math help

[space cadet]

Hey all, I need help with my Math 2020: Intro to Logic and Geometry class.

Can someone here explain to me the difference between an open set of numbers and a closed set? I'm pretty confused.

According to my professor, if you can add a pair of numbers in a set and the sum is also in the set, then it is closed under addition. Or if you multiply and the product is there, then it's closed under multiplication. But what if you multiply or add the two numbers and the answer is one of the numbers you already used? Can that count? Or what if it's zero? I guess I don't really understand what she's saying. So when is a set open? :(

Any insight would be much appreciated.


Thanks
Katie

[Kristophe]

If you add two numbers together, and get one of the numbers you started with, then yes, it's closed under addition since both of the numbers you started with are in the set. The same goes for multiplication.

a*b = a => b is the identity. If * = +, then b = 0. If * = x, then b = 1.

An open set would be something like {1 2 3 4 5}

5 + 1 = 6, but 6 is not in the set.

[Grey]

So, for example, the set of real numbers from zero to one inclusive is closed under multiplication, but open under addition.

[01101001]

For some examples of infinite sets, consider the set of even integers and the set of odd integers.

The even integers are closed under addition and closed under multiplication.

The odd integers are open under addition -- because two odd integers added together yield an even integer, not in the set -- and closed under multiplication.

[Cougar]

...if you can add a pair of numbers in a set and the sum is also in the set, then it is closed under addition.
Well, not just a pair but any pair. All combinations of additions have to yield members of the original set for it to be closed.

Or what if [the result is] zero?
It all depends on whether zero was in the original set.

Addition and multiplication are just familiar operators, but this idea of "open" and "closed" also pertains to any well-defined, abstract operator. Binary and other modulo numbers have their own special definitions of "addition" and "multiplication"....

[A Thousand Pardons]

Addition and multiplication are just familiar operators, but this idea of "open" and "closed" also pertains to any well-defined, abstract operator.
And, in the common notion of open interval (e.g., the interval 2 < x < 3) and closed interval (the interval 2 <= x <= 3, often denoted [2,3] ), the operation in that context is limits of sequences. If the limit of every sequence is contained in the set, then the set is closed. The limit of 2.9, 2.99, 2.999, 2.9999, etc. (2.999...) is 3.

But, that an interval is not closed with respect to addition or multiplication. That causes some confusion sometimes.

[jfribrg]

Here is a little hierarchy. Each number system is a superset of the one before it, and has a property that the one before it doesn't have

Natural numbers 1,2,3,4,5,....
Closed under addition and multiplication, exponentiation
Open under subtraction, division
Open with respect to zeroes of polynomials with natural number coefficients
Contains multiplicative identity
No additive identity
No additive inverse, no multiplicative inverse

Whole numbers 0,1,2,3,4,....
Closed under addition and multiplication
Open under subtraction, division, exponentiation
Open with respect to zeroes of polynomials with whole number coefficients
Contains multiplicative identity
Contains additive identity
No additive inverse, no multiplicative inverse

Integers ....,-3,-2,-1,0,1,2,3,4,....
Closed under addition ,subtraction, and multiplication,
Open under division, exponentiation
Open with respect to zeroes of polynomials with integral coefficients
Contains multiplicative identity
Contains additive identity
Contains additive inverse
Does NOT contain a multiplicative inverse

rational numbers a/b, where a,b are integers and b <> 0
Closed under addition ,subtraction, multiplication,division
Open under exponentiation
Open with respect to zeroes of polynomials with rational coefficients
Contains multiplicative identity
Contains additive identity
Contains additive inverse and multiplicative inverse (except for 0)

Real numbers
Closed under addition ,subtraction, multiplication,division
Closed under exponentiation (non-negative real exponents)
Open under exponentiation (negative real exponents)
Contains multiplicative identity
Contains additive identity
Contains additive inverse and multiplicative inverse (except for 0)
Open with respect to zeroes of polynomial equations with real coefficients

Complex numbers
Closed under addition ,subtraction, multiplication,division
Closed under exponentiation
Contains multiplicative identity
Contains additive identity
Contains additive inverse and multiplicative inverse (except for 0)
Closed with respect to zeroes of polynomial equations with complex coefficients


I like to think of each of these number systems as removing a limitation of the system preceding it. Of course there are other ways of looking at it. For instance, real numbers can be thought of as the union of all points on the number line of the two disjoint sets of transcendental numbers and non-trancendental numbers.

