# A Math on Twitter 5: Topologizing a Set by Continuous Functions

Let $f: X\longrightarrow Y$ be a function. The continuity of $f$ is then determined by the topologies on $X$ and $Y$. To be precise, $f: X\longrightarrow Y$ is continuous on $X$ if for every open set $U\subset Y$ $f^{-1}(U)$ is open in $X$. But if you have a clear idea of what class of functions, say $\{f_\alpha:X\longrightarrow Y|\alpha\in\mathscr{A}\}$, should be continuous, you can also define a topology on $X$ that makes $f_\alpha :X\longrightarrow Y$ continuous for all $\alpha\in\mathscr{A}$ as Sam Walters mentioned on his tweet. Here is how. Let $$\mathscr{S}=\{f_{\alpha}^{-1}(U)| \mbox{U is open in Y}, \alpha\in\mathscr{A}\}$$
Since $\bigcup\{G|G\in\mathscr{S}\}=X$, $\mathscr{S}$ is a subbase for a topology in $X$. (Note: $\mathscr{S}$ is not a base for a topology in $X$ unless $\mathscr{A}$ is a singleton set.) Denote by $\tau(\mathscr{S})$ the topology generated by the subbase $\mathscr{S}$. Then $\tau(\mathscr{S})$ is the smallest topology on $X$ that makes $f_\alpha :X\longrightarrow Y$ continuous for all $\alpha\in\mathscr{A}$. The reason is any topology on $X$ that makes $f_\alpha :X\longrightarrow Y$ continuous for all $\alpha\in\mathscr{A}$ would include $\mathscr{S}$. We have in fact seen a similar idea which is Tychonoff product topology. Consider the cartesian product
$$\prod_{\alpha\in\mathscr{A}}X_\alpha=\{c:\mathscr{A}\longrightarrow\bigcup_{\alpha\in\mathscr{A}}X_\alpha\}$$
of topological spaces $(X_\alpha,\tau_\alpha)$, $\alpha\in\mathscr{A}$. The topology we want is the one that makes the projection maps $\pi_\alpha :\prod_{\alpha\in\mathscr{A}}X_\alpha\longrightarrow X_\alpha$ for all $\alpha\in\mathscr{A}$ continuous in particular the smallest one. Let $$\mathscr{S}=\{\pi_\alpha^{-1}(U_\alpha)|U_\alpha\in\tau_\alpha,\forall\alpha\in\mathscr{A}\}$$
Then $\mathscr{S}$ is a subbase for a topology called the Tychonoff product topology. This is the smallest topology that makes the projection maps continuous. The projection maps are also open.

Given a surjective function $p: X\longrightarrow Y$ from a topological space $X$ onto a set $Y$, we can also define a topology on $Y$ that makes $p$ continuous: $U\subset Y$ is open if and only if $p^{-1}(U)$ is open in $X$. This topology on $Y$ automatically makes $p$ continuous. Moreover it is the largest topology on $Y$ that makes $p$ continuous. This topology is called the identification topology. The identification topology can be used to get a new topological space (identification space) from an old one. More specifically, it can be defined on a partition of $X$ or equivalently a quotient set of $X$ modulo an equivalence relation and it makes the canonical projection $\pi$ continuous. For example, let $X$ be the unit square $[0,1]\times[0,1]$ in $\mathbb{E}^2$ with the subspace topology. Partition $X$ into the following subsets:

1. $\{(0,0),(1,0),(0,1),(1,1)\}$ i.e. the set of four corner points.
2. $\{(x,0),(x,1)|0<x<1\}$
3. $\{(0,y),(1,y)|0<y<1\}$
4. $\{(x,y)\}$ where $0<x<1$, $0<y<1$.

The resulting identification space is the torus which is homeomorphic to $S^1\times S^1$ as shown in Figure 1.

Figure 1. Clifford Torus S^1 x S^1

References:

1. M. A. Armstrong, Basic Topology, Springer-Verlag, 1983
2. Benjamin T. Sims, Fundamentals of Topology, Collier Macmillan

# A Math on Twitter 4: Kepler’s Law

In his recent tweet, Sam obtained Kepler’s (second) law simply by using polar coordinates, integrals and conservation law of angular momentum. In this note I discuss basic physics about conservation law of angular momentum and Kepler’s second law as its consequence.

What is angular momentum?

Let $r$ be a vector from a fixed point (called the pivot).

