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Prove the following relations. $$ \text { If } E \subset F \text {, then } F^{c} \subset E^{c} $$

Short Answer

Expert verified
We proved that if \(E \subseteq F\), then \(F^c ⊆ E^c\) by choosing an arbitrary element x in \(F^c\), showing that \(x \notin F\) and thus \(x \notin E\), which means that \(x \in E^c\). Hence, \(F^c ⊆ E^c\).

Step by step solution

01

Definitions and notations

Let's begin by understanding the basic definitions and notations we will need in this proof. - E ⊆ F means that every element of E is also an element of F. - \(E^c\) denotes the complement of E, which includes all elements not in E. So for any element x, \(x \in E^c\) if and only if \(x \notin E\). We need to prove that if E ⊆ F, then \(F^c ⊆ E^c\).
02

Assume E ⊆ F

The problem statement says that E is a subset of F. So, we can take this as given: \(E \subseteq F\) This means that every element in E is also an element in F.
03

Choose an arbitrary element from \(F^c\)

To prove that \(F^c\) is a subset of \(E^c\), we need to show that any element in \(F^c\) is also in \(E^c\). Let x be an arbitrary element belonging to \(F^c\), that is: \(x \in F^c\)
04

Using the definition of complements

Since x is an element of \(F^c\), it means that x does not belong to F: \(x \notin F\)
05

Utilize E ⊆ F

Recall that E ⊆ F, meaning every element in E is also in F. Since x does not belong to F, x cannot be a part of E: \(x \notin E\)
06

Conclude that x belongs to \(E^c\)

As, \(x \notin E\), we can conclude that x is an element of the complement of E: \(x \in E^c\)
07

Prove that \(F^c ⊆ E^c\)

The above steps show that for any arbitrary element x in \(F^c\), x also belongs to \(E^c\). Hence, we can conclude that: \(F^c ⊆ E^c\) By completing these steps, the proof is now complete. If E ⊆ F, then \(F^c ⊆ E^c\).

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Key Concepts

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

Complement of a Set
The concept of the complement of a set is fundamental in set theory and helps students understand the relationship between a set and its 'opposite', or what it does not contain. When we discuss the complement of a set, we are referring to all the elements that are not in the set within a given universal set.

In mathematical terms, if we have a set called 'E,' then the complement of E, denoted by \(E^c\), includes every element that is not a member of E. If you consider the universal set, often noted as U, as the set of all possible elements, then formally we say \(E^c = U - E\). This concept is pivotal for many proofs and logical arguments in set theory.

Understanding complements is also useful in probability, logic, and various areas of mathematics, highlighting the significance of what is absent as much as what is present. Remember with complements, it's all about the elements that are left out of our original set E.
Subset Relation
The subset relation is another cornerstone of set theory and depicts the idea of one set being entirely contained within another. Symbolically, we write \(E \subseteq F\) to express that E is a subset of F. This means that every single element in the set E is also a part of set F, though F can have additional elements that are not in E.

This concept is similar to the notion that a square is a specific type of rectangle, so we might say the set of squares is a subset of the set of rectangles. Importantly, subset relations help us understand containment and hierarchy in many mathematical contexts, from geometry to algebra.

Additionally, the subset concept is crucial when discussing functions, ordering, and various structures in mathematics and a helpful way to categorize and organize information.
Proof Techniques in Mathematics
Mastering proof techniques in mathematics is essential for students to validate assumptions logically and rigorously. There are several techniques used for proving different mathematical statements, such as direct proof, proof by contradiction, proof by contrapositive, and induction, among others.

In the given exercise, a direct proof method is used. A direct proof typically begins by assuming the statement's conditions are true and then logically deducing the statement's conclusion. In our case, we started by assuming the subset relation \(E \subseteq F\), and we followed with logical steps to show that consequently \(F^c \subseteq E^c\).

Proper proof techniques are a mathematician's main tools for establishing truths and are essential not only in pure mathematics but also in applied fields where logical reasoning is paramount. With practice, these proof methods become second nature, equipping students to tackle more complex challenges in mathematics and beyond.

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Most popular questions from this chapter

A system is composed of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector \(\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)\), where \(x_{i}\) is equal to 1 if component \(i\) is working and is equal to 0 if component \(i\) is failed. (a) How many outcomes are in the sample space of this experiment? (b) Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components 1,3 , and 5 are all working. Let \(W\) be the event that the system will work. Specify all the outcomes in \(W\). (c) Let \(A\) be the event that components 4 and 5 are both failed. How many outcomes are contained in the event \(A ?\) (d) Write out all the outcomes in the event \(A W\).

Suppose that \(A\) and \(B\) are mutually exclusive events for which \(P(A)=.3\) and \(P(B)=.5\). What is the probability that (a) either \(A\) or \(B\) occurs; (b) \(A\) occurs but \(B\) does not; (c) both \(A\) and \(B\) occur?

A customer visiting the suit department of a certain store will purchase a suit with probability .22, a shirt with probability .30, and a tie with probability 28. The customer will purchase both a suit and a shirt with probability .11, both a suit and a tie with probability .14, and both a shirt and a tie with probability ,10. A customer will purchase all 3 items with probability \(.06 .\) What is the probability that a customer purchases (a) none of these items: (b) exactly 1 of these items?

Consider Example 5o, which is concerned with the number of runs of wins obtained when \(n\) wins and \(m\) losses are randomly permuted. Now consider the total number of runs-that is, win runs plus loss runs-and show that $$ \begin{gathered} P[2 k \text { runs }\\}=2 \frac{\left(\begin{array}{c} m-1 \\ k-1 \end{array}\right)\left(\begin{array}{l} n-1 \\ k-1 \end{array}\right)}{\left(\begin{array}{c} m+n \\ n \end{array}\right)} \\ P\\{2 k+1 \text { runs }\\}=\frac{\left(\begin{array}{l} m-1 \\ k-1 \end{array}\right)\left(\begin{array}{c} n-1 \\ k \end{array}\right)+\left(\begin{array}{c} m-1 \\ k \end{array}\right)\left(\begin{array}{l} n-1 \\ k-1 \end{array}\right)}{\left(\begin{array}{c} m+n \\ n \end{array}\right)} \end{gathered} $$

Compute the probability that a bridge hand is void in at least one suit. Note that the answer is not $$ \frac{\left(\begin{array}{l} 4 \\ 1 \end{array}\right)\left(\begin{array}{l} 39 \\ 13 \end{array}\right)}{\left(\begin{array}{l} 52 \\ 13 \end{array}\right)} $$ Why not?

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