I know this has been a long time coming, but alas, I've some time to kill on the computer. This one, as you can expect from the other posts, will be quite long and will contain a slew of definitions. Let's get started then, shall we?
In our last post, we defined a union of two(or more) sets, an intersection of two(or more) sets, a finite/infinite union/intersection of many sets and also introduced a symbolic definition. We also went over an important example which essentially built up \( \mathbb{N} \) using strictly our definitions. Now, let us delve further into set theory!
Definition 1.6. Let \( F \) be a non-empty set such that each member of \( F \) is itself a set. In other words, suppose \( A_{1}, A_{2}, A_{3} .... A_{n} \in F \). Then, we call \( F \) a family of sets.
Example. A good example -- a very good one, in fact -- is the power set. Let's define it now!
Definition 1.7. Let \( S \) be a non-empty set. We say the power set of \( S \), defined as \( P(S) \) is a family of \( S \), such that it contains each subset of \( S\) as a member.
Ex. Let \( S \) = {1, 2, 3} Then, \( P(S) \) = { {1}, {2}, {3}, {1,2}, {1, 3}, {2, 3}, {1, 2, 3}, \( \emptyset \) }.
Let's explain some of the members of the power set before we move forward to explain what that squiggly deformed circle is doing in there. For those of you who are unaware, that 'squiggly deformed circle' is the empty set.
Clearly, {1}, {2} and {3} are subsets of S. Definition 1.1 is not disobeyed at all here. Each set contains an element which is a member of the 'parent' set \( S \). The same logic follows when we look at {1, 2}, {1, 3}, {2, 3} and {1, 2, 3}. Now, let us discuss the foreign symbol which is listed last within the power set. First, what is it?
The empty set.
The empty set is one of the most trivial sets in set theory, and yet it continues to be a very dangerous set because most amateur mathematicians make the mistake of underestimating it or not accounting for it in certain proofs in which it is necessary that the empty set be included or excluded. First, let us define what the empty set is, although the name should be slightly obvious.
Definition 1.8. Suppose there is a set such that no element exists within the set. We define this set to be the empty set. Symbolically, we write the empty set as { } or, more commonly, \( \emptyset \).
Generally, the empty set acts as an elusive concept to those who are quite foreign to set theory and mathematics. Many who read the definition feel that they comprehend it, but it's only after a few problems do they feel somewhat overwhelmed by their true lack of understanding. Let's discuss this a bit more in depth.
Mathematically speaking -- as if we are discussing anything other than mathematics -- the empty set is not 'nothingness.' It is not a set which is just there for the sake of being there; it is a set which simply does not contain elements. Furthermore, the empty set is unique under any context. Thus, if we are discussing the empty set in \( \mathbb{N} \), we are also indirectly discussing the empty set in \( \mathbb{C} \). Let's also go through some thought provoking examples and prove certain properties associated with the empty set.
Example 1. What is {\( \emptyset \)} \( \cap \; \emptyset \)?
Solution. \( \emptyset \).
Example 2. What is {\( \emptyset \)} \( \cup \; \emptyset \)?
Solution. {\( \emptyset \)}.
Note the dichotomy here. This is very important.
Example 3. What is {\( \emptyset\)} \( \cap \) {\( \emptyset \)}?
Solution. {\( \emptyset \)}. If you answered those questions correctly, you've got a good grasp on the concept of the empty set. If not, please review the definition again or simply send me a private message. I'd be happy to explain the solution to you in a manner in which you find pleasing.
For example 1, let's fall back upon our definition of intersection: we must identify members in BOTH sets. In this case, the first set actually contains a member -- the empty set itself! The secondary set, on the other hand, contains no member since it is the empty set. As such, they've nothing in common. Thus, their intersection is the empty set. For example 2, if we follow the definition of a union, a similar conclusion follows -- this time, however, we realize that the union contains members from one set, two sets or BOTH sets. In this case, the union contains only one element -- The empty set. Example 3 follows distinctly from the definition of intersection.
Let's prove some interesting properties with the empty set then, shall we?
Theorem 1. Let A and B be two sets such that B = \( \emptyset \). Then, A \( \cup \) B = A.
Proof. To prove this, we must show that each side is a subset of the other side. Thus, we must show (A \( \cup \; \emptyset \)) \( \subseteq \) A and A \( \subseteq \) (A \( \cup \; \emptyset \)).
Suppose x \( \in \) A \( \cup \; \emptyset\). Then, x \( \in \) A or x \( \in \emptyset \). If x \( \in\) A, then it follows automatically that x \( \in\) A. If we consider x \( \in\ \emptyset \), then we automatically contradict the definition of the empty set since we're assuming the possibility that there even exists an element in the empty set. This is absurd, however. Hence, this is a vacuous truth. Vacuous truths are almost 'nonsensical' truth statements in a logical context. Thus, we have shown that A \( \cup \; \emptyset \) \( \subseteq \) A
Note: If you study set theory further, the word 'vacuous' will come quite often when referring to empty set. Feel free to research the word and the context in which it belongs. Although ProofWiki can be somewhat misleading at times, they have an excellent primer on it here: http://www.proofwiki.org/wiki/Definition:Vacuous_Truth
Showing A \( \subseteq \) A \( \cup \; \emptyset \) is quite trivial, and I leave it up to the viewer to try it out. At best, you will write two sentences; after supposing that x \( \in\) A, you'll find it very easy to showcase that A \( \subseteq \) A \( \cup \; \emptyset \). ,`.
Theorem 2. Let A and B be two sets such that B = \( \emptyset\). Then, A \( \cap \) B = \( \emptyset \).
Before we prove this, let us prove a lemma.
Lemma 1.1. Let \( S\) be any set. Then, \( \emptyset \subseteq \) \( S\).
Proof. Suppose this is false. Then, this implies that there exists an element x \( \in\ \emptyset \) that is not contained in \( S\). But, this is a contradiction -- the empty set contains no elements. Hence, the lemma follows. ,`.
Now, we can prove theorem 2.
Proof. Let x \( \in\) A \( \cap \; \emptyset\). Then, x \( \in \) A and x \( \in \emptyset\). This implies that there exists an element which can be found in both sets, but the empty set contains absolutely no element -- hence, how can x \( \in\) in both A and the empty set? Thus, it is vacuously true that A \( \cap \; \emptyset\) \( \subseteq \emptyset\).
The second part of the proof -- \( \emptyset \subseteq \) A \( \cap \; \emptyset\) -- follows from lemma 1.1. ,`.
Ah, I wanted to discuss a bit more, but alas ...
Feel free to message me with any inquiries regarding this post. I should hope to come back tomorrow to discuss more and I may even delve into a discussion on the axiom of choice, an extremely controversial axiom for many mathematicians, both amateur and professional alike.