Fall 2015 Sunset, Spring 2016 Dawn

Let me digress for a second: I delete more drafts than I publish. I guess that that is the writing process, as are many other things in life.

Last semester, I lightheartedly complained that algebra was my toughest class. Now that the semester has come to an end, in hindsight, all of my classes were extraordinarily fundamental. There certainly was a lot to take in, but to the advantage of the student, it was basic material; not much wit was necessary to answer the countless questions that were asked, but a solid knowledge and understanding of the definitions and results was.

As I enrolled in my next batch of classes for the coming semester and skimmed through the material that we will be covering, I decided that the most sensible thing that I can do right now is begin studying. I will be taking three courses, which I will arrange in increasing order of difficulty.
  1. Introduction to Real Analysis II
  2. Modern Algebra II
  3. Geometry of Manifolds
I am expecting analysis to be a smooth continuation of the not-too-difficult course that I took last semester. I felt tremendously confident with it and am expecting to feel the same way this time. Algebra will always be algebra: Tough. But since most of the fundamental definitions are already out of the way, I am expecting it to feel not as brutal as it felt last semester—Galois theory aside, of course (that will be insane). Now, geometry of manifolds will be an incarnate demon. It is the natural continuation of topology, a class in which I excelled, but the material is so alien to me right now and the professor so sophisticated that I am decidedly expecting it to be my toughest course by far, ever.

Until classes begin, I will dedicate the next few blog entries to geometry of manifolds, to accrue some basic definitions and results.

Geometry of Manifolds


A topological space $M$ is called a topological $n$-manifold if it has the following properties:
  • $M$ is Hausdorff,
  • $M$ is second countable, and
  • $M$ is locally Euclidean of dimension $n$.
Lemma. If $M$ is a second countable topological space, then every open cover of $M$ has a countable subcover.

A chart on $M$ is a pair $\left(U,\varphi\right)$, where $U$ is an open subset of $M$ and $\varphi:U\to\tilde U$ is a homeomorphism from $U$ to an open subset $\tilde U=\varphi\left(U\right)\subseteq\mathbb R^n$.

If $\left(U,\varphi\right)$ and $\left(V,\psi\right)$ are two charts such that $U\cap V\neq\varnothing$, then the composite map $\psi\circ\varphi^{-1}:\varphi\left(U\cap V\right)\to\psi\left(U\cap V\right)$, called the transition map from $\varphi$ to $\psi$, is a composition of homeomorphisms, and is therefore itself a homeomorphism. Two charts $\left(U,\varphi\right)$ and $\left(V,\psi\right)$ are said to be smoothly compatible if either $U\cap V=\varnothing$ or the transition map $\psi\circ\varphi^{-1}$ is a diffeomorphism.

We define an atlas for $M$ to be a collection of charts whose domains cover $M$. An atlas $\mathcal A$ is called a smooth atlas if any two charts in $\mathcal A$ are smoothly compatible with each other. A smooth atlas $\mathcal A$ on $M$ is called maximal if it is not contained in any strictly larger smooth atlas.

A smooth structure on a topological $n$-manifold $M$ is a maximal smooth atlas. A smooth manifold is a pair $\left(M,\mathcal A\right)$, where $M$ is a topological manifold and $\mathcal A$ is a smooth structure on $M$. Smooth structures are also called differentiable structures or $C^\infty$ structures by some authors.

Lemma. If $M$ is a topological manifold, then every smooth atlas for $M$ is contained in a unique maximal smooth atlas, and two smooth atlases for $M$ determine the same maximal smooth atlas if and only if their union is a smooth atlas.

I need to further peruse and dissect my book and the literature to continue...

Real Analysis: Riemann Integration and Beyond

Introduction


I blog for myself and hope that my public rantings will be useful to someone else.

They have!

In this entry, I will go over the material that my analysis exam will cover next week. This helps me to solidify what I already know.

To save time, I will not prove claims; for now, I will appeal to the proverbial it is left as an exercise to the reader.

This entry is a work in progress.

Riemann Integration


Definitions


Let $f$ be a bounded function defined on an interval $\left[a,b\right]$. A partition of $\left[a,b\right]$ is a finite set $P=\left\{a=x_0<x_1<\cdots<x_n=b\right\}$. Set $\Delta_j=x_j-x_{j-1}$ and define the mesh of a partition $P$ as $\text{mesh}\left(P\right)=\max_{1\leqslant j\leqslant n}\Delta_j$. For each interval $\left[x_{j-1},x_j\right]$ of this partition, we define the maximum and minimum of $f$ on this interval by
$$M_j\left(f,P\right)=\sup_{x_{j-1}\leqslant x\leqslant x_j}f\left(x\right)\qquad\text{and}\qquad m_j\left(f,P\right)=\inf_{x_{j-1}\leqslant x\leqslant x_j}f\left(x\right).$$
Then we define the upper and lower sums of $f$ with respect to the partition $P$ by
$$U\left(f,P\right)=\sum_{j=1}^nM_j\left(f,P\right)\Delta_j\qquad\text{and}\qquad L\left(f,P\right)=\sum_{j=1}^nm_j\left(f,P\right)\Delta_j.$$
If, in addition, we are given a set of points $X=\left\{x_j':1\leqslant j\leqslant n\right\}$, where $x_j'\in\left[x_{j-1},x_j\right]$ for $1\leqslant j\leqslant n$, we define the Riemann sum
$$I\left(f,P,X\right)=\sum_{j=1}^nf\left(x_j'\right)\Delta_j.$$
A partition $R$ is a refinement of a partition $P$ provided that $P\subseteq R$. If $P$ and $Q$ are two partitions, then $R$ is a common refinement of $P$ and $Q$ provided that $P\cup Q\subseteq R$.

