c$ .

Proof. Left as an exercise in bookkeeping.

If we scale things up again, this notion of distance boils down very much to the intuitive idea of closeness. Indeed, it is closeness that undergirds what it means for a function to be continuous, which is closely related (pun intended) to our ideas of convergence. The most general formulation of closeness in modern pure mathematics has to be that of a topology, which we briefly discuss now.

Definition 2. Let $K$ be a set. The collection $\mathcal B$ of subsets of $K$ forms a topological basis for $K$ if it satisfies the following properties:

For any $x \in K$ , there exists $B \in \mathcal B$ such that $x \in B$ .
For any $x \in K$ and any $B_1,B_2 \in \mathcal B$ , if $x \in B_1 \cap B_2$ , then there exists $B_3 \in \mathcal B$ such that $x \in B_3 \subseteq B_1 \cap B_2$ .

Theorem 2. Let $(K, d)$ be any metric space. For each $x \in K$ , define for any $r > 0$ the open ball $B_x(r) := \{y \in K : d(x,y) < r\}$ . Then $\mathcal B := \{B_x(r) : x \in K\}$ forms a topological basis for $K$ .

Proof. The first condition is trivially met, since for any $x \in K$ , $x \in B_x(1)$ . For the second condition, fix $x \in K$ , and suppose there exists $u,v \in K$ and $r_1, r_2 > 0$ such that $x \in B_u(r_1) \cap B_v(r_2)$ . This means $d(x,u) < r_1$ and $d(x,v) < r_2$ .

If we can find $r > 0$ such that $B_x(r) \subseteq B_u(r_1) \cap B_v(r_2)$ , then we are done, since $x \in B_x(r)$ trivially. This condition means that for any $y \in K$ ,

$d(y,x) < r \quad \Rightarrow \quad d(y,u) < r_1\wedge d(y,v) < r_2.$

If $d(y,x) < r$ , then the triangle inequality guarantees the estimate

$d(y,u) \leq d(y,x) + d(x,u) < r + d(x,u).$

This means we require $r < r_1 - d(x,u)$ . Similarly, we require $r < r_2 - d(x,v)$ . Hence, the choice $r := \frac 12 \min\{r_1 - d(x,u), r_2 - d(x,v)\}$ works.

Technically, we defined the basis of a topology. The full topology arises from combining this basis in unique ways, as follows.

Theorem 3. Let $K$ be a set and $\mathcal B$ be a topological basis for $K$ . Define

$\mathcal T := \{U \in \mathcal P(K):(\forall x \in U\ \exists B \in \mathcal B: x \in B \subseteq L)\}.$

Then $\mathcal T$ satisfies the following properties:

$\emptyset \in \mathcal T$ , $U \in \mathcal T$ .
For any $U_\alpha \in \mathcal T$ , $\displaystyle \bigcup_\alpha U_\alpha \in \mathcal T$ .
For any $U_1,\dots,U_n \in \mathcal T$ , $\displaystyle \bigcap_{i=1}^n U_i \in \mathcal T$ .

Proof. We prove the last property as it is the most interesting. We will prove the case $n =2$ , and the general case follows easily by induction. If $K_1 \cap K_2 = \emptyset$ then we are done.

Suppose otherwise and fix $x \in U_1 \cap U_2$ . For each $i$ , find $B_i \in U_i$ such that $x \in B_i \subseteq U_i$ . Then $x \in B_1 \cap B_2 \subseteq U_1 \cap U_2$ . Find $B_3 \subseteq B_1 \cap B_2$ that contains $x$ , so that $x \in B_3 \subseteq B_1 \cap B_2 \subseteq U_1 \cap U_2$ . Hence, $U_1 \cap U_2 \in \mathcal T$ , as required.

In this case, we call $\mathcal T$ a topology. Its elements are called open sets, and put together, the pair $(K, \mathcal T)$ is a topological space.

In fact, since we used $\mathcal B$ to create $\mathcal T$ , we say $\mathcal T$ is the topology generated by $\mathcal B$ . (It turns out that for any other topology $\mathcal S$ that contains $\mathcal B$ , $\mathcal T \subseteq \mathcal S$ , so that $\mathcal T$ is the smallest topology that contains $\mathcal B$ .)

