This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.

# A new look at continuity

### From Mathematics Is A Science

## Contents

## 1 From accuracy to continuity

The idea of continuity can be introduced and justified by considering the accuracy of a measurement.

Suppose we have a collection of square tiles of various sizes and we need to find the area $A$ of each of them in order to know how many we need to cover the whole floor.

The answer is, of course, to measure the side, $x$, of each tile and then compute $$A=x^2.$$ For example, we have: $$x=10\Rightarrow A=100$$ (in inches).

But what if the measurement isn't fully accurate? What if there is *always* some error? It's never $x=10$ but, say,
$$x=10 \pm .3.$$
As a result, the computed value of the area of the tile -- what we care about -- will also have some error! Indeed, the area won't be just $A=100$ but
$$A=(10 \pm .3)^2 .$$
Therefore,
$$\begin{array}{lll}
A&=10^2 \pm 2 \cdot 10 \cdot .3 +.3^2 \\
&=100.09 \pm 6.
\end{array}$$
The meaning of the result is that the actual area must be somewhere within the interval $(94.09,106.09)$.

Suppose next that we can always improve *the accuracy of the measurement* of the side of the tile $x$ -- as much as we like. The question is, can we also improve *the accuracy of the computed value* of $A$ -- to our, or somebody else's, satisfaction? The standard of accuracy might change...

What do we mean by that? Suppose $x=10$. The above computation shows that if the desired accuracy of $A$ is $\pm 5$, we haven't achieved it with the given accuracy of measurement $x$, which is of $\pm .3$. We can easily show, however, that $\pm .2$ would solve the problem:
$$\begin{array}{lll}
A&=(10 \pm .2)^2=10^2 \pm 2 \cdot 10 \cdot .2 +.2^2 \\
&=100.04 \pm 4.
\end{array}$$
It follows that the *actual* area must be within $4.04$ from $100$.

Let's rephrase this problem in order to solve it for all possible values of the desired accuracy of $A$.

Let's assume that the measurement of the side is $a$ and, therefore, the assumed area $A$ is $a^2$. Now suppose we want the accuracy of $A$ to be some small value $\varepsilon >0$ or better. What accuracy $\delta$ of $x$ do we need to be able to guarantee that?

Suppose the actual length is $x$ and, therefore, the actual area is $A=x^2$. Then we want to ensure that $A$ is within $\varepsilon$ from $a^2$ by making sure that $x$ is within $\delta$ from $a$. What should $\delta$ be?

To rephrase algebraically, we want to find $\delta$ such that $$|x-a| <\delta \Rightarrow |x^2-a^2|<\varepsilon.$$

The definition suggested by the above discussion is familiar from calculus.

**Definition.** A real-valued function $f$ is called *continuous* at $x=a$ if for any $\epsilon >0$ there is $\delta >0$ such that
$$|x-a| <\delta \Rightarrow |f(x)-f(a)|<\varepsilon.$$

Thus the answer to our question is:

- yes, we
*can*always improve the accuracy of the computed value of $A=x^2$ -- to anybody's satisfaction -- by improving the accuracy of the measurement of $x$.

The reason to be quoted is that $f(x)=x^2$ is continuous at $x=10$.

Note: The word “continuous” itself is justified on other grounds.

**Exercise.** Prove that $f(x)=x^2$ is continuous at $x=0,x=1,x=a$.

**Exercise.** Carry out this kind of analysis for: a thermometer put in a cup of coffee to find its temperature. Assume that the thermometer gives perfect readings. Hint: it'll take time for it to warm up.

To further illustrate this idea, consider a different situation. Suppose we don't care about the area anymore; we just want to fit these tiles into a strip $10$ inches wide. We take a tile and if it fits, it is used; otherwise it is discarded.

So, we still get a measurement $a$ of the side of the tile but our real interest is whether $a$ is less or more than $10$.