Observe that, by definition, $L\left(f,P\right)\leqslant I\left(f,P,X\right)\leqslant U\left(f,P\right)$.

Define $L\left(f\right)=\sup_PL\left(f,P\right)$ and $U\left(f\right)=\inf_PU\left(f,P\right)$. A bounded function $f$ on a finite interval $\left[a,b\right]$ is called Riemann integrable if $L\left(f\right)=U\left(f\right)$. In this case, we write
$$\int_a^bf\left(x\right)\,dx$$
for the common value.

Facts


Lemma. If $R$ is a refinement of $P$, then
$$L\left(f,P\right)\leqslant L\left(f,R\right)\leqslant U\left(f,R\right)\leqslant U\left(f,P\right).$$
Corollary. If $P$ and $Q$ are any two partitions of $\left[a,b\right]$, then
$$L\left(f,P\right)\leqslant U\left(f,Q\right).$$
In particular, we see that the set of numbers $\left\{L\left(f,P\right)\right\}$ is bounded above by any $U\left(f,Q\right)$. Hence by the completeness of $\mathbb R$, $\sup_PL\left(f,P\right)$ is defined. Moreover, $\sup_PL\left(f,P\right)\leqslant U\left(f,Q\right)$ for every partition $Q$. Therefore, $\inf_PU\left(f,P\right)$ is defined and $\sup_PL\left(f,P\right)\leqslant\inf_PU\left(f,P\right)$.

Riemann's Condition. Let $f$ be a bounded function on $\left[a,b\right]$. The following are equivalent:

1. $f$ is Riemann integrable.
2. For each $\varepsilon>0$, there is a partition $P$ so that $U\left(f,P\right)-L\left(f,P\right)<\varepsilon$.

Corollary. Let $f$ be a bounded real-valued function on $\left[a,b\right]$. If there is a sequence of partitions of $\left[a,b\right]$, $P_n$, so that
$$\lim_{n\to\infty}U\left(f,P_n\right)-L\left(f,P_n\right)=0,$$
then $f$ is Riemann integrable. Moreover, if $X_n$ is any choice of points $x_{n,j}'$ selected from each interval of $P_n$, then
$$\lim_{n\to\infty}I\left(f,P_n,X_n\right)=\int_a^bf\left(x\right)\,dx.$$
The evenly-spaced partition is often used. However, a strategically-chosen partition could be much better.

Theorem. Let $f$ be a bounded function on $\left[a,b\right]$. The following are equivalent:

1. $f$ is Riemann integrable.
2. For each $\varepsilon>0$, there is a partition $P$ so that $U\left(f,P\right)-L\left(f,P\right)<\varepsilon$.
3. For every $\varepsilon>0$, there is a $\delta>0$ so that every partition $Q$ such that $\text{mesh}\left(Q\right)<\delta$ satisfies $U\left(f,Q\right)-L\left(f,Q\right)<\varepsilon$.
4. For every $\varepsilon>0$, there is a $\delta>0$ so that every partition $Q$ such that $\text{mesh}\left(Q\right)<\delta$ and every choice of set $X=\left\{x_j':1\leqslant j\leqslant n\right\}$, where $x_j'\in\left[x_{j-1},x_j\right]$ satisfies
$$\left|I\left(f,Q,X\right)-\int_a^bf\left(x\right)\,dx\right|<\varepsilon.$$
Theorem. Every monotone function on $\left[a,b\right]$ is Riemann integrable.

Theorem. Every continuous function on $\left[a,b\right]$ is Riemann integrable.

The Fundamental Theorem of Calculus


Definitions


A function $f$ on $\left[a,b\right]$ has an antiderivative if there is a continuous function $F\left(x\right)$ on $\left[a,b\right]$ such that $F'\left(x\right)=f\left(x\right)$ for every point $x\in\left(a,b\right)$.

Facts


Theorem (Fundamental Theorem of Calculus). Let $f$ be a bounded Riemann integrable function on $\left[a,b\right]$, and let
$$F\left(x\right)=\int_a^xf\left(t\right)\,dt\qquad\text{for}\qquad a\leqslant x\leqslant b.$$
Then $F$ is a continuous function. If $f$ is continuous at a point $x_0$, then $F$ is differentiable at $x_0$ and $F'\left(x_0\right)=f\left(x_0\right)$.

