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Chapter 1: Problem 6

Prove that \((S \cap T)^{c}=S^{c} \cup T^{c}\) and \((S \cup T)^{c}=S^{c} \cap T^{c}\).

Short Answer

Expert verified
Question: Prove the following set identities using set theory definitions: \((S \cap T)^{c}=S^{c} \cup T^{c}\) and \((S \cup T)^{c}=S^{c} \cap T^{c}\). Answer: We proved the given identities by first showing that \((S \cap T)^{c}=S^{c} \cup T^{c}\) and \((S \cup T)^{c}=S^{c} \cap T^{c}\). In both cases, we showed this by proving two inclusions: 1. \((S \cap T)^{c} \subset S^{c} \cup T^{c}\) and \(S^{c} \cup T^{c} \subset (S \cap T)^{c}\) for the first identity. 2. \((S \cup T)^{c} \subset S^{c} \cap T^{c}\) and \(S^{c} \cap T^{c} \subset (S \cup T)^{c}\) for the second identity. By demonstrating these inclusions, we established the equality between the given sets in each identity.

Step by step solution

01

Prove \((S \cap T)^{c}=S^{c} \cup T^{c}\)

To show that \((S \cap T)^{c}=S^{c} \cup T^{c}\), we need to prove two things: 1. \((S \cap T)^{c} \subset S^{c} \cup T^{c}\) 2. \(S^{c} \cup T^{c} \subset (S \cap T)^{c}\) 1.1) Let \(x \in (S \cap T)^{c}\). This means that \(x \notin S \cap T\) or, equivalently, \(x \notin S\) or \(x \notin T\). So, \(x \in S^{c}\) or \(x \in T^{c}\). Thus, \(x \in S^{c} \cup T^{c}\), which means that \((S \cap T)^{c} \subset S^{c} \cup T^{c}\). 1.2) Let \(x \in S^{c} \cup T^{c}\). This means that \(x \in S^{c}\) or \(x \in T^{c}\) or, equivalently, \(x \notin S\) or \(x \notin T\). This implies that \(x \notin S \cap T\). Thus, \(x \in (S \cap T)^{c}\), which means that \(S^{c} \cup T^{c} \subset (S \cap T)^{c}\). Since we have proved both inclusions, we have shown that \((S \cap T)^{c}=S^{c} \cup T^{c}\).
02

Prove \((S \cup T)^{c}=S^{c} \cap T^{c}\)

To show that \((S \cup T)^{c}=S^{c} \cap T^{c}\), we need to prove two things: 1. \((S \cup T)^{c} \subset S^{c} \cap T^{c}\) 2. \(S^{c} \cap T^{c} \subset (S \cup T)^{c}\) 2.1) Let \(x \in (S \cup T)^{c}\). This means that \(x \notin S \cup T\) or, equivalently, \(x \notin S\) and \(x \notin T\). So, \(x \in S^{c}\) and \(x \in T^{c}\). Thus, \(x \in S^{c} \cap T^{c}\), which means that \((S \cup T)^{c} \subset S^{c} \cap T^{c}\). 2.2) Let \(x \in S^{c} \cap T^{c}\). This means that \(x \in S^{c}\) and \(x \in T^{c}\) or, equivalently, \(x \notin S\) and \(x \notin T\). This implies that \(x \notin S \cup T\). Thus, \(x \in (S \cup T)^{c}\), which means that \(S^{c} \cap T^{c} \subset (S \cup T)^{c}\). Since we have proved both inclusions, we have shown that \((S \cup T)^{c}=S^{c} \cap T^{c}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

De Morgan's Laws in Set Theory
De Morgan's Laws are fundamental rules in set theory and logic that describe the relationship between the complement of sets and the basic set operations of union and intersection. Just like a mathematical Rosetta Stone, they help us translate between the language of 'and' (intersection) & 'or' (union) when taking the complement of a set combination.

These laws state that the complement of the intersection of two sets is equal to the union of their complements, and vice versa for the union of two sets. The proper mathematical expressions for De Morgan's Laws are: \1. \((S \cap T)^{c} = S^{c} \cup T^{c}\)
2. \((S \cup T)^{c} = S^{c} \cap T^{c}\)Understanding these laws not only aids in abstract mathematical proofs but also shapes logical reasoning in fields such as computer science, digital circuit design, and real analysis.
Understanding Set Complement
The set complement concept plays a leading role in the grand theater of set theory. A complement of a set is akin to the flip side of a coin 鈥� it contains everything that the original set does not, within a given universal set. If you imagine a universal set U as an all-encompassing realm and a set S as a land within it, the complement of S, denoted as \(S^{c}\), represents all territories outside the borders of S in the realm of U.

In essence, to find the complement of a set is to answer the question: 'What's not in the set?' This concept is essential for grasping more complex theories and applications in mathematics, including logic and probability.
Proof by Inclusion
Proof by inclusion is a solid, step-by-step technique that resembles a handshake agreement between two sets 鈥� to show that two sets are indeed equal, one must demonstrate each is a subset of the other. This 'handshake' involves two assertions:
  • Every element of set A is also in set B, indicating \(A \subset B\)
  • Conversely, every element of set B is in set A, hence \(B \subset A\)

If both statements hold true, the sets A and B must be identical 鈥� they contain the exact same elements. This method fosters a deeper understanding of set relations and is a staple tool in proving set equalities, such as those seen in De Morgan's Laws.
Logic in Real Analysis
Logic serves as the backbone of real analysis, a branch of mathematics that delves into the behavior and properties of real numbers, sequences, series, and functions. It's like the rules of engagement for the continuous battlefield of mathematical concepts. Logical reasoning informs the way we approach proofs, like a seasoned detective piecing together clues to crack a case.

When you encounter a theorem in real analysis, its proof typically rests on logical pillars such as implication, equivalence, and negation, all of which are intrinsic to set theory. By understanding such logical foundations, students are better equipped to explore and appreciate the rigor and subtlety of real analysis.

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Most popular questions from this chapter

Show that \(\sqrt{2}\) is irrational. HINT: Show that if \(\sqrt{2}=m / n,\) where \(m\) and \(n\) are integers, then both \(\mathrm{m}\) and \(n\) must be even. Obtain a contradiction from this.

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