Just as in the previous example, we don't know the actual length $x$ exactly; it's always within some limits: $5.0 \pm 0.5$ or $a \pm \delta$. Here $\delta$ is the accuracy of measurement of $x$. The algebra is much simpler than before. For example, if the length is measured as $11$, we need the accuracy $\delta=1$ or better to make the determination. It's the same for the length $9$.

But what if the measurement is exactly $10$? Even if we can improve the accuracy, i.e., $\delta$, as long as $\delta > 0$, we can't know whether $x$ is larger or smaller than $10$.

Let's define a function $f$: $$f(x)=\begin{cases} 1 \quad\text{ (pass) } & \text{ if } x \le 10, \\ 0 \quad\text{ (no pass) } & \text{ if } x > 10. \end{cases}$$

Suppose we need the accuracy of $y = f(x)$ to be $\varepsilon = 0.5$. Can we achieve this by decreasing $\delta$? In other words, can we find $\delta$ such that $$|x-10|<\delta \Rightarrow |f(x)-1|<\varepsilon ?$$ Of course not: $$x>10 \Rightarrow |f(x)-1|=|0-1|=1.$$

Thus the answer to our question is:

- No, we
*cannot*always improve the accuracy of the computed value of $f(x)$ -- to anybody's satisfaction -- by improving the accuracy of the measurement of $x$.

The reason to be quoted is that $f$ is discontinuous at $x=10$.

**Exercise.** Carry out this kind of analysis for: the total test score vs. the corresponding letter grade. What if we introduce A-, B+, etc.?

Thus, the idea of continuity of the dependence of $y$ on $x$ is:

- We can ensure the desired accuracy of $y$ by increasing the accuracy of $x$.

**Exercise.** In addition to being continuous, $f(x)=x^2$ is also differentiable. How does that help with the accuracy issue? Hint: there is a simple dependence between $\varepsilon$ and $\delta$.

## 2 Continuity in a new light

The discussion in the last subsection reveals that continuity of function $f : {\bf R} \to {\bf R}$ at point $x=a$ can be introduced in terms of closeness (proximity) of its values, informally:

*if $x$ is close to $a$ then $f(x)$ is close to $f(a)$.*

This description applies to the function shown on the left:

On the right, even though $x$ is close to $a$, $f(x)$ does not have to be close to $f(a)$: the *discontinuity of the function does not allow us to finely control its output by controlling the input*.

Let's rephrase the second part of the continuity definition:

- for any $\varepsilon > 0$ there is a $\delta > 0$ such that
- $|x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon$.

It is advantageous to restate this part as:

Next, we use a concept that will help to keep things more compact. It is the concept of the *image of subset* $A \subset X$ *under function* $f:X \to Y$, defined as
$$f(A) := \{f(x) : x \in A \}.$$

Now the last formulation of continuity means simply that $f$ takes a certain subset of the $x$-axis to a subset of the $y$-axis: $$f\left( (a-\delta,a+\delta) \right) \subset \left( f(a)-\varepsilon, f(a)+\varepsilon \right).$$ These sets are open intervals:

We look at these intervals as $1$-dimensional “balls”, while in general, we define an open ball in ${\bf R}^n$ as: $$B(p,d)=\{u : || u-p || < d \}.$$ We simply used the norm, $$||x||=||(x_1,,...,x_n)||:=\sqrt { \sum _{i=1}^n |x_i|^2 },$$ to replace the absolute value.

In mathematics, we see a lot of examples of functions between higher-dimensional Euclidean spaces $f:{\bf R}^n \to {\bf R}^m$: parametric curves, parametric surfaces, vector fields, etc. We can accommodate the need to understand continuity of these functions by using the same idea of proximity: $$||x - a|| < \delta \Rightarrow || f(x) - f(a) || < \varepsilon.$$

Then our definition becomes even more compact, as well as applicable to all dimensions:

**Definition.** A function $f:{\bf R}^n \to {\bf R}^m$ is continuous at point $x=a$ if

- for any $\varepsilon > 0$ there is a $\delta > 0$ such that
- $f\left( B(a,\delta) \right) \subset B \left( f(a), \varepsilon \right) .$

The final simplification of the second part is: $$f( V ) \subset U,$$ where $$V=B(a, \delta), \quad U=B \left( f(a) , \varepsilon \right).$$ In dimension $2$, the relation between these sets is illustrated as follows:

**Exercise.** Using the definition, prove that constant functions $c:{\bf R}^n \to {\bf R}^m$ and the identity function are continuous.

