## DyingLoveGrape.## Introduction to Topology, Part 1: Set Theory.## Motivation.Topology, to me, has a flexible, somewhat-vague meaning; this isn't unreasonable, considering what topology To others (and as it is generally introduced in textbooks), topology essentially studies continuity of maps and connectedness of structures. For continuity of maps: there's a huge number of fields of mathematics which require continuity of maps, and, for each, continuity means something slightly different — topology standardizes these definitions and attempts to derive general information about continuity. For connectedness of structures: in geometry, shapes are rigid structures which have definite lengths, areas, and so forth; on the other hand, for topology, every shape is thought of to be made of rubber or clay: you can mold and reshape it so long as you don't tear it or poke holes in it. For example, as we will see, the square and the circle are the "same" shape topologically (just round out the edges of the square and reshape it into a circle). This might seem a bit strange: what could we gain from looking at shapes in such a way? The vague answer is: how connected (or "how many holes") a structure has. Neat. Note that this introduction is just a kind of vague, hand-wavy introduction: there's a ton of information in the field of topology, and a number of different tools that we can build up. One of the most exciting things about topology is that it can tell us about the world we live in or the universe around us just based on the general shape we think these things may be— ## Set Theory in a Nutshell.We need to get some definitions out of the way before we start doing topology. I won't exhaust the subject, so there may be things we need to note later. For now, let's just do some basic definitions and basic properties.
For example, the set of positive whole numbers less than 4 is equal to $\{1,2,3\}$. We can look at the subsets of this set, too. All the one-element subsets are: $\{1\}, \{2\}, \{3\}$. All the two element subsets are: $\{1,2\}, \{1,3\}, \{2,3\}$. All the three element subsets are: $\{1,2,3\}$. We also have a "special" subset, which is a subset of
Notationally, we can say that
\[1\in \{1,2,3\}\]
("1 is an element of $\{1,2,3\}$" or "1 is in $\{1,2,3\}$") and
\[\{1,2\}\subseteq \{1,2,3\}\]
("$\{1,2\}$ is a subset of $\{1,2,3\}$"). Note, though, that we cannot say $\{1,2\}\in \{1,2,3\}$ because this says the We also have operations we can do on sets. There are many, but we'll stick to just four. First, we'll draw our sets as circles which intersect (a Venn Diagram) to try to give an example of each of these operations.
For example, the union of the circles above is everything part of at least one circle: the red part, the blue part, and the purple part in the middle. For another example, given the sets $\{1,2,3\}$ and $\{3,4,5\}$ the union is \[\{1,2,3\}\cup \{3,4,5\} = \{1,2,3,4,5\}.\]
For example, the intersection of the circles above is only the purple part in the center. For another example, given the sets $\{1,2,3\}$ and $\{3,4,5\}$ the intersection is \[\{1,2,3\}\cap \{3,4,5\} = \{3\}.\]
Note that, in order to define the complement, we need a universe of elements for $A$ to be in. If we defined it to be literally everything not in $A$ then it wouldn't be so useful of a concept. For example, if we said, "The set $A = \{0,1,2,3\}$ is a subset of the integers," and asked for $A^{C}$, the solution would be $\{\dots, -2, -1, 4, 5, 6, 7, \dots\}$ since these are all of the integers not in $A$. On the other hand, if we said, "The set $A=\{0,1,2,3\}$ is a subset of the real numbers," then the complement would be every single real number except for $0,1,2,3$; in particular, $\pi, e, \sqrt{2}\in A^{C}$. Thus, the complement of $A$ depends on the space containing $A$. For example, the complement of the circle $A$ above is only the pure blue part on the right side, not including the middle purple part. This consists of every element not in the circle $A$. Before moving on, let's just introduce one last topic: one particularly nice way of "building" or "defining" sets.