Corollary. Let $f$ be a continuous function on $\left[a,b\right]$. Then $f$ has an antiderivative. Moreover, if $G$ is any antiderivative of $f$, then
$$\int_a^bf\left(x\right)\,dx=G\left(b\right)-G\left(a\right).$$
Lemma. Suppose that $f$ is an integrable function on $\left[a,b\right]$ bounded by $M$. Then
$$\left|\int_a^bf\left(t\right)\,dt\right|\leqslant M\left(b-a\right).$$
Remark. A jump discontinuity in the integrand $f$ can result in a point where the integral is not differentiable. Nor is it the case that every differentiable function is an integral.

Normed Vector Spaces, Inner Product Spaces, and $L^p$ Norms


Definitions


Let $V$ be a vector space over $\mathbb R$. A norm on $V$ is a function $\left\|\cdot\right\|$ on $V$ taking values in $\left[0,+\infty\right)$ with the following properties:

1. (positive definite) $\left\|x\right\|=0$ if and only if $x=0$,
2. (homogeneous) $\left\|\alpha x\right\|=\left|\alpha\right|\left\|x\right\|$ for all $x\in V$ and $\alpha\in\mathbb R$, and
3. (triangle inequality) $\left\|x+y\right\|\leqslant\left\|x\right\|+\left\|y\right\|$ for all $x,y\in V$.

We call the pair $\left(V,\left\|\cdot\right\|\right)$ a normed vector space.

Let $x=\left(x_1,x_2,\dots,x_n\right)\in\mathbb R^n$. Then
$$\left\|x\right\|_1=\sum_{i=1}^n\left|x_i\right|\qquad\qquad\qquad\qquad\left\|x\right\|_2=\left(\sum_{i=1}^n\left|x_i\right|^2\right)^{1/2}\qquad\qquad\qquad\qquad\left\|x\right\|_\infty=\max_{1\leqslant i\leqslant n}\left|x_i\right|$$
The above norms are well known.

For any normed vector space $\left(V,\left\|\cdot\right\|\right)$, the unit ball of $V$ is the set $\left\{x\in V:\left\|x\right\|\leqslant1\right\}$.

Let $K$ be a compact subset of $\mathbb R^n$, and let $C\left(K\right)$ denote the vector space of all continuous real-valued functions on $K$. The most natural and important norm on this vector space is the uniform norm given by
$$\left\|f\right\|_\infty=\sup_{x\in K}\left|f\left(x\right)\right|.$$
In a normed vector space $\left(V,\left\|\cdot\right\|\right)$, we say that a sequence $\left\{v_n\right\}_{n=1}^\infty$ converges to $v\in V$ if $\lim_{n\to\infty}\left\|v_n-v\right\|=0$. Equivalently, for every $\varepsilon>0$, there is an integer $N>0$ so that $\left\|v_n-v\right\|<\varepsilon$ for all $n\geqslant N$. This is written $\lim_{n\to\infty}v_n=v$.

Call $\left\{v_n\right\}_{n=1}^\infty$ a Cauchy sequence if for every $\varepsilon>0$, there is an integer $N>0$ so that $\left\|v_n-v_m\right\|<\varepsilon$ for all $n,m\geqslant N$.

Say that $\left(V,\left\|\cdot\right\|\right)$ is complete if every Cauchy sequence in $V$ converges to some vector $v\in V$. A complete normed space is called a Banach space.

For a normed vector space $\left(V,\left\|\cdot\right\|\right)$, we define the open ball with center $a\in V$ and radius $r>0$ to be $B_r\left(a\right)=\left\{v\in V:\left\|v-a\right\|<r\right\}$. A subset $U$ of $V$ is open if for every $a\in U$, there is some $r>0$ so that $B_r\left(a\right)\subseteq U$. A subset $C$ of $V$ is closed if it contains all of its limit points. That is, whenever $\left\{x_n\right\}_{n=1}^\infty$ is a convergent sequence of points in $C$ with limit $x=\lim_{n\to\infty}x_n$, then $x$ belongs to $C$.

A subset $K$ of a normed vector space $V$ is compact if every sequence $\left\{x_n\right\}_{n=1}^\infty$ of points in $K$ has a sub-sequence $\left\{x_{n_i}\right\}_{i=1}^\infty$ which converges to a point in $K$.

An inner product on a vector space $V$ is a function $\langle x,y\rangle$ on pairs $\left(x,y\right)$ of vectors in $V\times V$ taking values in $\mathbb R$ satisfying the following properties:

1. (positive definiteness) $\langle x,x\rangle\geqslant0$ for all $x\in V$ and $\langle x,x\rangle=0$ only if $x=0$.
2. (symmetry) $\langle x,y\rangle=\langle y,x\rangle$ for all $x,y\in V$.
3. (bilinearity) For all $x,y,z\in V$ and scalars $\alpha,\beta\in\mathbb R$, $\langle\alpha x+\beta y,z\rangle=\alpha\langle x,z\rangle+\beta\langle y,z\rangle$.