You can also continue, generalizing the complex numbers into quaternions, where instead of a complex number a + bi, where i*i=-1, you have A + BI, where A,B,I are real valued 3d vectors, and the dot product I*I = -1.

You can take an abstract view, and build up a hierarchy of algebraic structures. Instead of natural ->whole->rational->real->complex, you have semigroup->group->ring->field. This does not mean that natural numbers are a semigroup, etc., but it is a different and more general hierarchy. FYI: Integers and rationals are rings, reals and complex numbers are both rings and fields, but this is already OT, since the original question is about logic, not algebra.

[A Thousand Pardons]

Natural numbers 1,2,3,4,5,....
Closed under addition and multiplication
Open under subtraction, division, exponentiation
Closed under exponentiation, no?

PS: I'm not sure why you left off the "(except for 0)" for the Integers

[Bad jcsd]

An open set is a subset A of a metric space in which for every point there is a neighborhood which is a subset of A. So for example on the real line the open inetrval (1,2) (all the real numbers inbetween 1 and 2 but not including 1 and 2) is an open set. A subset B of a metric space is a closed set iff the complement of B is an open set, so for example on the rela number line the clossed interval [1,2] (all the real numbers from 1 to 2 including 1 and 2), is a closed set.

The concept of set closure is different. A set A is closed under some function * iff *:AxA-->A describes * (i.e. for all a,b in A if a*b = c then c is also in A), * is what is known as a binary operator.

[A Thousand Pardons]

An open set is a subset A of a metric space in which
"Open set" is a topological term--all topologies have open sets, they are not restricted to metric spaces.

[Bad jcsd]

I like to think of each of these number systems as removing a limitation of the system preceding it. Of course there are other ways of looking at it. For instance, real numbers can be thought of as the union of all points on the number line of the two disjoint sets of transcendental numbers and non-trancendental numbers.

The problem though is once you start to extend beyond the rela numbers you also start losing properties, so for example the rela numbers are an ordered field, but the complex numbers are not and then as you extend to other Cayley-Dickson constructions (the quaternions, the octonions and so on) you rapidly lose useful algebraic qualties (i.e. the quartenions are non-commuative, the octnions are non-commutative and non-associative and so on). Real numbers don't contain trancendetal numbers, I think you mena the extension of the reals called the surreal numbers (which infact contian more numbers than just the union of the transfinite numbers and the real numbers).

[qupte]You can also continue, generalizing the complex numbers into quaternions, where instead of a complex number a + bi, where i*i=-1, you have A + BI, where A,B,I are real valued 3d vectors, and the dot product I*I = -1.[/quote]

Actually quaternions form a four-dimensional real vector space.

You can take an abstract view, and build up a hierarchy of algebraic structures. Instead of natural ->whole->rational->real->complex, you have semigroup->group->ring->field. This does not mean that natural numbers are a semigroup, etc., but it is a different and more general hierarchy. FYI: Integers and rationals are rings, reals and complex numbers are both rings and fields, but this is already OT, since the original question is about logic, not algebra.

The natural numbers form a semigroup under both additon and mulplication, semigroups and groups are types of objects called magmas which have a single binary operation defined on them, rings and fields are types of ringoids whioch have two binary opertaions defined on them so:

So:
groups are monoids which are semigroups which are magmas

Fields are skew fields which are unit rings which are rings which are semirings which are ringoids.

(note : there are many other simlair structures not mentioned).

[jfribrg]

Natural numbers 1,2,3,4,5,....
Closed under addition and multiplication
Open under subtraction, division, exponentiation
Closed under exponentiation, no?

PS: I'm not sure why you left off the "(except for 0)" for the Integers

Re: exponentiation, that has been fixed. Sometimes my cut and paste fingers work faster than my brain #-o

I left off the "except for 0" from the integers because they do not have a multiplicative inverse.

[Bad jcsd]

An open set is a subset A of a metric space in which
"Open set" is a topological term--all topologies have open sets, they are not restricted to metric spaces.

Nevertheless that is the defintion of an open set. I'd guess thta for a topological space the mertic space is only a subset (certainly this is the case when delaing with manifolds which don't have a metric defined on them).

[A Thousand Pardons]

Real numbers don't contain trancendetal numbers, I think you mena the extension of the reals called the surreal numbers
No, the transcendentals are a subset of the reals, unless the term means something different--pi, for instance, is a transcendental.

