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Calculus as a part of topology

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Three pillars of calculus -- three structures of the Euclidean space

Topology, Algebra, and Geometry are disciplines within Mathematics. In calculus we use them freely, when necessary, without a second thought. Let's try to break this down...

The early calculus is about the derivative, i.e., the rate of change of a function. This doesn't seem like a part of Topology, Algebra, and Geometry. However, take a look at where this is all happening. The locus is the Euclidean space. Such a space has three different types of structures present at the same time.

Each is responsible for its own issues, distilled to the most basic below.


ALGEBRA
Algebra of vectors.png
  • For every two vectors, what is their sum?
  • For every vector and a number, what is their scalar product?


TOPOLOGY
Topology convergent sequence.png
  • For every sequence of points, what is its (if any) limit?


GEOMETRY
  • For every two points, what is the distance between them?
  • For every two lines, what is the angle (if any) between them?


These structures also interact with algebra leading the way. First, we can rephrase the last three questions as:

  • For every sequence of vectors, what is its (if any) limit?
  • For every vector, what is its length?
  • For every two vectors, what is the angle between them?

Then, the effect of algebraic operations on limits, distances, and angles is discussed.

How much of each do we really need?

We do need all the algebra, but how much of the geometry is really necessary?

Let's recall what happened to the original definition of continuity: $$||x-a||<\delta \Rightarrow ||f(x)-f(a)||<\epsilon.$$ Here we see the following used very explicitly:

  • the algebraic: subtraction of vectors, and
  • the geometric: the magnitudes of the resulting vectors.

As a result of our analysis though, the algebra and the geometry have disappeared; it was about topology all along!

Note: the order structure of the reals is also used; we choose to file this under topology.

We will see later how some of the issues in calculus are purely topological in nature. Below, we will see how far we can get in elementary calculus without geometry.

The plan is simple: every time we see “$|\ |$” in a proof, we'll try to get rid of it.

Limits of sequences

Suppose also $\{x_n:n=1,2,...\}$ is a sequence of points in $X={\bf R}$ and $a\in X$. What does it mean for the sequence to converge to a point: $$\lim_{n \to \infty} x_n = a;$$ $$x_n \to a \text{ as } n\to \infty.$$ The elements of the sequence are supposed to be eventually within any distance -- no matter how small -- from its limit $a$. Geometrically, the sequence converges to $a$ if for any $\epsilon >0$ there is $N>0$ such that $$n>N \Rightarrow |x_n-a| <\epsilon.$$ Topologically, we realize that this simply means that $x_n$ belongs to a standard Euclidean neighborhood: $$n>N \Rightarrow x_n \in B(a,\epsilon).$$ We then replace this neighborhood with one of arbitrary nature, as follows. We assume that $X$ is a set with a basis of neighborhoods $\gamma$. Then $x_n\to a$ as $n\to \infty$, if for any $U\in \gamma$ there is $N>0$ such that $$n>N \Rightarrow x_n\in U.$$

How do algebraic operations affect limits?

Theorem (Algebraic Rules of Limits). Suppose these two limits exist, first, and, second, are equal to the numbers on the right: $$ \begin{aligned} \lim_{n \to \infty} x_n & = a, \\ \lim_{n \to \infty} y_n & = b. \end{aligned} $$ Then, these limits on the left exist, first, and, second, are equal to the numbers on the right: $$\begin{array}{lllllll} \text{Sum:}\quad &\lim_{n \to \infty} (x_n + y_n) & = a + b; & \\ \text{Difference:}\quad &\lim_{n \to \infty} (x_n - y_n) & = a - b; & \\ \text{Constant:}\quad &\lim_{n \to \infty} (cx_n) & = ca, & \text {for any real }c;\\ \text{Product:}\quad &\lim_{n \to \infty} (x_n\cdot y_n) & = a\cdot b; & \\ \text{Quotient:}\quad &\lim_{n \to \infty} \left(\dfrac{x_n}{y_n}\right) & = \dfrac{a}{b}, & \text{provided }b \ne 0. \\ \end{array} $$

Proof. Let's recall the proof of the sum rule. It is short: we choose $n$ large enough so that $$|x_n-a| <\epsilon/2 \text{ and } |y_n-b| <\epsilon/2;$$ then: $$|(x_n+y_n)-(a+b)| < |x_n-a|+|y_n-b| <\epsilon/2 +\epsilon /2 =\epsilon.$$ But that's the triangle inequality! Restated: $$x_n \in B(a,\epsilon/2) \text{ and } y_n \in B(b,\epsilon/2) \Rightarrow x_n+y_n \in B(a+b,\epsilon).$$ $\blacksquare$

Exercise. Provide a similar analysis for series.

