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Let's start by defining a surjective function. Can anyone tell me what it means for a function to be surjective?
Is it when every element in the codomain has at least one pre-image in the domain?
Exactly! A surjective function covers every element in its codomain. Now, if we have a surjective function and the sets involved are finite, what can we say about its injectivity?
Oh, so it has to be bijective, right? Because every element would uniquely map.
Correct! In finite sets, surjectivity implies bijectivity. Let's remember this with the acronym SIB: Surjective Implies Bijective for Finite sets. But what about infinite sets?
I read that in infinite sets, you can have a surjective function that's not injective.
Exactly right! Can anyone give me a counterexample where a surjective function fails to be injective?
Like mapping all even numbers from the set of integers to the same even number?
Yes! Great example! This highlights how critical it is to check the nature of our sets when dealing with functions. In summary, SIB works for finite sets, but infinite sets need careful analysis.
Let's dive into equivalence relations now. Who can remind me what defines an equivalence relation?
It must be reflexive, symmetric, and transitive.
Right! And any equivalence relation partitions a set. Imagine you have a set with 30 elements partitioned into three equal subsets. How many ordered pairs do we create in the equivalence relation?
Wouldn't it be 10 pairs for each subset? So, 300 total?
Exactly! 10 squared for each subset gives us 300. Remember, the order matters since we’re dealing with ordered pairs. This is essential for combinatorial reasoning.
This feels a lot like counting numbers; Is there a formula we can use?
Yes, it relates closely to the combinatorial view of set partitions. Good insights, everyone! Keep this in mind as we transition to Stirling numbers and their applications!
Now let’s introduce Stirling numbers. Who can explain what they are in relation to partitions?
They count the number of ways to partition a set into a specific number of non-empty subsets, right?
Precisely! The Stirling function of the second kind, denoted as S(n, k), gives us that count. Can anyone tell me how we might apply this to find the number of surjective functions?
We can partition the domain into k non-empty subsets, and each partition will define a unique surjective mapping, right?
Exactly! Once we have these partitions, each way those subsets can be assigned to elements in the codomain gives us a surjection. Remember, S(n, k) counts those partitions, and we also factor in the permutations of how those subsets map to the elements in the codomain. Bravo!
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The section elaborates on surjective functions, providing proof and counterexamples that link surjective functions to bijective functions in finite and infinite sets. It also discusses equivalence relations, counting ordered pairs related to these relations, and introduces the Stirling function concerning the number of ways to partition a set. Through concrete examples, the concepts of injective and bijective functions are clarified, emphasizing the importance of cardinality.
In this section, we're introduced to essential concepts in discrete mathematics concerning functions and their characteristics. The discussion begins with examining surjective functions (onto functions) where every element in the codomain has a pre-image in the domain. Notably, it establishes that if a surjective function maps a finite set to itself, it must also be bijective (both injective and surjective). We explore a counterexample with infinite sets, identifying conditions where the function remains surjective but loses its injectivity.
Next, we delve into equivalence relations and their partitions, focusing on how a set can be divided into disjoint subsets, and we calculate the number of ordered pairs in these partitions. This leads to the introduction of Stirling numbers, particularly Stirling numbers of the second kind, which count the ways to partition a set, offering insights into combinatorial problems. The exploration of injective functions follows, clarifying how unique mappings impact the number of potential functions between sets, reinforced by counting principles for bijective mappings.
Overall, this section provides foundational understandings critical for more advanced topics in discrete mathematics, particularly in areas involving functions, relations, and combinatorial counting.
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In question number 6 you are asked to either prove or disprove the following. You are given a function f: A → A and you are given that the function is surjective. Then the question is, is it necessary that the function is a bijective as well? It turns out that the statement is true provided your set A is a finite set.
A function is referred to as surjective if every element in the codomain (target set) has at least one pre-image in the domain (source set). In this section, we determine whether surjectivity implies bijectivity when the set A is finite. A function is bijective if it is both surjective and injective (one-to-one). Thus, if A is finite, and every element in A is being 'hit' by the function, then no two elements in A could map to the same element without exceeding the finite constraint — which confirms the function must also be injective, thereby making it bijective.