The $L^p$ norms on $C\left[a,b\right]$ for $1\leqslant p<\infty$ are defined by
$$\left\|f\right\|_p=\left(\int_a^b\left|f\left(x\right)\right|^p\,dx\right)^{1/p}.$$
For $1\leqslant p<\infty$, $\ell^p$ consists of the set of all finite sequences $a=\left\{a_n\right\}_{n=1}^\infty$ such that
$$\left\|a\right\|_p=\left(\sum_{n=1}^\infty\left|a_n\right|^p\right)^{1/p}<\infty.$$
Let $w\left(x\right)$ be a strictly positive piecewise continuous function on $\left[a,b\right]$. Then define a norm $\left\|\cdot\right\|_{L^p\left(w\right)}$ on $C\left[a,b\right]$ or even on the space $PC\left[a,b\right]$ of piecewise continuous functions by
$$\left\|f\right\|_{L^p\left(w\right)}=\left(\int_a^b\left|f\left(x\right)\right|^pw\left(x\right)\,dx\right)^{1/p}.$$

Facts


Claim. A sequence $\left\{x_n\right\}_{n=1}^\infty$ in a normed vector space $V$ converges to a vector $x$ if and only if for each open set $U$ containing $x$, there is an integer $N$ so that $x_n\in U$ for all $n\geqslant N$.

Cauchy-Schwarz Inequality. For all vectors $x,y$ in an inner product space $V$,
$$\left|\langle x,y\rangle\right|\leqslant\left\|x\right\|\left\|y\right\|.$$
Equality holds if and only if $x$ and $y$ are collinear.

Corollary. For $f,g\in C\left[a,b\right]$, we have
$$\left|\int_a^bf\left(x\right)g\left(x\right)\,dx\right|\leqslant\left(\int_a^bf\left(x\right)^2\,dx\right)^{1/2}\left(\int_a^bg\left(x\right)^2\,dx\right)^{1/2}.$$
Corollary. An inner product space $V$ satisfies the triangle inequality
$$\left\|x+y\right\|\leqslant\left\|x\right\|+\left\|y\right\|\qquad\text{for all}\qquad x,y\in V.$$
Moreover, if equality occurs, then $x$ and $y$ are collinear.

Corollary. Let $V$ be an inner product space with induced norm $\left\|\cdot\right\|$. Then the inner product is continuous (i.e., if $x_n$ converges to $x$ and $y_n$ converges to $y$, then $\langle x_n,y_n\rangle$ converges to $\langle x,y\rangle$).

Lemma. Let $A,B>0$. Then
$$A^tB^{1-t}\leqslant tA+\left(1-t\right)B\qquad\text{for all}\qquad0<t<1.$$
Moreover, equality holds for some (or all) $t$ only if $A=B$.

Hölder's Inequality. Let $w$ be a positive function on an interval $\left[a,b\right]$. Let $f\in L^p\left(w\right)$ and $g\in L^q\left(w\right)$ where $1<p<\infty$ and $1/p+1/q=1$. Then
$$\left|\int_a^bf\left(x\right)g\left(x\right)w\left(x\right)\,dx\right|\leqslant\left\|f\right\|_{L^p\left(w\right)}\left\|g\right\|_{L^q\left(w\right)}.$$
Minkowski's Inequality. The triangle inequality holds for $L^p\left(w\right)$, that is,
$$\left(\int_a^b\left|f\left(x\right)+g\left(x\right)\right|^pw\left(x\right)\,dx\right)^{1/p}\leqslant\left(\int_a^b\left|f\left(x\right)\right|^pw\left(x\right)\,dx\right)^{1/p}+\left(\int_a^b\left|g\left(x\right)\right|^pw\left(x\right)\,dx\right)^{1/p}.$$

Fundamentals of Point-Set Topology, Part 1

In anticipation of a course that I will soon be taking as a part of my first semester as a math Ph.D. student, I will be starting a straightforward series of blog entries containing fundamental results of general (point-set) topology (as well as other courses). Mind you, they will contain nothing extraordinary (I might introduce something interesting every once in a while, though); this is purely to keep track of the minutiae that I might easily forget down the road.



A topology on a set $X$ is a collection $\mathscr T$ of subsets of $X$ having the following properties:
  1. $\varnothing$ and $X$ are in $\mathscr T$.
  2. The union of the elements of any subcollection of $\mathscr T$ is in $\mathscr T$.
  3. The intersection of the elements of any finite subcollection of $\mathscr T$ is in $\mathscr T$.
A set $X$ for which a topology $\mathscr T$ has been specified is called a topological space.