Then the angular momentum is given by $$L=r\times p$$ where $p=mv$ is the linear momentum of the mass $m$. \begin{align*}\frac{dL}{dt}&=\frac{d}{dt}(r\times mv)\\&=\frac{dr}{dt}\times mv+r\times\frac{d(mv)}{dt}\\&=v\times mv+r\times\frac{dp}{dt}\\&=r\times\frac{dp}{dt}\end{align*} since $v\times mv=0$. That is, $\frac{dL}{dt}=r\times F$ and this is called torque. If torque $r\times F=0$ then $L$ is constant. This is conservation law of angular momentum. $r\times F=0$ if and only if $r$ and $F$ are parallel or antiparallel except for the trivial cases $r=0$ or $F=0$. A force that acts exclusively parallel or antiparallel to the position vector is called a central force. That is to say, central forces obey conservation law of angular momentum.

Conservation law of angular momentum implies Kepler’s second law

The area $dA$ spanned by $r$ and $dr$ is $$dA=\frac{1}{2}|r\times dr|$$

Figure 2. The area dA spanned by r and dr.

\begin{align*}\frac{dA}{dt}&=\frac{1}{2}|r\times v|\\&=\frac{1}{2m}|r\times mv|\\&=\frac{1}{2m}|L|\end{align*} $\frac{dA}{dt}$ is the area velocity of the radial vector $r$. It measures how fast area is covered per unit time. For the planetary motion gravitational force is a central force so $L$ is constant which means $\frac{dA}{dt}$ is constant. Hence conservation law of angular momentum implies the second Kepler law: The radial vector $r$ of a planet sweeps equal areas in equal time.

References:

Walter Greiner, Classical Mechanics, Point Particles and Relativity, Springer-Verlag, 2004

# A Math on Twitter 3

Problem: Let $A$ and $B$ be $n\times n$ matrices such that their sum $A+B$ is invertible. Then show that $$A(A+B)^{-1}B=B(A+B)^{-1}A$$ (Hat tip: Sam Walters)

Solution. \begin{aligned}I&=(A+B)(A+B)^{-1}\\&=A(A+B)^{-1}+B(A+B)^{-1}\end{aligned}\label{eq:matrix} Multiply \eqref{eq:matrix} by $B$ from the right $$\label{eq:matrix2}B=A(A+B)^{-1}B+B(A+B)^{-1}B$$ Also multiply \eqref{eq:matrix} by $A$ from the left $$\label{eq:matrix3}A=A(A+B)^{-1}A+B(A+B)^{-1}A$$ Subtract \eqref{eq:matrix3} from \eqref{eq:matrix2}. $$\label{eq:matrix4}B-A=A(A+B)^{-1}B-B(A+B)^{-1}A+B(A+B)^{-1}B-A(A+B)^{-1}A$$ In a similar manner from $I=(A+B)^{-1}(A+B)$, we obtain $$\label{eq:matrix5}A-B=A(A+B)^{-1}B-B(A+B)^{-1}A+A(A+B)^{-1}A-B(A+B)^{-1}B$$ \eqref{eq:matrix4}+\eqref{eq:matrix5} results $$A(A+B)^{-1}B=B(A+B)^{-1}A$$

A mathematician who Twitter username is Manifoldless beat me to it by a few minutes :). But not just that. His solution is shorter and smarter than mine: \begin{align*}A(A+B)^{-1}B&=(A+B-B)(A+B)^{-1}(A+B-A)\\&=[I-B(A+B)^{-1}](A+B-A)\\&=A+B-A-B+B(A+B)^{-1}A\\&=B(A+B)^{-1}A\end{align*}

# A Math on Twitter 2

Problem: Given $x,y\geq 0$ satisfying $$\label{eq:ellipse}x+y+\sqrt{2x^2+2xy+3y^2}=4$$ prove $x^2y<4$. (Hat tip: Sam Walters)

Solution. First rewrite \eqref{eq:ellipse} as $$\label{eq:ellipse2}\sqrt{2x^2+2xy+3y^2}=4-x-y$$ Squaring \eqref{eq:ellipse} we obtain an equation of ellipse $$\label{eq:ellipse3}(x+4)^2+2(y+2)^2=40$$ (Figire 1)

Figure 1

Graphically we see that the inequality holds as shown in Figure 2.

