## Fifty-Six Days of Research & Poster Sessions

It has been quite some time since the last time I blogged and I apologize for the moment where I broke my what I think was a consistent posting schedule. With the past fifty-two days I have had nearly no sleep what-so-ever because I was on a time crunch for my research paper “Category-Theoretic Generalization of Higman’s Lemma Admits Applicability and Constructive Proof.” The first-third of the fifty-two days was used to just figure out how to prove the final statement of my main result, which turned out to be a quite perfect for how elegantly all the concepts ended up fitting into place. The second-third was all about TeX’ing my paper day-in and day-out and getting it torn apart by two professors from here and Massachusetts, and my mentor in Ohio. They are the biggest reasons why I have accomplished so much.  After all the words and diagrams that I have written and drawn, I finally had something that I was ready to present to undergraduate and graduate students, and professors. So for the final-third of this research marathon, I presented my paper in poster sessions. One in particular, the Spring 2016 Meeting of the Southern California-Nevada Section of the Mathematical Association of America, was a wonderful experience because I gained so many connections and absorbed so much advice that I hope to discuss in the future. In addition I was awarded the Meritorious Award for my research and poster session presentation at the meeting.

In terms of the blogging schedule, I will keep with the same schedule at the best of my ability. The we will resume our expository posts on Thursday and go from there.

## Closed Sets Defined Relative to Properties of Open Sets in Topolgical Spaces

We form our topologies around the “basis” of open sets. However, we must also take a look at the closed sets that make the “holes” in our topology. So what is a closed set? A closed set is a set $A$ such that given an open set $X$ of a topology $\mathcal{T}_X$, we can say that $X-A$ is open. Let us now apply this to the real numbers $\mathbb{R}$. Our closed sets in this case are closed intervals $[a,b]$. This is valid because the complement $\mathbb{R}-[a,b]=(-\infty,a)\cup(b,\infty)$ which is an open set.

With this notion of closed sets in topologies, we may develop some conditions involve in any topological space $X$:

1. $\emptyset$ and $X$ are closed,
2. arbitrary intersections of closed sets are closed, and
3. finite unions of closed sets are closed.

We may prove this with DeMorgan’s law, however we will leave that to the reader as an exercise. However now we move on to what we call the interior and closure of a given subset $A$ of a topological space $X$. Essentially, the interior is the union of all open sets contained in $A$ and the closure is the intersection of all closed sets in $A$. We denote the interior of $A$ as $\text{Int} \ A$ and the closure of $A$ as $\text{Cl} \ A$. Automatically you should be thinking that $\text{Int} \ A$ is open and $\text{Cl} \ A$ is closed because the finite union of open sets are open and the arbitrary intersection of closed sets are closed. Therefore relative to $A$, we get

$\text{Int} \ A\subset A \ \text{Cl} \ A$.

Since we have been defining topologies through closed sets, we should define how bases would work. So given a topological space $X$ and a a subset $A$, we get the following properties pertaining to the basis of $X$:

1. $x\in\text{Cl} \ A$ iff every open set $U$ containing $x$ intersects $A$, and
2. $x\in\text{Cl} \ A$ iff every basis element $B$ containing $x$ intersects $A$.

And this brings back connections to our definition of a basis with respect to the open sets of the topological space $X$.

## Understanding Three “Standard” Topologies

Today we are going to be discussing three topologies that are seemingly “standard” among all others: the Order Topology, Product Topology, and Subspace Topology. They are easily predictable by their names and individual properties. We start with the Order Topology.

The Order Topology is a topology defined by some order relation like $<$ on the set $X$. The only subsets of $X$ are intervals given $a,b\in X$:

• Open interval
• $(a,b)$
• Half-open intervals
• $(a,b]$
• $[a,b)$
• Closed interval
• $[a,b]$.

To form our desired topology, we must form a basis collection $\mathcal{B}$. This collection $\mathcal{B}$ contains all open intervals $(a,b)$ in set $X$, if it exists, the smallest element $a_0$ of the half-open interval $[a_0,b)$, and, if it exists, the largest element $b_0$ of the half-open interval $(a,b_0]$. The subbasis for this topology is the finite intersection of open rays:

$\mathcal{B}=\bigcap_{n=1}^i$ (open rays)

Let us consider the Order Topology on $\mathbb{R}\times\mathbb{R}$. Then we get the basis for the topology to be the collection of all open intervals of the for $(a\times b, c\times d)$. We can also think of the product or the real numbers to form a Product Topology.

