Question

Sets Tutorial for beginiers!!! (reflesh the page if you have question marks i used math symbols there)

\(\huge\color{darkgreen}{\text{Sets Tutorial for beginires!!!}}\)

Answer 1

Recall that the sets are finite and infinite. Set not containing any element is called empty set. The empty set is indicated by 0. If every element of B is an element of set A, the set B is a subset of A and write B⊂A. If B⊂A and B ≠ A, B ≠ Ø, then B is called proper subset of A. Let us now consider the concepts of intersection and union of sets. Let A - set of simple two-digit numbers, B - set of two-digit numbers ending in 1: A = {11, 13, 17, 19, ..., 79, 83, 89, 97} B = {11, 21, 31, 41, 51, 61, 71, 81, 91}. Consider the set C consisting of the common elements these sets. Common elements of the sets A and B are simple two-digit numbers ending in 1, t. e .: C = {11, 31, 41, 61, 71}. They say that C is the intersection of sets A and B. The figure sets A and B are shown by Euler diagram. The igure formed by the intersection Circles shown in the figure by hatching represents set C.

Definition. The intersection of two sets is the set consisting of all the common elements of these sets.

The intersection of sets A and B is denoted: A∩B. From the definition it follows that the intersection of A and B contains only those elements that belong to both set A and the set B, t. e., any element xy belonging to the intersection of sets A and B, has the property: x € x € A and B. Using the task set by means of its characteristic property, we can write: A∩B = {x | x € A and x € B}. If the sets A and B have no common elements, their intersection is the empty set, ie. e. in this case A∩B = Ø. It is obvious that A∩A = A, A∩Ø = Ø. Note also that if A⊂B, then A⊂B, and if it B⊂A A⊂B = B. Let us return to the consideration of sets A and B, where A - set of simple two-digit numbers, B - set of double numbers ending in 1. Consider now the set K, which owns all of the elements of A and all
elements of the set B. The set K is composed of all two-digit numbers that are simple or terminate number 1, t. e.
K = {11, 13, 17, 19, 21, ..., 83, 89, 91, 97}. A set K is union of sets A and B. The figure sets A and B are shown by circles of Euler.
The figure, which unites all the points belonging to the first range and all points belonging to the second round, as shown in Figure hatching represents the set K.

Definition. The union of two sets is the set consisting of all elements that belong to at least one of these sets.

Union of sets A and B is denoted by: A∪B. From the definition it follows that the union of the sets A and B contains only those elements which belong to at least one of these sets, t. e., any element x of A∪B sets has the property that x € x € A or B. This can be written as follows: A ∪ B = {x | x € x € A or B}.

It is obvious that A ∪ Ø = A, A ∪ A = A. Note also that if A ⊂ B, then A ∪ B = B, and if B⊂A, then A ∪ B = A.

Here are some examples.

Example 1. Find the intersection and union sets of numbers that appear in the record numbers 162655 and 255424. Let P - a plurality of digits which are used in the recording the first day, a L - the set of numbers that are used in recording the second number. Then P = {1, 2, 5, 6}, L = {2, 4, 5}. Intersection of the sets P and L is the set of M, composed of digits that appear in the records of both numbers, m. f.

M = {2, 5}.

Example 2. Find the intersection and union
sets C and D, if C - the set of points (x, y), where x - any
number, 2 ≤ y ≤ 3 - the set of points (x, y), where 1 ≤ x ≤ 4,
y - any number.

The set C - horizontal bar, limited lines y = 2 and y = 3, and the set D - vertical strip bounded by straight lines x = 1 and x = 4. The intersection sets C and D is many of their points in common, t. e. a rectangle formed by the intersection of these bands. Unionsets C and D is set consisting of all points horizontal stripes all points vertically strip.
We now explain how interconnected the number of elements two finite sets and the number of elements and their intersections association.
Let A and B - a finite set, n (A) n (B) - the number of these sets of elements, n (A ∩ B) and n (A ∪ B) - number elements of union and intersection of sets. If A and B have no common elements, ie. E. A ∪ B = Ø, it is obvious that n (A ∪ B) = n (A) + n (V). If A ∩ B ≠ Ø, then, in the amount of n (A) + n (B) includes twice the number of elements of the intersection, m. f. n {A ∩ B}. herefore, in this case n {A ∪ B} = n (A) + n (B) - n {A ∩ B}. Note that this formula includes the first time that A ∩ B = Ø, as in this case, n (A ∩ B) = 0.

Example Z. In tourist group, go to the
foreign trip, consists of 28 people, each of whom
fluent in English or German, with some
proficient in both languages. It is known that 18 people own
English, and 15 people - the German language. How Much
people in the group owns two of these languages?
Let A - many tourists-English
language, B - a lot of tourists, German-speaking.

Then n (A) = 18, n (B) = 15, n {A ∪ B) = 28. From formula
n (A ∪ B) = n (A) + n (B) - n (A ∪ B) we find that 28 = 18 + 15 - n (A ∩ B), m. e. n (A ∩ B) = 5.
Hence, in a group of 5 tourists own two specify the language.