Properly speaking, a topological space is an ordered pair $(X,\mathscr T)$ consisting of a set $X$ and a topology $\mathscr T$ on $X$.

If $X$ is a topological space with topology $\mathscr T$, we say that a subset $U$ of $X$ is an open set of $X$ if $U$ belongs to the collection $\mathscr T$.

Using this terminology, one can say that a topological space is a set $X$ together with a collection of subsets of $X$, called open sets, such that $\varnothing$ and $X$ are both open, and such that arbitrary unions and finite intersections of open sets are open.



If $X$ is any set, the collection of all subsets of $X$ is a topology on $X$; it is called the discrete topology. The collection consisting of $X$ and $\varnothing$ only is also a topology on $X$; we shall call it the indiscrete topology, or the trivial topology.

Let $X$ be a set; let $\mathscr T_f$ be the collection of all subsets $U$ of $X$ such that $X-U$ either is finite or is all of $X$. Then $\mathscr T_f$ is a topology on $X$, called the finite complement topology.



Suppose that $\mathscr T$ and $\mathscr T'$ are two topologies on a given set $X$. If $\mathscr T'\supset\mathscr T$, we say that $\mathscr T'$ is finer than $\mathscr T$; if $\mathscr T'$ properly contains $\mathscr T$, we say that $\mathscr T'$ is strictly finer than $\mathscr T$. We also say that $\mathscr T$ is coarser than $\mathscr T'$, or strictly coarser, in these two respective situations. We say that $\mathscr T$ is comparable with $\mathscr T'$ if either $\mathscr T'\supset\mathscr T$ or $\mathscr T\supset\mathscr T'$.



If $X$ is a set, a basis for a topology on $X$ is a collection $\mathscr B$ of subsets of $X$ (called basis elements) such that
  1. For each $x\in X$, there is at least one basis element $B$ containing $x$.
  2. If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3\subset B_1\cap B_2$.
If $\mathscr B$ satisfies these two conditions, then we define the topology $\mathscr T$ generated by $\mathscr B$ as follows: A subset $U$ of $X$ is said to be open in $X$ (that is, to be an element of $\mathscr T$) if for each $x\in U$, there is a basis element $B\in\mathscr B$ such that $x\in B$ and $B\subset U$. Note that each basis element is itself an element of $\mathscr T$.



Let us prove that the collection $\mathscr T$ generated by the basis $\mathscr B$ is, in fact, a topology on $X$. If $U$ is the empty set, it satisfies the defining condition of openness vacuously. Likewise, $X$ is in $\mathscr T$, since for each $x\in X$ there is some basis element $B$ containing $x$ and contained in $X$. Now let us take an indexed family $\left\{U_\alpha\right\}_{\alpha\in J}$ of elements of $\mathscr T$ and show that$$U=\bigcup_{\alpha\in J}U_\alpha$$belongs to $\mathscr T$. Given $x\in U$, there is an index $\alpha$ such that $x\in U_\alpha$. Since $U_\alpha$ is open, there is a basis element $B$ such that $x\in B\subset U_\alpha$. Then $x\in B$ and $B\subset U$, so that $U$ is open, by definition.

Now, let us take two elements $U_1$ and $U_2$ of $\mathscr T$ and show that $U_1\cap U_2$ belongs to $\mathscr T$. Given $x\in U_1\cap U_2$, choose a basis element $B_1$ containing $x$ such that $B_1\subset U_1$; choose also a basis element $B_2$ containing $x$ such that $B_2\subset U_2$. The second condition for a basis enables us to choose a basis element $B_3$ containing $x$ such that $B_3\subset B_1\cap B_2$. Then $x\in B_3$ and $B_3\subset U_1\cap U_2$, so $U_1\cap U_2$ belongs to $\mathscr T$, by definition.

Finally, we show by induction that any finite intersection $U_1\cap\cdots\cap U_n$ of elements of $\mathscr T$ is in $\mathscr T$. This fact is trivial for $n=1$; we suppose it true for $n-1$ and prove it for $n$. Now$$(U_1\cap\cdots\cap U_n)=(U_1\cap\cdots\cap U_{n-1})\cap U_n.$$By hypothesis, $U_1\cap\cdots\cap U_{n-1}$ belongs to $\mathscr T$; by the result just proved, the intersection of $U_1\cap\cdots\cap U_{n-1}$ and $U_n$ also belongs to $\mathscr T$.



Another way of describing the topology generated by a basis is given in the following lemma:

Lemma. Let $X$ be a set; let $\mathscr B$ be a basis for a topology $\mathscr T$ on $X$. Then $\mathscr T$ equals the collection of all unions of elements of $\mathscr B$.