The Product Topology is defined on the cartesian product $X\times Y$ where $X$ and $Y$ are both topological spaces. To form our basis for this topology, we consider the collection of all products $U\times V$ such that $U$ and $V$ are open subsets of $X$ and $Y$ respectively. So, we can think of each topological space $X$ and $Y$ having their own bases themselves. Therefore, given the basis $\mathcal{C}$ for $X$ and the basis $\mathcal{D}$ for $Y$, we get the following basis $\mathcal{B}$ for the Product Topology $X\times Y$:

$\mathcal{B}=\{C\times D:C\in\mathcal{C} \ \text{and} \ D\in\mathcal{D}\}$.

To consider the $\mathbb{R}\times\mathbb{R}$-example again, we essentially have the product of two open intervals $(a,b)\times(c,d)$ which serves as the basis for the topology. Graphically, it will show the intersections between these open intervals forming the topology on $X\times Y$. However, to form a subbasis for this topology, we must consider functions called projections.

The projection of $X\times Y$ are the functions $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$. These functions can be defined by equations such that $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$. If we take $U$ to be an open subset of $X$, then we get the set $\pi_1^{-1}(U)=U\times Y$. The same is with $V$ being an open subset of $Y$: $\pi_2^{-1}(V)=X\times V$. As this might be already obvious, both of these sets are open subsets of $X\times Y$ and their intersection is $U\times V$:

$U\times V=\pi_1^{-1}(U)\cap\pi_2^{-1}(V)$.

Therefore, we define the subbasis collection $\mathcal{S}$ for $X\times Y$ to be

$\mathcal{S}=\{\pi_1^{-1}(U):U \ \text{open in} \ X\}\cup\{\pi_2^{-1}(V) \ \text{open in} \ Y\}$.

We now move to the final “standard” topology: the Subspace Topology.

The Subspace Topology is essentially a subspace $Y$ of a topology $X$ such that

$\mathcal{T}_Y=\{Y\cap U:U\in\mathcal{T}\}$

which defines the topology on $Y$.

We can now define the basis $\mathcal{B}$ for $Y$ such that

$\mathcal{B}_Y=\{B\cap Y:B\in\mathcal{B}\}$.

From here, it should be quite easy to form our subbasis for this Subspace Topology $Y$ of $X$. So I will leave it as an exercise for the reader.

## The Factorization of Functors + Update

(Huge thanks to Jon Beardsley for this wonderful discussion.)

Let $\text{Seq}$ be a posetal category, $[n]$ be a finite chain category, and $\text{Ab}$ be an abelian category. Then let there be a functor $F:\text{Seq}\to\text{Ab}$ such that

$F(n)=\{\text{free abelian group on} \ n \ \text{generators}\}$

if $n\leq10$ and

$F(n)=\{\text{free abelian group on} \ 10 \ \text{generators}\}$

if $n>10$.

We take the morphism $[n]\to[n+1]$ to either the subgroup inclusion of the free group on $n$-generators into the free group on $n+1$-generators (in the case that $n+1$ is less than or equal to $10$), and the identity otherwise. So so far, we have defined a functor $F:\text{Seq}\to\text{Ab}$ by telling what each object and achieve morphism go to (and you can check that $F$ preserves compositions of morphisms).

By factoring $[n]$ into $F$, we are given a functor $G:\text{Seq}\to[10]$, that takes $\{1\mapsto1, 2\mapsto2, \cdots, 9\mapsto9, 10\mapsto10, 11\mapsto10, 12\mapsto10, \cdots\}$ is formed, and another functor $H:[10]\to\text{Ab}$ the takes $k\in\{1,\cdots,10\}$ to the free abelian group on $k$-generators.

From this, we are given

$\text{Seq}\rightarrow^G[10]\rightarrow^H\text{Ab}$.

And therefore, when taking a look at what $F$ looks like in comparison to the factorization, you will see that $H\circ G=F$.

My apologies for no post yesterday as planned. I am currently on the road going to some place. Which puts me in a position to be typing this post on my phone in a very cramped vehicle. However that is not stopping me from fulfilling my two-post-every-other-week plan.

A major update that I would like to make is that Mondays will now be a discussion on what I learned and what I thought about in reading Munkres’s Topology.

## (DISCONTINUED) #3: Munkres Problem Set 2.13.5

2.13.5

Problem: Show that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$. Prove the same if $\mathcal{A}$ is a subbasis.