Answer 2

Good job! :)

Answer 3

Wow

Answer 4

lol thx guys 1st time i hadn't tageed anyone immediatly

Answer 5

Wow!! Very helpful tutorial @AlexandervonHumboldt2 u did an awesome job, as usual! :)

Answer 6

Next part of sets Tutorial:



Suppose you want to compare the number of elements of X and Y, where X - set of two-digit numbers have - a lot of three-digit numbers ending in 9. Use directly counting the number of elements can be determined that n (x) = n (Y). However, this conclusion can not come resorting to counting. With each two-digit number is three-digit number, which is btained from it by assigning. Wherein each of the plurality of three-digit number Y would be appropriate for a two-digit number, and it is for the number, which is obtained from a strikethrough the last digit. It is said that between the sets X and Y is set to-one correspondence. Generally, if every element of A is put in corresponds to a single element of B and any element of B is corresponds to some single element of the set A, say that between the sets A and B set up a one to-one correspondence. Setting in the above example are mutually one correspondence between the sets X and Y, we hereby showed that n (x) = n (Y). We now give an example of one-one correspondences between infinite sets. Consider the set N of natural numbers and the set P even natural numbers. Natural number n we put in matching the number 2n. Then each element of N will correspond to a single element of R. If Each element of the set P will be appropriately for a single element of the plurality of N, as any natural even number can be written as 2n, where n € N, and in a unique way. We have shown that between the sets N and P can be establish a one-to-one correspondence. This example could a surprise, since the set P is a proper subset of N. However, it is clear that familiar to us representations relating to finite sets, can not be extrapolated to infinite sets. We now consider the example of one-one correspondence between the non-numeric sets. Given two circle with a common center O. For an arbitrary point the inner circumference assign the point K ' the outer circumference lying on the line OK. Obviously, for any point inside the circle there only the corresponding point and the outer circumference, Conversely, any point on the outer circumference is corresponding to a point inside the circle. In this way one correspondence between the set of points inside
circle and the outer set of points circumference. Establish a one-to-one correspondence between the sets of points circles, we thus demonstrated that the same number of circumferential points, although it is obvious that the length the inner circumferential length less external. This surprising fact once again shows us the features of the relations between the endless sets.

Answer 7

Great job :) You're on a roll with the tutorials

Answer 8

yeah you can go to history and physics and find my 2 other tutorial there

Answer 9

your job is complete and very good! @AlexandervonHumboldt2

Answer 10

thx

Answer 11

lol a qh replied to my post. this will give another chance for my tutorial to be the third tutorial on my tutotirla list in the os blog.

Answer 12

\(\huge\color{darkgreen}{\text{End of Sets Tutorial for beginires!!!}}\)

Answer 13

Nice condensed tutorial :) I am still reading second part but I have a question from part 1... are below sets same ?

{1, 2, 3}

{1, 1, 2, 3}

Answer 14

@rational By "same", do you mean "equal"?

Answer 15

yes if

A = {1, 2, 3}

B = {1, 1, 2, 3}

is it correct to say set A = set B

Answer 16

"Two sets are equal if and only if they have the same elements."
Since both set A and set B have the same elements, namely 1, 2 and 3, A=B.

Answer 17

Ahh yes, the definition clears up thnks Callisto !

Answer 18

Welcome :)

Answer 19

yeah i forgot about it

Answer 20

We can make the expression "Two sets are equal if and only if they have the same elements" more clear.
What does it mean to say two sets have the 'same elements'.

$$\rm \Large A \subset B~ and~ B \subset A \Rightarrow A = B $$

Answer 21

Saying two sets have the same elements, intuitively this means that if an element is found in A, then it must also be found in B.
And conversely, if a particular element is found in B, then it must also be found in A.
But then what you just did was use subset idea :)

Answer 22

your tutorial is great i was just commenting on a comment , callisto's comment

Answer 23

i like comments that makes me feels my tutorial needs impovements

Answer 24

this quote i was commenting on:
\(\color{blue}{\text{Originally Posted by}}\) @Callisto
"Two sets are equal if and only if they have the same elements."
Since both set A and set B have the same elements, namely 1, 2 and 3, A=B.
\(\color{blue}{\text{End of Quote}}\)

Answer 25

lol

Answer 26

Suppose you have the sets,
A = { 1 , 1, 2, 3 }
B = { 1, 2, 3 }

In a sense they are not exactly the same sets. Set A has an extra element 1.
We get rid of this ambiguity by saying if A subset B and B subset A, then A = B.
Now there is no doubt that A = B , by set equality.

It is possible that we could create a new definition of "=" and say that
X\( \neq \) Y if they have unlike elements or when X or Y has an 'extra' same element.
Therefore by this definition the sets {1,1,2,3} and {1,2,3} are not equal.
But in set theory this is not how we define set equality.

Answer 27

Didn't know you want to get rigorous here .
In that case, it should be \[\rm \Large A \subseteq B~ and~ B \subseteq A \Rightarrow A = B\]

Instead of using the symbol "\(\subset \)", it would be more appropriate to use the symbol "\(\subseteq\)" as \(A\subset B \Rightarrow (\forall x\in A, x\in A \Rightarrow x\in B ) \text{ and } (\exists y \in B \text{ s.t. } y\in B \text{ and } y\notin A)\). If there exists y in B and at the same time, y is not in A, then A and B cannot be equal.

Answer 28

Wow that is so amazing!!

Answer 29

@Miracrown

Answer 30

@iambatman @Nnesha @ZeHanz @wio

Answer 31

@EclipsedStar

Answer 32

Many books use \(\subset\) just as he did. Many also use \(\underset{\neq}{\subset}\)

Answer 33

Now show the bijection between \(\mathbb{Q}\) and \(\mathbb{Z}\). And cantors argument!

Answer 34

@zzr0ck3r right, i was using subset to include the possibility of being equal

Answer 35

Great for "beginires" LOL

Answer 36

Cool awesome :)

Answer 37

\(\Huge\color{blue}{\text{Thx!!!}}\)

Answer 38

can u tag u for math help

Answer 39

\(\huge\color{purple}{\text{Wow!}}\)

Answer 40

WOW, Fascinating job Alex! Keep it up!
HOW can you not have 3000 medals? xD
You should have the title above this... :) (y)

Answer 41

thx

Answer 42

i need 982 more meldas to 3k medla title

Answer 43

@jabez177