**Exercise.** Prove that the norm is continuous.

**Exercise.** Prove that the composition of continuous functions is continuous.

**Exercise.** Provide an analog of this definition for “continuity from the left and right” of a function the domain of which is a closed interval such as $[0,1]$. For a function of two variables, what if the domain is a closed square, a closed disk?

## 3 Continuity on subsets

Our interest is mainly algebraic topology. Consequently, apart from pointing out monstrosities to avoid, we will keep our attention limited to

- “nice” subsets of Euclidean spaces, such as
*cells*, and - those subsets “nicely” glued together.

We have already seen an example: realizations of graphs. When we come back to algebra, it will explain how these pieces fit together *globally* to form components, holes, voids, etc.. Right now, in a way, we are interested in *local* topological properties, such as continuity.

Now, an example of an especially “nice” set is
$$\bar{B}(a,\delta)= \{u\in {\bf R}^n : ||u-a|| \le \delta \}, $$
the *closed ball* centered at $a\in {\bf R}^n$ of radius $\delta$. In dimension $1$, this is simply a closed interval, yet functions defined on such a set won't fit into the last definition of continuity.

For example, continuity from the left and right has to be treated as an exception, i.e., as a separate definition. Suppose we are considering continuity of a function $f$ at $x=a$ when its domain is $[a,b]$. Then a half of the interval $(a-\delta,a+\delta)$ lies outside the domain! Our answer is to verify the proximity condition only for the points inside $[a,b]$ and ignore the rest. This way it doesn't matter how the domain fits into some bigger space.

To accommodate this idea, we retrace the analysis in the last subsection, for a function $$f:X \to Y$$ with $$X \subset {\bf R}^n, Y \subset {\bf R}^m.$$ Recall that this notation simply means that $$x\in X \Rightarrow f(x)\in Y.$$

We will need to adjust the second part of the definition a tiny bit more:

- $||x - a|| < \delta,x \in X \Rightarrow || f(x) - f(a) || < \epsilon.$

With this little addition, $x \in X$, we have sidestepped the issue of continuity from the left/right and, in higher dimensions, the case when a point $x$ is close to $a$ but outside the domain of the function.

As before, the inequalities indicate simply that $f$ takes $x$ that lies in a certain subset of the domain of $f$ to $y=f(x)$ in another subset of its range:

- if $x \in B(a, \delta)$ and $x \in X$ then $f(x) \in B(f(a), \epsilon)$.

Of course, it was assumed from the beginning that $f(x)\in Y$. The condition can be further re-written as

- $f \left( B(a, \delta) \cap X \right) \subset B \left( f(a) , \epsilon \right) \cap Y.$

Adding “$\cap Y$” is not a matter of convenience; we just want to be clear that we don't care about anything outside $X$ and $Y$.

Let's state this most, for now, compact definition.

**Definition.** A function $f:X \to Y$ with $X \subset {\bf R}^n, Y \subset {\bf R}^m$ is called *continuous at* $x=a$ if

- for any $\epsilon >0$, there is $\delta$ such that
- $f( V ) \subset U,$

where $$V=B(a, \delta) \cap X, \quad U=B \left( f(a) , \epsilon \right) \cap Y.$$

In dimension $2$, the relation between these sets is illustrated as follows:

**Exercise.** Show how this new definition incorporates the old definition of one-sided continuity.

The following is familiar.