Re: exponentiation, that has been fixed. Sometimes my cut and paste fingers work faster than my brain #-o

cool

I left off the "except for 0" from the integers because they do not have a multiplicative inverse.
Missed that! LOL!
Integers ....,-3,-2,-1,0,1,2,3,4,....
Closed under addition ,subtraction, and multiplication,
Open under division, exponentiation
Open with respect to zeroes of polynomials with integral coefficients
Contains multiplicative identity
Contains additive identity
Contains additive inverse and multiplicative inverse.
Something else to fix. :)

[Bad jcsd]

[quote=Bad jcsd]Real numbers don't contain trancendetal numbers, I think you mena the extension of the reals called the surreal numbers

oops for some reason I read that as transfinite??? THough I still wrote transcendental for some reason.

[A Thousand Pardons]

An open set is a subset A of a metric space in which
"Open set" is a topological term--all topologies have open sets, they are not restricted to metric spaces.

Nevertheless that is the defintion of an open set. I'd guess thta for a topological space the mertic space is only a subset (certainly this is the case when delaing with manifolds which don't have a metric defined on them).
That's one way of defining an open set in a metric space, and hence a topology (http://mathworld.wolfram.com/TopologicalSpace.html). The definition of open set is just any set in the topology.

[Bad jcsd]

An open set is a subset A of a metric space in which
"Open set" is a topological term--all topologies have open sets, they are not restricted to metric spaces.

Nevertheless that is the defintion of an open set. I'd guess thta for a topological space the mertic space is only a subset (certainly this is the case when delaing with manifolds which don't have a metric defined on them).
That's one way of defining an open set in a metric space, and hence a topology (http://mathworld.wolfram.com/TopologicalSpace.html). The definition of open set is just any set in the topology.

Ah right IO see I'm not over;ly fmalir with toplogy outside th ecncept of a metric space or manifolds which have a local metric.

[space cadet]

You people are so cool. I wish one of you was my professor because you actually seem to know what you're talking about.

It can be rather discouraging when your math professor teaches a lesson like this: ". . . so when you're dealing with numbers in base 12, the digits for eleven and twelve are expressed E and T. So E is ten plus two and T is ten plus one. Or something like that. So if you're multiplying numbers like this: 14E2T * 13E um you um you add eleven and twelve together I mean multiply them and um then carry the first two digits and... uhhh yeah, and then add two and three and the first two digits and carry that and then E and one... oh wait, that's addition. Okay, never mind. Forget everything I just said." *erases board covered with stuff we've been copying for the past half hour*


entire class full of elementary ed majors: :-s #-o :o :roll: :-k ](*,) #-o :-s

I think it's time for someone to retire.

[jfribrg]

Integers ....,-3,-2,-1,0,1,2,3,4,....
Closed under addition ,subtraction, and multiplication,
Open under division, exponentiation
Open with respect to zeroes of polynomials with integral coefficients
Contains multiplicative identity
Contains additive identity
Contains additive inverse and multiplicative inverse.
Something else to fix. :)

That's fixed as well. A thousand thanks to a thousand Pardons for being my not-so-anonomous referee. Next time I think i will just plagarize a math book. That way I won't have to worry about proofreading.

[Disinfo Agent]

Hey all, I need help with my Math 2020: Intro to Logic and Geometry class.

Can someone here explain to me the difference between an open set of numbers and a closed set? I'm pretty confused.

According to my professor, if you can add a pair of numbers in a set and the sum is also in the set, then it is closed under addition. Or if you multiply and the product is there, then it's closed under multiplication. But what if you multiply or add the two numbers and the answer is one of the numbers you already used? Can that count? Or what if it's zero? I guess I don't really understand what she's saying. So when is a set open? :(

Any insight would be much appreciated.


Thanks
Katie
Katie, as you can see from the posts above, there are several definitions of "closed set" in mathematics, depending on the context. You say you need to understand this for a class called Logic and Geometry, but your examples suggest that what you're struggling with is the algebraic notion of a set closed under a certain operation (which, admittedly, has some connection to logic).

Think of the set of all nonnegative integers, N = {0, 1, 2, 3, ...}. If we take a pair of numbers from this set and add them, the result is still in the same set:

1 + 1 = 2 still in N
2 + 5 = 7 still in N
2 + 0 = 2 still in N
0 + 0 = 0 still in N
etc.

So we say that N is closed under addition: the result of adding any pair of nonnegative integers is still a nonnegative integer (nevermind which one).
It's also closed under multiplication. However, for subtraction we have:

4 - 2 = 2
1 - 1 = 0

but

0 - 1 = -1 not in N!
5 - 2 = -3 not in N!

So, we say that N is not closed under subtraction, because the subtraction of two nonnegative integers may no longer be a nonnegative integer. (I have to say that I'd never seen the word "open" be used in this context, though...)