Proof. Given a collection of elements of $\mathscr B$, they are also elements of $\mathscr T$. Because $\mathscr T$ is a topology, their union is in $\mathscr T$. Conversely, given $U\in\mathscr T$, choose for each $x\in U$ an element $B_x$ of $\mathscr B$ such that $x\in B_x\subset U$. Then $U=\bigcup_{x\in U}B_x$, so $U$ equals a union of elements of $\mathscr B$. $\blacksquare$

This lemma states that every open set $U$ in $X$ can be expressed as a union of basis elements. This expression for $U$ is not, however, unique.



Here is one way of obtaining a basis for a given topology:

Lemma. Let $X$ be a topological space. Suppose that $\mathscr C$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathscr C$ such that $x\in C\subset U$. Then $\mathscr C$ is a basis for the topology of $X$.

Proof. We must show that $\mathscr C$ is a basis. The first condition for a basis is easy: Given $x\in X$, since $X$ is itself an open set, there is by hypothesis an element $C$ of $\mathscr C$ such that $x\in C\subset X$. To check the second condition, let $x$ belong to $C_1\cap C_2$, where $C_1$ and $C_2$ are elements of $\mathscr C$. Since $C_1$ and $C_2$ are open, so is $C_1\cap C_2$. Therefore, there exists by hypothesis an element $C_3$ in $\mathscr C$ such that $x\in C_3\subset C_1\cap C_2$.
Let $\mathscr T$ be the collection of open sets of $X$; we must show that the topology $\mathscr T'$ generated by $\mathscr C$ equals the topology $\mathscr T$. First, note that if $U$ belongs to $\mathscr T$ and if $x\in U$, then there is by hypothesis an element $C$ of $\mathscr C$ such that $x\in C\subset U$. It follows that $U$ belongs to the topology $\mathscr T'$, by definition. Conversely, if $W$ belongs to the topology $\mathscr T'$, then $W$ equals a union of elements of $\mathscr C$, by the preceding lemma. Since each element of $\mathscr C$ belongs to $\mathscr T$ and $\mathscr T$ is a topology, $W$ also belongs to $\mathscr T$. $\blacksquare$



When topologies are given by bases, it is useful to have a criterion in terms of the bases for determining whether one topology is finer than another. One such criterion is the following:

Lemma. Let $\mathscr B$ and $\mathscr B'$ be bases for the topologies $\mathscr T$ and $\mathscr T'$, respectively, on $X$. Then the following are equivalent:
  1. $\mathscr T'$ is finer than $\mathscr T$.
  2. For each $x\in X$ and each basis element $B\in\mathscr B$ containing $x$, there is a basis element $B'\in\mathscr B'$ such that $x\in B'\subset B$.
Proof. $(2)\implies(1)$. Given an element $U$ of $\mathscr T$, we wish to show that $U\in\mathscr T'$. Let $x\in U$. Since $\mathscr B$ generates $\mathscr T$, there is an element $B\in\mathscr B$ such that $x\in B\subset U$. Condition $(2)$ tells us there exists an element $B'\in\mathscr B'$ such that $x\in B'\subset B$. Then $x\in B'\subset U$, so $U\in\mathscr T'$, by definition.
$(1)\implies(2)$. We are given $x\in X$ and $B\in\mathscr B$, with $x\in B$. Now $B$ belongs to $\mathscr T$ by definition and $\mathscr T\subset\mathscr T'$ by condition $(1)$; therefore, $B\in\mathscr T'$. Since $\mathscr T'$ is generated by $\mathscr B'$, there is an element $B'\in\mathscr B'$ such that $x\in B'\subset B$. $\blacksquare$



We now define three topologies on the real line $\mathbb R$, all of which are of interest.

If $\mathscr B$ is the collection of all open intervals in the real line,$$(a,b)=\{x|a<x<b\},$$the topology generated by $\mathscr B$ is called the standard topology on the real line. Whenever we consider $\mathbb R$, we shall suppose it is given this topology unless we specifically state otherwise.

If $\mathscr B'$ is the collection of all half-open intervals of the form$$[a,b)=\{x|a\leq x<b\},$$where $a<b$, the topology generated by $\mathscr B'$ is called the lower limit topology on $\mathbb R$. When $\mathbb R$ is given the lower limit topology, we denote it by $\mathbb R_l$.

Finally, let $K$ denote the set of all numbers of the form $1/n$, for $n\in\mathbb Z_+$, and let $\mathscr B''$ be the collection of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. The topology generated by $\mathscr B''$ will be called the $K$-topology on $\mathbb R$. When $\mathbb R$ is given this topology, we denote it by $\mathbb R_K$.



Lemma. The topologies of $\mathbb R_l$ and $\mathbb R_K$ are strictly finer than the standard topology on $\mathbb R$, but are not comparable with one another.

