21.1.2. Tutorial 3

Hello everyone welcome to tutorial number 3.

So, let us start the question number 1, here you are given that arbitrary sets A and B. And the sets A and B as such that, this condition holds namely (A ∩ C) ⊆ (B ∩C) for any set C that you consider. If that is the case and you have to show that A ⊆ B, so you are given the premise that the (A ∩ C) ⊆ (B ∩ C) for any set C.

So, since this condition holds for any set C if I substitute C = A in this condition then I get that A ∩ A ⊆ B ∩ A, but I know that A ∩ A is nothing but the set A, so that means I can say that my premise, which I obtained by substituting C = A is that is A ⊆ A ∩ B.

In this section, we start by examining the implications of a condition involving arbitrary sets A and B. The premise states that for any set C, if the intersection of A with C is a subset of the intersection of B with C, then we aim to prove that A is a subset of B. By substituting C with A, we find that A is contained in the intersection of A and B, hence establishing that A must be included in B.

Imagine two classrooms, A and B, where A contains all students who play soccer. The condition states that if any additional group of students (C) that overlaps in both classrooms shows that soccer players are also students in B, we need to prove that all soccer players in A are also in B. By examining overlaps, we confirm that every soccer player (A) is also enrolled in the second classroom (B).

Now my goal is to show that A ⊆ B, so for showing that A ⊆ B; I have to show that you take any element x in the set A, it should be present in the set B as well. So, I am taking an arbitrary element x and I am assuming it is present in the set A and since, I have established the premise that A ⊆ (A ∩ B), that means if since the element x is present in A same element x will be present in the A ∩ B as well.

Now, as per the definition of A ∩ B it means the element x is present in A and simultaneously it is present in B as well, that means definitely it is present in B as well, so what I have established here is that if I start with an arbitrary element x which is present in the set A, I have established that it is present in the set B as well where and since x was chosen arbitrarily here.

Here, we demonstrate that any arbitrary element x in the set A can be shown to also exist in B, using the earlier established premise that A is a subset of A ∩ B. By showing that if an element x belongs to A, it must also belong to A ∩ B, it implies that x, being in both A and B set, thus confirming A ⊆ B.

Consider a company where set A represents employees who work in the marketing department. If we establish through a company's policy (A ∩ B) that these employees also have access to a database (set B), we conclude that any marketing employee (x) has access to the database, reinforcing that all marketing employees are included in the database group.

In question two you are given the following; you are given 3 sets A, B, C such that this predicate holds for every element x in the set A and the property here is that if x ∈ A then the implication that x ∈ B → x ∈ C is true and this is a universally quantified statement, that means this condition holds for every element x in the set A.

Then you have to show, you have to either prove or disprove whether the A ∩ B ⊆ C or not, so in fact we are going to prove this statement we will prove that you take any element x which is arbitrarily chosen, and if it is present in the A ∩ B then it is present in C as well.

This section introduces a situation involving three sets. The condition asserts that for all elements x in A, if x belongs to B, then it must also be in C. The task is to either validate or refute the assertion that A ∩ B is a subset of C. The next steps involve proving that any element x from A ∩ B is indeed contained within C based on the given logical relationships.

Think about the relationship between students (set A), who are enrolled in specific courses (set B), and those who pass a specific exam (set C). The assertion states that if a student is in a course (B), then they also pass the exam (C). Thus, proving that all students enrolled (A ∩ B) would indeed pass that examination (C) encourages a structured approach to education.

So, we start with an arbitrary we chosen an element x and we start by reworking the premise that is given to be true here.

So, the premise here is that this universally quantified statement is true for every element x. So, what I am doing here is I am rewriting this implication here, so remember the statement p → q is logically equivalent to the disjunction of negation p and q. So, that is why I have splitted this implication into a disjunction and I get this equivalent form, then I apply the same rule again over this implication.

So, this implication is now replaced by this disjunction and I put a negation here. Now you see everywhere I have disjunction, I can apply the associativity property of the disjunction.

The focus here is on the logical manipulation of a statement derived from the universal property of sets. By rewriting implications into logical disjunctions using p → q equivalent transformations, we proceed to show that the properties translate through negation and disjunction effectively, thus confirming the original statement.

Think of a locked room analogy where students can only enter if they possess a key (B). The statement asserts that having a key implies they have permission (C). Reorganizing the statement provides insights into how keys and permissions interconnect, bringing clarity to access protocols in systems.

Now in question 3, there are several parts and, you are given a set S, consisting of n elements where n is not 0. So, implicitly I am assuming here that n is greater than equal to 1. And now I have to count various number possible relations satisfying some properties. So, the first property here is I am interested to count the number of relations on this set S which are symmetric.

This part lays out a task of calculating the number of different types of relations formed among 'n' elements, specifically focusing first on symmetric relations. A symmetric relation involves an ordered pair, and if (i, j) exists, then (j, i) must also be included. The goal is structured count of how these pairs can be organized within the total relationships across the set.

Consider a friendship circle where friendships can be mutual. If person A is friends with person B, then B is also friends with A. Hence, the various ways in which friendships can form versus friendship pairs can be counted symmetrically to reflect these relationships.

Now in part b of the question I want to find out the number of relations over the set S, which are anti-symmetric and just to recap, this is the requirement from a anti-symmetric relation. If both (a, b) and (b, a) are present in the relation then that is allowed only if a is equal to b that means a is not equal to b contra-positively if a is not equal to b then you cannot have simultaneously both (a, b) as well as (b, a) present in the relation R.

Here we define anti-symmetric relations, where reciprocal pairs of distinct members of the set are not allowed to coexist. The challenge is then to calculate how many of these anti-symmetric relations can exist in a set of 'n' elements, taking into account both diagonal pairs (which can appear) and the restrictions on distinct pairs.

Imagine a family hierarchy where a person can inherit a position. If person A is designated as the successor over B, B cannot overturn that designation by also naming that relationship, thus emphasizing the anti-symmetrical nature of hierarchy where one can lead but two can't equal the path.

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