Figure 2. Ellipse (x+4)^2+2(y+2)^2=40, x=0..-4+sqrt(40) (red) and y=4/x^2 (blue)

Suppose $x>0$ (for $x=0$ the inequality $x^2y<4$ is trivial). Since $x>0,y>0$ then, $$x^2y<4\Longleftrightarrow (y+2)^2<\left(\frac{4}{x^2}+2\right)^2$$ Solve \eqref{eq:ellipse3} for $(y+2)^2$. $$\label{eq:ellipse4}(y+2)^2=20-\frac{(x+4)^2}{2}$$ Now subtract $\left(\frac{4}{x^2}+2\right)^2$ from the RHS of \eqref{eq:ellipse4}. $$20-\frac{(x+4)^2}{2}-\left(\frac{4}{x^2}+2\right)^2=\frac{-x^6-8x^5+16x^4-32x^2-32}{2x^4}<0$$ since $-x^6-8x^5+16x^4-32x^2-32<0$ for $0<x<-4+\sqrt{40}$ as shown in Figure 3.

Figure 3. The graph of f(x)=-x^6-8x^5+16x^4-32x^2-32, x=0..-4+sqrt(40)

Update: Republic of Math graphically came up with a sharper inequality $x^2y<1$. The graphics can be seen here. As you can see in the graphics, there is still room for even (slightly) more sharp inequality. In fact $x^2y<0.9$ as you can see in Figure 4 below.

Figure 4. Ellipse (x+4)^2+2(y+2)^2=40, x=0..-4+sqrt(40) (red) and y=0.9/x^2 (blue)

Update: While I could not analytically find the smallest value of $a>0$ such that $x^2y<a$, I found graphically that $a$ can be as small as $0.789$. Figure 5 and Figure 6 are the graphs of $f(x)=-x^6-8x^5+16x^4-8ax^2-2a^2$ for $0\leq x\leq -4+\sqrt{40}$ for $a=0.789$ and for $a=0.788$, respectively. For $a=0.789$, $f(x)<0$ on $[0,-4+\sqrt{40}]$.

Figure 5. The graph of f(x)=-x^6-8x^5+16x^4-8ax^2-2a^2 (for a=0.789), x=0..-4+sqrt(40)

But with $a=0.788$ $f(x)$ is no longer negative for all $x$ in $[0,-4+\sqrt{40}]$.

Figure 5. The graph of f(x)=-x^6-8x^5+16x^4-8ax^2-2a^2 (for a=0.788), x=0..-4+sqrt(40)

# The Relation Between Volume and Surface Area

The derivative of the volume $V=\frac{4}{3}\pi r^3$ of a sphere with radius $r$ with respect to $r$ is the surface area $S=4\pi r^2$. Is this a coincidence? It is not. There is a good reason why it happened that way. To understand it more easily, let us take a look at it’s lower dimensional analogue, namely the derivative of the area $\pi r^2$ of a circle with radius $r$ is the circumference $2\pi r$. Let me first explain why we get that. Let $A(r)=\pi r^2$. Then
$$\frac{dA}{dr}=\lim_{\Delta r\to 0}\frac{\pi(r+\Delta r)^2-\pi r^2}{\Delta r}.$$
$\pi(r+\Delta r)^2-\pi r^2$ is the area (shape of washer) between two circles both centered at the origin with radii $r+\Delta r$ and $r$, respectively as seen in the following figure:

As $\Delta r$ gets smaller, you see that the washer gets thiner. So if $\Delta r\to 0$, the washer becomes circle of radius $r$ resulting $\frac{dA}{dr}$ its circumference. By doing the same analysis, you can see why the derivative of the volume of sphere is its surface area.

Do any other objects share the relationship? Sure! For instance, the derivative of the area $x^2$ of a square with side $x$ is its circumference $2x$. The derivative of the area $\pi r^2h$ of a cylinder with radius $r$ and height $h$ is its lateral surface area $2\pi rh$. How about a cube with volume $V=x^3$? In this case the derivative is $3x^2$ so it is not the surface area. Why is this? It is due to symmetry i.e. it depends on whether the volume increases symmetrically when you increase your variable such as length of side, radius, or height, etc. In case of a cube, increasing length from $x$ to $x+\Delta x$ results increase of volume from only three faces of the cube i.e. the volume of cube does not increase symmetrically in that case. When $\Delta x\to 0$, the volume increment becomes three faces resulting the derivative the area of those three faces $3x^2$.

How about a box with volume $V=xyz$? In this case, divergence, not gradient, would give a similar relationship. In fact, $\nabla\cdot V=yz+xz+xy$ which is the surface area of box with length $x$, width $y$ and height $z$. Imagine that a box is filled with fluid and assume that volume increase amounts to the fluid flowing into the box through its faces. $\frac{\partial V}{\partial x}$ would measure the rate of fluid flowing into the $yz$ face per unit time. That would indeed be the same as the area of the face $yz$.