Proof. Let $\mathscr T$, $\mathscr T'$, and $\mathscr T''$ be the topologies of $\mathbb R$, $\mathbb R_l$, and $\mathbb R_K$, respectively. Given a basis element $(a,b)$ for $\mathscr T$ and a point $x$ of $(a,b)$, the basis element $[x,b)$ for $\mathscr T'$ contains $x$ and lies in $(a,b)$. On the other hand, given the basis element $[x,d)$ for $\mathscr T'$, there is no open interval $(a,b)$ that contains $x$ and lies in $[x,d)$. Thus $\mathscr T'$ is strictly finer than $\mathscr T$.
A similar argument applies to $\mathbb R_K$. Given a basis element $(a,b)$ for $\mathscr T$ and a point $x$ of $(a,b)$, this same interval is a basis element for $\mathscr T''$ that contains $x$. On the other hand, given the basis element $B=(-1,1)-K$ for $\mathscr T''$ and the point $0$ of $B$, there is no open interval $(a,b)$ that contains $0$ and lies in $B$.
To show that $\mathbb R_l$ and $\mathbb R_K$ are not comparable with one another, observe that given a basis element $[x,d)$ for $\mathscr T'$, there is no set for $\mathscr T''$ that contains $x$ and lies in $[x,d)$ because they are all open. On the other hand, given the basis element $B=(-1,1)-K$ for $\mathscr T''$ and the point $0$ of $B$, there is no half-open interval $[a,b)$ for $\mathscr T'$ that contains $0$ and lies in $B$. $\blacksquare$



Since the topology generated by a basis $\mathscr B$ may be described as the collection of arbitrary unions of elements of $\mathscr B$, what happens if you start with a given collection of sets and take finite intersections of them as well as arbitrary unions? This question leads to the notion of a subbasis for a topology.

A subbasis $\mathscr S$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by the subbasis $\mathscr S$ is defined to be the collection $\mathscr T$ of all unions of finite intersections of elements of $\mathscr S$.

I will later check that $\mathscr T$ is indeed a topology...

Understanding Analysis, Stephen Abbott, Chapter 1.2 Exercises

This is a fairly popular analysis book with many, many questions and no answers. So I will gradually attempt to answer some of them here. I may be wrong sometimes, so feel free to correct me if you can. If you are using this to cheat, then shame on you (this blog entry became the second most viewed after just one week; what's up with that?)!


Claim: $\sqrt3$ is irrational.

Let $\sqrt3$ be rational. Then $\sqrt3=p/q$, where $p,q\in\mathbb Z$, $q\neq0$, and $\text{gcd}(p,q)=1$. This last assumption can be made without loss of generality. It follows that

$$3=\frac{p^2}{q^2}\Longrightarrow3q^2=p^2.$$

$p^2$ cannot be even because it would imply that $p$ is even and that $q^2$ is even, which would in turn imply that $q$ is even. However, $p$ and $q$ cannot both be even because we assumed that $\text{gcd}(p,q)=1$. Therefore, $p^2$ must be odd, which implies that $p$ is odd and that $q^2$ is odd, which in turn implies that $q$ is odd. Let $p=2m+1$, and $q=2n+1$, as per the definition of an odd number, where $m,n\in\mathbb Z$. Then

$$\begin{align}3(2n+1)^2&=(2m+1)^2\Longrightarrow\\3(4n^2+4n+1)&=4m^2+4m+1\Longrightarrow\\12n^2+12n+2&=4m^2+4m\Longrightarrow\\6n^2+6n+1&=2m^2+2m\Longrightarrow\\2(3n^2+3n)+1&=2(m^2+m).\end{align}$$

Nevertheless, this last expression is absurd; an odd number cannot be equal to an even number. Therefore, $\sqrt3$ is irrational. This proves the claim.



Claim: If for every $n\in\mathbb N$, $A_n\supseteq A_{n+1}$ and $A_n$ contains infinitely-many elements, then $\bigcap_{n\in\mathbb N}A_n$ contains infinitely-many elements.

Let $A_n=\{n,n+1,n+2,\dots\}$ for every $n\in\mathbb N$. Then the hypothesis holds. Moreover, let $x\in\bigcap_{n\in\mathbb N}A_n$. However, $x\notin A_{x+1}$. Therefore, $\bigcap_{n\in\mathbb N}A_n=\emptyset$. This dismisses the claim.



Claim: If for every $n\in\mathbb N$, $A_n\supseteq A_{n+1}$ and $A_n$ is a nonempty set containing a finite number of real numbers, then $\bigcap_{n\in\mathbb N}A_n$ is a nonempty set containing a finite number of real numbers.

Let, for every $n\in\mathbb N$, $A_n\supseteq A_{n+1}$, $A_n$ be a nonempty set containing a finite number of real numbers, and $|A_1|=m$. Then there exists an $N\in\mathbb N$ such that $A_n=A_{n+1}$ for every $n\geq N$. Therefore, $\bigcap_{n\in\mathbb N}A_n=A_N$, which is a nonempty set containing a finite number of real numbers. This proves the claim.



Claim: $A\cap(B\cup C)=(A\cap B)\cup C$.