It's much easier to give examples of this than to define it. For example, \[A = \{x\in {\mathbb N}\,|\, x \lt 5\}\] is said, "$A$ is the set consisting of the $x$'s in the natural numbers with the property that $x \lt 5$." In this case, $A = \{1,2,3,4\}$ because these are the only natural numbers less than 5. [Note: I take ${\mathbb N} = \{1,2,3,4,5,\dots\}$ to be the set of natural numbers, but some sources will include $0$ in the set of natural numbers. There are good arguments for and against including $0$, but I prefer it without $0$.] The universe is important in the set-builder notation, since taking nearly the same set as above with a different universe, \[B = \{x\in {\mathbb Z}\,|\, x \lt 5\}\] where \[{\mathbb Z} = \{\dots, -2,-1,0,1,2,\dots\}\] is standard shorthand for the set of all integers, gives us a different set altogether: \[B = \{\dots, -3,-2,-1,0,1,2,3,4\}\] so $B$ contains each negative whole number and $0$ in addition to the elements that were in $A$. If we were to say, \[C = \{x\in {\mathbb R}\,|\, x\lt 5\}\] this would have even more elements than $B$ does! What are some of the elements in $C$ that aren't in $B$ or $A$? Here's some: $\pi, -2\pi, -e^{\pi}, \sqrt{2}, \dots$ We can have more complicated sets as well. For example, \[D = \{x\in {\mathbb N}\,|\, x\mbox{ is a prime number.}\}\] is the set $D = \{2,3,5,7,11,\dots\}$. Or, we could combine properties, as in \[E = \{x\in {\mathbb N}\,|\, x > 1 \mbox{ and } x \leq 5\}\] which is the set $E = \{2,3,4,5\}$. How would this be different if, when defining $E$, instead of $x\in {\mathbb N}$ we asked for $x\in {\mathbb R}$? ## Properties and Proofs.In general, there's a few common techniques one can use to prove statements in set theory. Let's just go through some simple ones. For all of these, $A,B$ are sets and $x,y$ are elements. Usually it is the case that we use capital letters for sets and lower-case letters for elements.
These last two methods are so common that I'm going to take some time to do some examples.
We need to show that each set is a subset of the other. We take an arbitrary element in the left-hand set and show it's in the right-hand set: so, take $x\in \{n\in {\mathbb N}\,|\, n\mbox{ is odd}.\}$. Since $x$ is odd, we can either note that $\mbox{odd } \times \mbox{ odd } = \mbox{ odd}$, or we can prove it explicitly; let's do the latter. If $x$ is odd, then $x = 2k + 1$ for some $k\in {\mathbb Z}$; then we note
\[x^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^{2} + 2k) + 1\]
which shows that $x^2$ is odd. Hence, $x\in \{n\in {\mathbb N}\,|\, n^{2}\mbox{ is odd.}\}$. This implies the left-hand set is a subset of the right-hand set; we now need to show the reverse containment.
First, note that it's relatively easy to show these sets aren't equal by showing that there is an element in one which is not in the other; for example, $6$ is in the left-hand set, but not the right-hand set. Despite their inequaity, we might want to find any relations between the sets: note that, for example, the right-hand set is a subset of the left-hand set (why?) and so we have \[\{2, 4, 8, 16, 32, \dots\}\subseteq \{x\in{\mathbb N}\,|\,x\mbox{ is even.}\}.\]
As before, we need to find an element in one set which is not in the other set. Be careful when reading these sets, also: the second set is talking about the elements $n$ such that $n^2$ is positive; hence, the second set is actually all of ${\mathbb Z}$ except for the element $0$ (why is this?). The left-hand set only has the elements $\{1,2,3,4,\dots\}$. Hence, for example, $-1$ is in the left-hand set but not the right hand set. Indeed, it is actually the case that \[\{n\in {\mathbb Z}\,|\, n\mbox{ is positive}.\}\subseteq \{n\in {\mathbb Z}\,|\, n^{2}\mbox{ is positive.}\},\] which you should be able to show. At this point, we should be able to be clever enough to prove certain properties about sets and their operations. We'll prove one, and then we'll just note some of the rest.
This looks a bit intimidating at first glance, but let's go through it like we did the others: we need to show each side is a subset of the other.
There's a few other equalities that we might need, and a few that we'll prove on the spot when we need them. These equalities are from the For $A,B,C$ sets, we have that... \[(A\cup B)^{C} = A^{C} \cap B^{C}\] \[(A\cap B)^{C} = A^{C} \cup B^{C}\] \[A\cap (B\cup C) = (A\cap B)\cup (A\cap C)\] \[A\cup (B\cap C) = (A\cup B)\cap (A\cup C)\] \[\left(\bigcup_{i} A_{i}\right)^{C} = \bigcap_{i} A_{i}^{C}\] \[\left(\bigcap_{i} A_{i}\right)^{C} = \bigcup_{i} A_{i}^{C}\]
In general, there is a neat duality between $\cup$ and $\cap$ via the complement, but we won't explore this here.
## Next time...In the next post, we'll discuss what a topology is, what it looks like, and what some common properties of spaces are. |