Answer: Let $\{\mathcal{T}_\alpha\}$ be a family of all topologies on $X$ that contains the basis $\mathcal{A}$ and $\mathcal{T}_\mathcal{A}$ be the topology generated by the basis $\mathcal{A}$. Note $\bigcap T_\alpha$ is a topology on $X$ and $\bigcap T_\alpha\subset T_\mathcal{A}$ since $\mathcal{A}\subset\mathcal{T}_\mathcal{A}$. On the other hand, any $U\in\mathcal{T}_\mathcal{A}$ is a union of elements in $\mathcal{A}$, so $U\in T_\alpha$ for all $\alpha$ and thus $\mathcal{T}_\mathcal{A}\subset\bigcap T_\alpha$. $\blacksquare$

Happy February! Sorry for the short Mondays. The time is ticking for my research paper and I need to have the abstract and the main portions of my paper finished in exactly three weeks. I also do not want to just skim the next section (2.13.5 is the last problem of Section $13$) and then try to pull out dumb responses to the problems in the text. I will be leaving Section $16$ problems for next Monday where I will be doing at least two to three problems to make up for my “lesser” days. There will also be another interesting discussion this Thursday so look out for that!

## The Lucas-Lehmer Theorem

The largest prime was discovered eighteen days ago by a mathematician named Curtis Cooper. The specific prime that he found was a Mersenne prime $2^{74207281}-1$ which contains exactly $22338618$ digits. Although found by chance, there was some mathematics involved that computers used to determine if a certain number was prime or not.

It is called the Lucas-Lehmer “Theorem” (better off as a “sequence”) in which we start from the number $4$ with the next number being $n^2-2$. So the sequence would go $\{4, 14, 194, 37634,\cdots\}$ and it will continue to rapidly increase exponentially.

Let us consider the example $2^3-1=7$ which is indeed prime. To verify, we take the exponent, $3$, and subtract $1$ to get $2$ ($3-1=2$). We then take a look at the second term in the Lucas-Lehmer “sequence” which is $14$. We need to show that $14$ is divided with remainder $0$ by the “prime” number $7$. So

$14 \ \text{mod} \ 7=0.$

And indeed we receive a remainder of $0$ therefore confirming that $7$ is prime.

So let $a=2^n-1$ such that $n\in J$ and the Lucas-Lehmer “sequence” be denoted by $\{b_{n-1}\}_{n\in J}$. Then for a given “testable” number, it is prime iff

$(b_{n-1}) \ \text{mod} \ (a)=0.$

Otherwise, not prime.

Today’s post is short because I have finals this week. More to come next week!

## Schedule! (Subject to Change)

This is going to be short, however I have decided to commit to a schedule that can change due to some unforseen circumstance. I want it to be a schedule that I can keep constant for a very long time! 🙂

Schedule (last updated: 2/5/15)

1. Every Monday: Discussion pertaining to what I learned in Munkres’s Topology
2. Every OTHER (sometimes do more than “every other”) Thursday: Expositions or write-ups on my research papers currently being written up and open to suggestions and edits

There will be a post on Thursday pertaining to my current research paper so look out for that. It will start the every-other-Thursday cycle.

## (DISCONTINUED) #2: Munkres Problem Set 2.13.4

2.13.4

Problem:

1. If $\{\mathcal{T}_\alpha\}$ is a family of topologies on $X$, show that $\bigcap\mathcal{T}_\alpha$ is a topology on $X$. Is $\bigcup\mathcal{T}_\alpha$ a topology on $X$?
2. Let $\{\mathcal{T}_\alpha\}$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all collections $\mathcal{T}_\alpha$, and a unique largest topology contained in all $\mathcal{T}_\alpha$.
3. If $X=\{a,b,c\}$, let $\mathcal{T}_1=\{\emptyset,X,\{a\},\{a,b\}\}$ and $\mathcal{T}_2=\{\emptyset,X,\{a\},\{b,c\}\}$. Find the smallest topology containing $\mathcal{T}_1$ and $\mathcal{T}_2$, and the largest topology contained in $\mathcal{T}_1$ and $\mathcal{T}_2$.