Let $A=\{1,2,3\}$, $B=\{2,3,4\}$, and $C=\{3,4,5\}$. Then $A\cap(B\cup C)=\{2,3\}$, and $(A\cap B)\cup C=\{2,3,4,5\}$, which are different. Therefore, the claim is dismissed.



Claim: $A\cup(B\cup C)=(A\cup B)\cup C$.

Let $x\in A\cup(B\cup C)$. Then $x\in A$ or $x\in B\cup C$. $x\in B\cup C$ implies that $x\in B$ or $x\in C$. $x\in A$ or $x\in B$ implies that $x\in A\cup B$. $x\in A\cup B$ or $x\in C$ implies that $x\in(A\cup B)\cup C$. Therefore, $A\cup(B\cup C)\subseteq(A\cup B)\cup C$. The converse can be proved in the same manner.



Claim: If $A,B\subseteq\mathbb R$ and $g:\mathbb R\to\mathbb R$, then $g(A\cap B)\subseteq g(A)\cap g(B)$.

Lemma: $A\subseteq B\implies f(A)\subseteq f(B)$, $A,B\subseteq\mathbb R$, $f:\mathbb R\to\mathbb R$.

$A\cap B\subseteq A\implies f(A\cap B)\subseteq f(A)$.
$A\cap B\subseteq B\implies f(A\cap B)\subseteq f(B)$.
$f(A\cap B)\subseteq f(A)\cap f(B)$.

Given a function $f:D\to\mathbb R$ and a subset $B\subseteq\mathbb R$, let $f^{-1}(B)$ be the set of all points from the domain $D$ that get mapped into $B$; that is, $f^{-1}(B)=\left\{x\in D:f(x)\in B\right\}$. This set is called the preimage of $B$.

Let $f(x)=x^2$. If $A$ is the closed interval $[0,4]$ and $B$ is the closed interval $[-1,1]$, find $f^{-1}(A)$ and $f^{-1}(B)$.

$$\begin{align}f^{-1}(A)&=[-2,2]\text{, and}\\f^{-1}(B)&=[-1,1].\end{align}$$

Does $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$ in this case?

$$\begin{align}A\cap B&=[0,1],\\f^{-1}(A\cap B)&=[-1,1]\text{, and}\\f^{-1}(A)\cap f^{-1}(B)&=[-1,1].\end{align}$$

Therefore, they are equivalent.

The good behavior of preimages just demonstrated is completely general. Show that for an arbitrary function $g:\mathbb R\to\mathbb R$, it is always true that $g^{-1}(A\cap B)=g^{-1}(A)\cap g^{-1}(B)$ and $g^{-1}(A\cup B)=g^{-1}(A)\cup g^{-1}(B)$ for all sets $A,B\in\mathbb R$.

Let $x\in g^{-1}(A\cap B)$. Then $g(x)\in A\cap B$. This implies that $g(x)\in A$ and $g(x)\in B$. This implies that $x\in g^{-1}(A)$ and $x\in g^{-1}(B)$. This implies that $x\in g^{-1}(A)\cap g^{-1}(B)$.



To be continued...

The Cantor Set

Let $C_0=\left[0,1\right]$. In other words, $C_0$ is the set of all real numbers greater than or equal to zero and less than or equal to one. Let the following be an illustration of $C_0$:


Now, let $C_1$ be equal to $C_0$ with its middle third removed. Mathematically, $C_1=\left[0,\frac13\right]\cup\left[\frac23,1\right]$. Let the following be an illustration of $C_1$:


If we continue in this manner, always removing the middle thirds of the remaining intervals, then we will obtain the following sequence:

$C_2$:


$C_3$:


$C_4$:


Let $n$ be any number in $\left\{0,1,2,3,\dots\right\}$. Define the Cantor set $C$ as the limit of $C_n$ as $n$ approaches infinity. Mathematically, $C=\lim_{n\to\infty}C_n$.

Clearly, $C_n$ comprises $2^n$ intervals of length $3^{-n}$ each (convince yourself of this fact). It therefore follows that the Cantor set $C$ has a length of

$$\lim_{n\to\infty}\frac{2^n}{3^n}=\lim_{n\to\infty}\left(\frac23\right)^n=0.$$

Intuitively, if an interval has a length of zero, then said interval must contain nothing at all; it must be the empty set, right?

Nevertheless, observe that the endpoints of the intervals are preserved through the sequence. In other words, observe that the points zero and one are preserved; all $C_n$ contain zero and one. It therefore follows that $C_0$ preserves two points, $C_1$ preserves four points, $C_2$ preserves eight points, and so on. Generally, $C_n$ preserves $2^{n+1}$ points. Therefore, the Cantor set $C$ preserves

$$\lim_{n\to\infty}2^{n+1}=\infty$$

points. In other words, a set with a length of zero can have infinitely many points. How can this be?

This exercise clearly demonstrates that our intuition of the mathematical infinity is insufficient to understand it, and that the need for rigorous tools and careful approaches to dissect it are of the utmost essence.