One could also say that the operations of addition and multiplication are closed in N, but subtraction is not. If we want to make subtraction closed, we need to include all the possible results of a subtraction of nonnegative integers. In other words, we need to join the negative integers to set N.

Doing that, we get the set of the relative integers, Z = {..., -2, -1, 0, 1, 2, ...}. This set is closed under addition, multiplication, and subtraction.

However, it is not closed under division, since, for instance:

15 : 5 = 3, which is still in Z
(-14) : 7 = -2, which is still in Z

but

3 : 2 = 1.5, which is not in Z!

In order to make division a closed operation, we must bring in the fractional numbers, and work within the set Q of the rational numbers. This set is closed under addition, multiplication, subtraction, and division (with a nonzero divisor).

And we could go on to R and C, by requiring that other operations be closed, as jfribrg said above...

Edit: "closed", not "open"... :roll:

[Bad jcsd]

At the heart of the cocnept of set closure is the concept of the binary operator, which as I said earlier is a function. We don't usually think of the binary operator '+' as a function, but it is i.e. f(x,y) = x + y, so a binary operator on a set A is a said a function f:AxA-->A the fact that the domain AxA is the set of all ordered pairs of members of A and that range is A is enough to guratee that a binary operator on a set A is always closed over A.

[A Thousand Pardons]

the fact that the domain AxA is the set of all ordered pairs of members of A and that range is A is enough to guratee that a binary operator on a set A is always closed over A.
for sure, but in space cadet's case, A would be the real numbers (or in jfribrg's case, the complex numbers, or the quaternions), and multiplication and addition is defined over A. The question then is whether subsets of A, like the integers, or {1,2,3,4,5}, are closed.

[Bad jcsd]

the fact that the domain AxA is the set of all ordered pairs of members of A and that range is A is enough to guratee that a binary operator on a set A is always closed over A.
for sure, but in space cadet's case, A would be the real numbers (or in jfribrg's case, the complex numbers, or the quaternions), and multiplication and addition is defined over A. The question then is whether subsets of A, like the integers, or {1,2,3,4,5}, are closed.

the defintion of a binary operator automatically implies closure that is the point. Sometimes we want to know if a subset forms a subgroup, subfield, etc, and closure is one of the conditons that needs to be checked.

[A Thousand Pardons]

the defintion of a binary operator automatically implies closure that is the point.
Not just any old binary operator though, so I'm not sure what you mean. The range of the operator can be outside the domain--or, if you're looking at a domain AxA, outside of the set A.

[Bad jcsd]

the defintion of a binary operator automatically implies closure that is the point.
Not just any old binary operator though, so I'm not sure what you mean. The range of the operator can be outside the domain--or, if you're looking at a domain AxA, outside of the set A.
In the most formal sense binary operator is a function f:AxA-->A where A is an arbitary set, so the domain is always AxA and the range is always A otherwise it's not technically a binary operator on A.

[A Thousand Pardons]

In the most formal sense binary operator is a function f:AxA-->A where A is an arbitary set, so the domain is always AxA and the range is always A otherwise it's not technically a binary operator on A.
Depends (http://www.faqs.org/faqs/sci-math-faq/numbers/), some books include closure in the definition, some don't. Here's a cite from Mathworld (http://mathworld.wolfram.com/BinaryOperation.html)[/url] that disagrees with yours, and here's a [url=http://mathworld.wolfram.com/BinaryOperator.html]cite from Mathworld (http://planetmath.org/encyclopedia/BinaryOperation.html) that agrees. :)

A binary operation could be just an operation that is applied to two quantities. What you are talking about is the definition where the quantities of the domain and the range are both restricted to A. Of course, that guarantees closure, over A--if you include it in the definition.

But you can have binary operations applied with restrictions such that the range is outside of the domain. The addition operation applied to the odd numbers, for instance, is a binary operation defined on integers (and hence, odd numbers) which would have the range (even numbers) entirely outside of the domain (odd numbers, or technically the product). Your definition would say that it is not a binary operation on the odd numbers, but the other defintion would say that it is a binary operation, but it is not closed.

From the context, when we talk about "the binary operation of addition applied on the odd numbers" it is clear that the definition we are using does not include closure.

[Bad jcsd]

Just about all maths books will formally define a binary operation as a function f:AxA-->A, it is important that closure is part of the defintion of a binary operaor as closure is rarely included as a group, field, etc, axiom. (and when closure is incluided as an axiom it is becasue the cocnept of a bianry operator is only vaguely defined). Links on the internet are not authoritive not even mathworld, in just about all cases whenerever a binary operatortion is formally defined the defintion automatically implies closure.