Limits of functions

Theorem (Algebraic Rules of Limits). Suppose that $\lim\limits_{x \to a} f(x)$ and $\lim\limits_{x \to a} g(x)$ both exist and $$ \begin{aligned} \lim_{x \to a} f(x) & = L, \\ \lim_{x \to a} g(x) & = M. \end{aligned} $$ Then, these limits on the left exist, first, and, second, equal to the numbers on the right: $$\begin{array}{lllllll} \text{Sum:}\quad &\lim_{x \to a} (f(x) + g(x)) & = L + M; & \quad & \\ \text{Difference:}\quad &\lim_{x \to a} (f(x) - g(x)) & = L - M; & \quad & \\ \text{Constant:}\quad &\lim_{x \to a} (cf(x)) & = cL; & \quad & \\ \text{Product:}\quad &\lim_{x \to a} (f(x)\cdot g(x)) & = L\cdot M; & \quad &\ \\ \text{Quotient:}\quad &\lim_{x \to a} \left(\dfrac{f(x)}{g(x)}\right) & = \dfrac{L}{M} & \quad &&\text{provided }M \ne 0. \\ \end{array} $$

Theorem (Substitution Rule). Suppose $g$ is continuous at $y = \lim_{x \to a} f(x)$, which exists. Then $$\lim_{x \to a} g(f(x)) = g(\lim_{x \to a} f(x)),$$

Theorem (Limit Comparison). Suppose $f(x) \leq g(x)$ for all $x$ "close" to $a$ (i.e., for all $x $in $(b,a)$ and $(a,c)$). Suppose $\lim\limits_{x \to a} f(x), \lim\limits_{x\to a} g(x)$ exist. Then $$ \lim_{x \to a} f(x) \leq \lim_{x \to a} g(x). $$

Page 40.png

$$ \lim_{x \to 0} x \sin \dfrac{1}{x} = 0.$$

Why the oscillations don't matter? The graph of $h$ is between the two diagonals; and these diagonals meet at 0. So, $h(x)$ has no where else to go but 0. Here how $h$ is “squeezed” between two functions:

  1. $ –x \leq h(x) \leq x$ for $x > 0 $
  2. $\lim\limits_{x \to 0} –x = 0$
  3. $\lim\limits_{x \to 0} x = 0$
  4. $0 = 0$, equal limits!

So, $h$ is "sandwiched" between two functions that approach the same value as $x$ approaches 0. Then, $h$ approaches that value as $x$ approaches 0. This is the "squeeze theorem", or "sandwich theorem", or "theorem about two policemen", etc.

Theorem (The Squeeze Theorem). Suppose $$f(x) \leq h(x) \leq g(x),$$ for all $x$ close to $a$. Suppose the limits of $f$ and $g$ at $x = a$, $\lim\limits_{x \to a} f(x)$ and $\lim\limits_{x \to a} g(x)$,

  1. exist and
  2. are equal to each other.

Then, $\lim\limits_{x \to a} h(x)$ exists and equal to that number.


Our original understanding of continuity of functions in calculus was as ones with no gaps in their graphs. It was confirmed (without proof) by the following.

Theorem (Intermediate Value Theorem). Suppose $f :[a,b] \to {\bf R}$ is continuous. Then for any $c$ between $f(a)$ and $f(b)$ there is $d \in [a,b]$ such that $f(d) = c$.

Intermediate value theorem.png

The derivative

Looking back we discover that so far we haven't explicitly used any geometry to develop calculus!

Of course we have relied on the mathematics from calculus courses that may rely on geometry. But did we really need the geometry? And was it avoidable?

First, we rely on limits that use the absolute value:

$\lim_{x\to a}f(x)=L$ if for any $\epsilon >0$ there is a $\delta >0$ such that $0<|x-a|<\delta$ implies that $|f(x)-L|<\epsilon$.

However, the definition can be easily rewritten without need for measuring:

$\lim_{x\to a}f(x)=L$ if for any open neighborhood $\epsilon$ of $L$ there is an open neighborhood $\delta$ of $a$ such that $x\in\delta$ implies $f(x)\in\epsilon$.

It only requires purely topological ideas such as basis of topology. This also covers the derivative.

The Riemann integral

Do we need geometry for integration?

A typical construction of the Riemann integral is as follows. The integral over an interval $[a,b]$ is the limit of its Riemann sums $R(P)$: $$I=\int_a^b f(x)dx=\lim _{||P||\rightarrow 0} R(P),$$ where $P$ is a partition of the interval $[a,b]$ and $||P||$ is its "mesh" (the largest length among the subintervals in $P$). In other words,

  • for any $\epsilon >0$ there is $\delta >0$ such that $|I-R(P)|<\epsilon$ whenever $||P||<\delta$.

