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Show that if \(f\) is a function from \(S\) to \(T,\) where \(S\) and \(T\) are nonempty finite sets and \(m=\lceil|S| /|T|\rceil,\) then there are at least \(m\) elements of \(S\) mapped to the same value of \(T .\) That is, show that there are distinct elements \(s_{1}, s_{2}, \ldots, s_{m}\) of \(S\) such that \(f\left(s_{1}\right)=f\left(s_{2}\right)=\cdots=f\left(s_{m}\right)\)

Short Answer

Expert verified
By the Pigeonhole Principle, at least \(m\) elements of \(S\) are mapped to the same element in \(T\).

Step by step solution

01

Understand the Problem

Given a function \(f\) from set \(S\) to set \(T\), where both sets are nonempty and finite, we need to show that there are at least \(m\) elements in \(S\) that are mapped to the same element in \(T\). Here, \(m\) is defined as \(m=\lceil \frac{|S|}{|T|} \rceil\).
02

Use the Pigeonhole Principle

The Pigeonhole Principle states that if \(n\) items are put into \(m\) containers, with \(n > m\), then at least one container must contain more than one item. Apply this principle to the sets \(S\) and \(T\).
03

Apply Pigeonhole Principle to Sets

Let \(|S| = n\) and \(|T| = k\). By the Pigeonhole Principle, if we distribute \(n\) elements into \(k\) boxes, there will be at least one box that contains at least \(\lceil \frac{n}{k} \rceil\) elements because \(n/k\) may not be an integer.
04

Show That \(m = \lceil \frac{|S|}{|T|} \rceil\)

Since \(m = \lceil \frac{|S|}{|T|} \rceil\), there exists a value \(t \in T\) such that at least \(m\) elements of \(S\) are mapped to \(t\). This follows directly from the Pigeonhole Principle applied to \(S\) and \(T\).
05

Conclude the Proof

Thus, there are at least \(m\) elements \(s_1, s_2, \ldots, s_m\) in \(S\) such that \(f(s_1) = f(s_2) = \cdots = f(s_m)\). Therefore, the given condition is satisfied.

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Key Concepts

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

Finite Sets
Finite sets are collections of distinct elements that have a specific, countable number of elements (denoted as \(|S|\) for a set S). For example, the set of vowels in the English alphabet \(S = \{a, e, i, o, u\}\) is finite because it contains exactly 5 elements.
When dealing with functions and mappings, it’s crucial to know the size of these finite sets, denoted as \(|S|\) and \(|T|\) for sets \(S\) and \(T\), respectively.
This concept helps in understanding how elements in two sets can relate to each other and is foundational to the Pigeonhole Principle.
Function Mapping
A function is a rule that assigns each element in one set (called the domain) to an element in another set (called the codomain). If \(S\) is the domain and \(T\) is the codomain, a function \(f:S \rightarrow T\) maps every element in \(S\) to an element in \(T\).
Imagine you have set \(S\) with 7 elements and set \(T\) with 3 elements. A function \(f\) from \(S\) to \(T\) means you assign each of the 7 elements in \(S\) to one of the 3 elements in \(T\).
Understanding how many elements of \(S\) map to a single element in \(T\) is key to applying the Pigeonhole Principle. This principle says that if there are more 'pigeons' than 'holes,' at least one 'hole' must contain more than one 'pigeon.' This helps in proving statements about the existence of certain types of mappings, as discussed in the original exercise.
Ceil Function
The ceil function, denoted as \(\lceil x \rceil\), rounds a number \(x\) up to the nearest integer. For instance, \(cel{x} \rceil = 2\) if \(1 < x \leq 2\) and \(3\) if \(2 < x \leq 3\).
In the context of the Pigeonhole Principle, the ceil function helps determine how many elements from set \(S\) should map to each element in set \(T\).
If you have \(|S| = 7\) and \(|T| = 3\), then you need at least \(\lceil \frac{|S|}{|T|} \rceil = \lceil \frac{7}{3} \rceil = 3\) elements from \(S\) mapping to the same element in \(T\). This mathematical tool is very useful in ensuring that the minimum condition required by the Pigeonhole Principle is met and can be easily calculated.

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