Unless stated otherwise, all metric spaces will be assumed to be equipped with the topology generated by the basis of open balls defined by the underlying metric.

Corollary 1. Let $(K, d)$ be any metric space. Then $(K, \mathcal T)$ is a topological space, where the topology $\mathcal T$ is generated by the basis in Theorem 2.

It is Corollary 1 that motivates a unified study of metric and topological spaces, where universal properties in the latter apply immediately to the former.

Topological spaces form the simplest and broadest framework for us to the study continuous functions.

Definition 3. Let $(K, \mathcal S), (L, \mathcal T)$ be topological spaces. We say that the function $f : K \to L$ is topologically continuous if for any $V \in \mathcal T$ , $f^{-1}(V) \in \mathcal S$ .

Lemma 1. Let $(K, d)$ be a metric space. Then $U$ is open in $K$ if and only if for any $x \in U$ , there exists $r > 0$ such that $x \in B_x(r) \subseteq U$ .

Proof. The direction $(\Leftarrow)$ is trivial. For $(\Rightarrow)$ , fix $x \in U$ and find $x_0 \in K, r_0 > 0$ such that $x \in B_{x_0}(r_0) \subseteq U$ . If we can find $r > 0$ such that $B_x(r) \subseteq B_{x_0}(r_0)$ , then we are done. This subset relation is equivalent to the implication

$d(y,x) < r \quad \Rightarrow \quad d(y,x_0) < r_0.$

Given $r$ , the triangle inequality yields

$d(y, x_0) \leq d(y,x) + d(x,x_0) < r + d(x,x_0).$

Setting $r := r_0 - d(x,x_0) > 0$ does the trick.

Theorem 4. Let $(K, d_1)$ and $(L, d_2)$ be metric spaces. Then the function $f : K \to L$ is topologically continuous if and only if it is metrically continuous.

Proof. We first rephrase metrical continuity in terms of open sets as follows:

$x \in B_c(\delta) \subseteq K \quad \Rightarrow \quad f(x) \in B_{f(c)}(\epsilon) \subseteq L.$

This can be further simplified to the subset relation $f(B_c(\delta)) \subseteq B_{f(c)}(\epsilon)$ as subsets of $L$ , which is equivalent to $B_c(\delta) \subseteq f^{-1}(B_{f(c)}(\epsilon))$ as subsets of $K$ .

To prove $(\Rightarrow)$ , fix $c \in K$ and $\epsilon > 0$ . Since $B_{f(c)}(\epsilon)$ is an open set in $L$ , $f^{-1}( B_{f(c)}(\epsilon))$ is an open set in $K$ . Furthermore, $f(c) \in B_{f(c)}(\epsilon)$ implies that $c \in f^{-1}( B_{f(c)}(\epsilon))$ . By Lemma 1, there exists $\delta > 0$ such that $c \in B_{c}(\delta) \subseteq f^{-1}( B_{f(c)}(\epsilon))$ , as required.

To prove $(\Leftarrow)$ , fix any set $V$ that is open in $L$ . We need to prove that $f^{-1}(V)$ is open in $K$ . To that end, fix $x \in f^{-1}(V) \iff f(x) \in V$ . Since $V$ is open in $L$ , by Lemma 1, there exists $\epsilon > 0$ such that $f(x) \in B_{f(x)}(\epsilon) \subseteq V$ . By assumption, there exists $\delta > 0$ such that $B_{x}(\delta) \subseteq f^{-1}( B_{f(x)}(\epsilon)) \subseteq f^{-1}(V)$ , as required.

To study these objects in more detail and verify that they do agree with our standard notions of continuity shoots us into a whole new stratosphere of study called topology, which lies far beyond the scope of introductory real analysis. These ideas take extend intuitions about $\mathbb R$ to any space where closeness (i.e. a topology) is defined, especially in probability theory, and unsurprisingly finds many use cases across the STEM fields through it.

—Joel Kindiak, 25 Jan 25, 1525H

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Generalising Continuity

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