**Definition.** A function $f$ is called *continuous* if it is continuous at every point of its domain.

The question, “What do you know about continuity of $1/x$?” might be answered very differently by different students:

- a calculus 1 student: “$1/x$ is discontinuous... well, it's discontinuous at $x=0$ and continuous everywhere else.”
- a topology student: “The function $f:(-\infty,0)\cup (0,\infty) \to {\bf R}$ given by $f(x)=1/x$ is continuous.”

...because we don't even consider continuity at a point where the function is undefined.

**Exercise.** Show that continuity of a function defined on the $x$-axis (its domain) as a function of $(x,y)$ is equivalent to its continuity as a function of $x$ only.

There is still an even more profound generalization to be made. We already test for continuity of $f$ *from the inside* of its domain $X$, without mention of any $x \not\in X$. Now would like to take this idea one step further and avoid referring to open balls that come from the ambient Euclidean spaces.

## 4 The intrinsic definition of continuity

Let's start with the case of $X={\bf R}^n,\ Y={\bf R}^m$. Suppose that

- $\gamma _X$ is the set of all open balls in $X={\bf R}^n$, and
- $\gamma _Y$ is the set of all open balls in $Y={\bf R}^m$.

We can restate the definition in a more abstract form:

**Exercise.** Prove that this definition is equivalent to the original.

Now, once again, what if the domain is a *subset* $X$ of the Euclidean space?

Then, we just choose for $\gamma _X$, instead of the open balls, the *intersections of the open balls with the subset* $X\subset {\bf R}^n$:
$$\gamma _X := \{ B(x,\delta)\cap X:\ x\in X,\delta>0 \}.$$
The definition above remains the same.

This approach fits into our strategy of studying the *intrinsic properties* of objects as if anything outside $X$ simply doesn't exist!

Moreover, if our interest is to study $X$, we need to consider all and only functions with domain $X$. For the new definition, we will make no reference to the Euclidean space that contains $X$. For that, we just need to determine what collections $\gamma _X$ are appropriate.

First we observe that these balls can be easily replaced with sets of any other shape:

The importance of these intervals, disks, and balls for continuity is the reason why we start studying topology by studying these collections of sets. We will call them *neighborhoods*. For example, $B(a, \delta)$ is a *ball neighborhood* of $a$.

Second, we notice that continuity is a local property and, therefore, we don't need a single yardstick to measure closeness (i.e., distances between points) throughout $X$ (or $Y$). We just need the ability for each $a \in X$, to answer the question: “How close is $x$ to $a$?” And the new answer doesn't look that different:

- Calculus 1: “It's within $\delta$”;
- Topology: “It's within $V$”.

The news is, all references to numbers are gone!

In order to have a meaningful concept of continuity, we need these collections of neighborhoods to satisfy certain properties...

First of course, we want to be able to discuss continuity of $f:X\to Y$ at *every point* $a\in X$. So, these neighborhoods have to cover the whole set:

Second, we want to be able to *zoom in* on every point -- to detect breaks in the graphs of functions. So, we should be able to “refine” the collection:

This is how we understand the refining process illustrated with disks:

To summarize, we require:

- (B1) $\cup \gamma = X$;
- (B2) for any two neighborhoods $U, V\in \gamma$ and any point $c\in U\cap V$ in their intersection, there is a neighborhood $W\in \gamma$ of $c$ such that $W \subset U \cap V$.

We will demonstrate that this is *all* the extra structure we need to have on an arbitrary set $X$ to be able to define and study continuity!

**Definition.** Suppose $X$ is any set. Any collection $\gamma$ satisfying properties (B1) and (B2) is called a *basis of neighborhoods* in $X$.

And the new definition of continuity is wonderfully compact:

**Definition.** Given two sets $X,Y$ with bases of neighborhoods $\gamma_X,\gamma_Y$, function $f:X\to Y$ is *continuous* if for any $U \in \gamma _Y$ there is $V \in \gamma _X$ such that $f( V ) \subset U$.

**Exercise.** Prove that the definition of continuity in the last subsection satisfies this definition.