Imagine a classroom with a finite number of students (elements in set A). If each student has a unique desk, and every desk is occupied (surjection), then it follows that no two students can occupy the same desk (injective). Hence, all desks are assigned (bijective). However, if there were an infinite number of desks and a finite number of students, not all desks can be occupied, demonstrating that surjectivity does not guarantee bijectivity in an infinite setting.
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But the statement need not be true if the set A is an infinite set. So, here is a counterexample. So, imagine the function f given from the set 0 to infinity to the set 0 to infinity. So, that is my set A and the function is defined as follows. The mapping 0 → 0 and the mapping of f(n) → 0.
In this scenario, we provide a concrete example where the function is surjective but not injective. The definition of the function f suggests that the value 0 is assigned to more than one element (simply put, any number n greater than 0 maps to 0). Since not all elements from the set are being sent to unique values, the function cannot be deemed bijective. This demonstrates a situation where infinite sets allow for multiple pre-images steering us towards non-injectivity.
Think of a party where every guest is assigned a coat hook labeled with numbers ranging from 0 to infinity. If everyone throws their coats on hook number 0 regardless of how many guests there are (mapping 0 to each guest), then that hook becomes overly crowded (non-injective). In contrast, if only a finite number of guests were there, it would be unmanageable for them to share that hook without each having a unique place for their coat (injective).
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In question number 7, you are given an equivalence relation over a set A, where the set A has 30 elements. Since it is an equivalence relation, the relation partitions the set A into three subsets each of equal size. So, the question asks you how many ordered pairs are there in that equivalence relation?
Equivalence relations create a partition in the set, dividing the original set into distinct groups where every element in each subset is related to every other element in that subset. Here, with 30 elements divided into 3 equal-sized subsets of 10 elements each, each subset contributes pairs whose creation follows the logic of 'if i and j are in the same subset, then (i, j) is included in the relation.' Therefore, since we have 10 elements in each subset, the number of ordered pairs for each subset would be 10 squared, leading to a total of 300 pairs across all three subsets.
Imagine students grouped into three classrooms with 10 students each. For every pair of students in the same class, you can mark off a friendship noted in a notebook. If every student in the classroom interacts with every other student, the total friendships can be calculated simply by squaring the number of students in each room. So, with 10 students in a classroom, each room of friendships would result in 100 recorded interactions!
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Key Concepts
Surjective Function: A function where every element in the codomain has at least one pre-image.
Injective Function: A function where each element of the domain corresponds to a unique element in the codomain.
Bijective Function: A function that is both injective and surjective.
Equivalence Relation: A relation that is reflexive, symmetric, and transitive.
Stirling Numbers: The number of ways to partition a set into non-empty subsets.
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A surjective function from the set {1, 2, 3} to {A, B} could map 1 to A, 2 to B, and 3 to A, ensuring every element in the codomain has at least one pre-image.
An example of equivalence relation can be 'is friends with' in a social network, where the relation is reflexive, symmetric, and transitive.
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For surjection, every set must receive, at least one giving without deceive.
Imagine a party where each guest (codomain) must have a buddy (pre-image). Each buddy ensures everyone's invited.
Remember SIB for surjective implies bijective in finite sets.
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Review the Definitions for terms.
Term: Surjective Function
Definition:
A function where every element in the codomain is the image of at least one element from the domain.
Term: Bijective Function
Definition:
A function that is both injective (one-to-one) and surjective (onto).
Term: Injective Function
Definition:
A function where each element of the domain maps to a unique element in the codomain.
Term: Equivalence Relation
Definition:
A relation that is reflexive, symmetric, and transitive, which partitions a set into classes.
Term: Stirling Numbers
Definition:
Numbers that count the ways to partition a set into a specific number of non-empty subsets.