Even if you are going to use the term binary operator in an informal sense it is still wrong to call it an arbiatry function g:AxB-->C a binary operator, the term 'operator' (actually the term operator usually implies a map between two function spaces, or more specifically a map between two vector spaces) associates the function with a single set, for example the term unary operator refers to a function h:A-->A.

[A Thousand Pardons]

Just about all maths books will formally define a binary operation as a function f:AxA-->A, it is important that closure is part of the defintion of a binary operaor as closure is rarely included as a group, field, etc, axiom. (and when closure is incluided as an axiom it is becasue the cocnept of a bianry operator is only vaguely defined). Links on the internet are not authoritive not even mathworld, in just about all cases whenerever a binary operatortion is formally defined the defintion automatically implies closure.
I am aware of the non-authoritative links--the Mathworld links I gave more or less contradicted each other. :)

Still, that other link was to the sci.math FAQ--if you can support your contention, they'll certainly change it. All it says though, is that some math books do not include closure in the definition. I just picked up a math book, The Theory of Matrices by Peter Lancaster and it says "A binary operation from A X B to S is a rule which associates a unique member of S with each ordered pari of A X B." It does not say that A has to equal B much less has to equal S.

Closure is never automatic implied--it is either explicitly part of the definition of binary operator, or it is explicitly part of the definition of whatever structure one is defining.

[Bad jcsd]

All I can do is offer another example:

http://planetmath.org/encyclopedia/BinaryOperation.html

Out of my books all that define binary operations define it as functions of the type f:A-->AxA and in all of them closure is implicit in the use of the term binary operator (this applies to semigroups, groups, fileds and vector spaces)., even when no defintion is given for the term (i.e. the cocnept of closure is not even mentioned).

The defintion f:AxA-->A is so commonly used I'd go as far as to say that saying otherwise is wrong. Certainly few would agree with the idea that a binary operation is a function f:AxB-->S where A and B and S are not equal as that's just the definition of a binary function, of which a binary operation is just a special case.

Closure is automatically implied by defining a bianry opeartion as a function f:AxA-->A

[A Thousand Pardons]

All I can do is offer another example:

http://planetmath.org/encyclopedia/BinaryOperation.html

and, an opposing example: :)
http://planetmath.org/encyclopedia/Operation.html

[Bad jcsd]

All I can do is offer another example:

http://planetmath.org/encyclopedia/BinaryOperation.html

and, an opposing example: :)
http://planetmath.org/encyclopedia/Operation.html

Intereseting, but it is not a counterexample as it does not contradict the defitnion of a binary operation.

[A Thousand Pardons]

All I can do is offer another example:

http://planetmath.org/encyclopedia/BinaryOperation.html

and, an opposing example: :)
http://planetmath.org/encyclopedia/Operation.html

Intereseting, but it is not a counterexample as it does not contradict the defitnion of a binary operation.
Planetmath may not be so sure--they include closure in their definition of a group:
http://planetmath.org/encyclopedia/Group.html

[Disinfo Agent]

At the heart of the cocnept of set closure is the concept of the binary operator, which as I said earlier is a function. We don't usually think of the binary operator '+' as a function, but it is i.e. f(x,y) = x + y, so a binary operator on a set A is a said a function f:AxA-->A the fact that the domain AxA is the set of all ordered pairs of members of A and that range is A is enough to guratee that a binary operator on a set A is always closed over A.
Minor nitpick: the range need not be equal to A. Often, in the notation f:B-->C, C is the codomain (http://planetmath.org/encyclopedia/Codomain.html) of f, which means that it contains the range of f, but may have other elements.

There is no universal agreement as to the definition of the range of a function. Some authors define the range of a function to be equal to the codomain, and others define the range of a function to be equal to the image.
You are still right that the notation f:AxA-->A implies that the binary operation f is closed on A. However, we are often more interested in investigating whether f is closed on a subset S of A.
(Of course, you could always say that that's the same as investigating whether the restriction of f to that subset is closed...)

[A Thousand Pardons]

You are still right that the notation f:AxA-->A implies that the binary operation f is closed on A. However, we are often more interested in investigating whether f is closed on a subset S of A.
(Of course, you could always say that that's the same as investigating whether the restriction of f to that subset is closed...)
Bad jcsd would say that is the same as saying that f is a binary operator on S.

Some books certainly do follow that convention--but it is always "on S".