1. $\bigcap\mathcal{T}_\alpha$ is a topology on $X$ due to the fact that each $\mathcal{T}_\alpha$ is a topology on $X$ and if $\{\mathcal{T}_\alpha\}$ is empty, then $\bigcap\mathcal{T}_\alpha$ is the discrete topology on $X$. For $\bigcup\mathcal{T}_\alpha$, let $X=\{a,b,c\}$ and define the topologies $\mathcal{T}_1=\{\emptyset,X,\{a\},\{a,b\}\}$ and $\mathcal{T}_2=\{\emptyset,X,\{a\},\{b,c\}\}$. If we take an intersection of $\mathcal{T}_1$ and $\mathcal{T}_2$, then we are presented with a situation where $\{a,b\}\cap\{b,c\}=\{b\}\notin\mathcal{T}_1\cup\mathcal{T}_2$. Therefore showing that $\bigcup\mathcal{T}_\alpha$ is not a topology on $X$. $\blacksquare$
2. We must note that there exists a topology on $X$ containing all the collections $\mathcal{T}_\alpha$ (discrete topology). Thus, there exists a topology on $X$ contained in all $X$. Therefore denoted as $\mathcal{T}_l=\bigcap\mathcal{T}_\alpha$ to be the largest topology on $X$. For the smallest topology on $X$ that contains all the collections $\mathcal{T}_\alpha$, we take the intersection of all topologies that satisfy the property of containing the union of all the $\mathcal{T}_\alpha$ such that $\mathcal{T}_s=\bigcap_{\bigcup\mathcal{T}_\alpha\subset F}F$ where there exists at least one $F$. Therefore I claim that $\mathcal{T}_l$ is the largest topology and $\mathcal{T}_s$ is the smallest topology on $X$. To show the uniqueness of $\mathcal{T}_s$, let $F$ be a topology that contains all the $\mathcal{T}_\alpha$. $\mathcal{T_s}$ is the intersection of $X$ and therefore $\mathcal{T}_s\subset F$. Then suppose $\mathcal{T}'_s$ fits the criteria of $\mathcal{T}_s$, then $\mathcal{T}'_s\subseteq\mathcal{T}_s$ and $\mathcal{T}_s\subseteq\mathcal{T}'_s$. And therefore $\mathcal{T}_s=\mathcal{T}'_s$, thus $\mathcal{T}_s$ is unique. The uniqueness proof of $\mathcal{T}_l$ is similar. $\blacksquare$
3. The smallest topology containing $\mathcal{T}_1$ and $\mathcal{T}_2$ is the union of $\mathcal{T}_1$ and $\mathcal{T}_2$ such that $\mathcal{T}_1\cup\mathcal{T}_2=\{\emptyset,X,\{a\},\{a,b\},\{b,c\},\{b\}\}$ for which $\{b\}$ has to be thrown in where $\{b\}=\{a,b\}\cap\{b,c\}$. The largest topology contained in $\mathcal{T}_1$ and $\mathcal{T}_2$ is the intersection such that $\mathcal{T}_1\cap\mathcal{T}_2=\{\emptyset,X,a\}$. $\blacksquare$

## (DISCONTINUED) #1: Munkres Problem Set 2.13.1, 2.13.3

As an introductory remark for this being my first set of problems from the Munkres text, I have a few things to note.

• The numbering system corresponds as follows: “<chapter>.<section>.<problem number>”.
• Please feel free to correct any mistakes that I make in the comments and I will update them, as I am doing this for my benefit and the readers benefit as well.
• The schedule for these posts will solidify for every Monday at $3$AM PST.
• You can find Munkres’s Topology in the References section on my blog if you need a quick link to the text at your convenience.

For this problem set, I am sorry it has only two problems. I will try to post a few more following up in the middle of the week so that can everyone see more. If not, we will start fresh next week.

2.13.1

Problem: Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x\in A$ there is an open set $U$ containing $x$ such that $U\subset A$. Show that $A$ is open in $X$.

Answer: Let $U_x$ denote the open set where $x\in U_x\subset A$. For each $x\in A$, there will be an open set $U_x$ associated with it. Therefore $\bigcup_{x\in A}U_x=A$. By the definition of an open set, $A$ is an open set by the fact that the union of opens sets forms an open set. $\blacksquare$

2.13.3

Problem: Show that the collection $\mathcal{T}_C$ given in Example $4$ of Section $12$ is a topology on set $X$. Is the collection

$\mathcal{T}_\infty=\{U|X-U$ is infinite or empty or all of $X\}$

a topology on $X$?

Answer: By the definition of a topology,

1. $X-X=\emptyset$ and $X-\emptyset=X$; therefore $X$ and $\emptyset$ are in $\mathcal{T}_C$.
2. Let $\{U_\alpha\}_{\alpha\in J}$ be an indexed family on nonempty elements of $\mathcal{T}_C$. ThenTherefore this shows that $\bigcup U_\alpha\in\mathcal{T}_C$ for all $\alpha\in J$.
3. And Therefore this shows that $\bigcap_{i=1}^n U_i\in\mathcal{T}_C$ for some $i$.

Thus $\mathcal{T}_C$ is a topology on $X$. $\blacksquare$