This can be made purely topological:

  • for any $\epsilon >0$ there is open cover $\delta$ of $[a,b]$ such that $|I-R(P)|<\epsilon$ whenever all elements of $P$ are covered by the elements of $\delta$.

Exercise.

We have realized:

  • finding the position is equivalent to computing the area under the graph of the velocity,
  • this computation will probably have to be carried out by approximating the area by vertical bars.

Next we consider the general setup for Riemann sums.

We will approximate by "sampling" our function at several values of $x$, $n$ of them.

Given $y = f(x)$, defined on $[a,b], a < b$.

Given $n \geq 0$, $\text{integer } > 0$. Divide $[a,b]$ into $n$ intervals of equal length.

Interval.png

Then the intervals are: $$ [a_{0},a_{1}], [a_{1},a_{2}],... , [a_{n-1},a_{n}].$$ Compute the length of each: $$\Delta x = \frac{b-a}{n}$$ Note: Recall and compare to: $$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}. $$

Given the sample points, one in each interval, $$ c_{1} \text{ in } [a_{0},a_{1}], \, c_{2} \text{ in } [a_{1},a_{2}],... , c_{n} \text{ in } [a_{n-1},a_{n}].$$ Then the Riemann sum is $$f(c_{1})\Delta x + f(c_{2}) \Delta x + ... + f(c_{n})\Delta x. $$

Observe that in each term of this sum: $$\underbrace{f(c_{k})}_{\text{Height of the 1st rectangle}} \cdot \overbrace{\Delta x}^{\text{width of rectangle}} $$ is the area of the corresponding rectangle

This is what the construction looks like with $n=6$ and arbitrary sample points:

Riemann.png

Further, these are the rectangles:

AreaRectangles.png

$$\text{Riemann sum } = \text{ Total area of the rectangles}$$

Then what's the meaning of $f(c_{2})\Delta x$? It's the algebraic area of the rectangle.

Conclusion: the Riemann sum, and the integral, captures not the distance covered but rather the displacement.

NegativeDisplacement.png

Riemann sums approximate the algebraic area "under" the graph or, more precisely, the area between the graph and the $x$-axis.

Finally, what if we "refine" the approximation and keep refining, i.e., $n \to \infty$?

The Riemann integral $\int_{a}^{b} f(x) dx$ is the limit of the Riemann sums as $n \to \infty$.

$$ \underbrace{\int_{a}^{b} f(x)\, dx}_{\text{the exact area under the graph}} = \lim_{n \to \infty}\underbrace{\sum_{i=1}^{n} f(c_{i})\Delta x}_{\text{RS, areas of bars}} $$

Observe that we have more and more terms here ($n \to \infty$). Meanwhile, $\Delta x \to 0$ as $\Delta x = \frac{b-a}{n} \to 0$ as $n \to \infty$.

If the limit exists, $f$ is called integrable.

Integrable.png

Then "the area under the graph" makes sense!

Recall, $\int_{a}^{b} f(x) dx$ is the limit of the Riemann sums as $n \to \infty$. $$ \int_{a}^{b} f(x)\, dx = \lim_{n \to \infty}\sum_{i=1}^{n} f(c_{i})\Delta x. $$

Sums have very simple properties. Meanwhile, limits behave well with respect to addition. So, it's easy to derive properties of integrals.

Properties.

  1. Constant function rule:
    $$\int\limits_{a}^{b} c\, dx = c \left( b-a \right) $$
  2. Constant multiple rule:
    $$ \int_{a}^{b} c\, f(x)\, dx = c \, \int_{a}^{b} f(x)\, dx$$
  3. Sum Rule:
    $$\int_{a}^{b} \left( f(x) + g(x) \right) \, dx = \int_{a}^{b} f(x)\, dx + \int_{a}^{b} g(x)\, dx $$

These correspond to the rules of limits. (But there is no product rule or quotient rule.)

Additivity: $$\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx. $$

The area interpretation of additivity:

SumAreas.png

Or, $$\int_{a}^{b} f(x) \, dx (orange)+ \int_{b}^{c} f(x) \, dx (green)= \int_{a}^{c} f(x) \, dx (purple). $$ In other words, $$\text{Sum of the two areas } = \text{ The total area from } a \text{ to } c$$

Comparison Properties.

1. If $f(x) \geq 0$ on $[a,b]$ then $$ \int_{a}^{b} f(x) \, dx \geq 0 $$ (To prove, look at the RS, all terms $\geq 0$, $\Delta x \geq 0$, then use a comparison theorem for limits).

2. If $f(x) \geq g(x)$ on $[a,b]$ then $$\int_{a}^{b} f(x) \, dx \geq \int_{a}^{b} g(x) \, dx $$ Here: green LHS and orange RHS.

Comparison2.png