Commutative and Associative Addition in Closed Sets: A Conceptual Explanation

[tirwit]

I'm reading Riley's "Mathematical Methods for Physics and Engineering" and I came across this expression about vector spaces:

"A set of objects (vectors) a, b, c, ... is said to form a linear vector space V if the set is closed under commutative and associative addition (...)"

What I don't understand is: what does commutative and associative addition have to do with a closed set?!

 

[VeeEight]

It doesn't have to do with closed sets.
In a topological space, a set is closed if the complement is open.

A space, V, is closed with respect to some operation, . , if for all x and y in V, we have that xy is in V. The definition of a vector space requires that all linear combinations of vectors is in the set, as well as a bunch of other axioms.

Take a look at the field axioms if you need some reference.

 

[mathman]

The essential point is that the term "closed" is in use in two completely different contexts.

In topology we talk about closed sets, which means sets containing all limit points - there may not be any arithmetic at all.

In arithmetic (generalized to such things as vector spaces) we mean that any result of an operation is contained within the space, while the topology is usually defined independently.

 

[tirwit]

Ah ok ;)

Thanks both :)

 

[ice109]

The essential point is that the term "closed" is in use in two completely different contexts.

In topology we talk about closed sets, which means sets containing all limit points - there may not be any arithmetic at all.

In arithmetic (generalized to such things as vector spaces) we mean that any result of an operation is contained within the space, while the topology is usually defined independently.

that's not completely true since in a metric space a set is closed iff it's closed under "taking limits" i.e. closed with respect to convergent sequenes.

 

[Studiot]

Riley is an excellent book by the way.

When saying that closure means that for a binary operation eg + on a set S

If A is in S and B is in S then A + B is in S

We mean for any finite number of such operations.

Taking an infinite number may result in another member of S or it may not. Many interesting cases that occur are the ones that do not.

 

[HallsofIvy]

The essential point is that the term "closed" is in use in two completely different contexts.

In topology we talk about closed sets, which means sets containing all limit points - there may not be any arithmetic at all.

In arithmetic (generalized to such things as vector spaces) we mean that any result of an operation is contained within the space, while the topology is usually defined independently.

that's not completely true since in a metric space a set is closed iff it's closed under "taking limits" i.e. closed with respect to convergent sequenes.

No, it's still "completely true". In a metric space, a closed set can be defined that way but in a metric space the two definitions are equivalent.

 

[ice109]

No, it's still "completely true". In a metric space, a closed set can be defined that way but in a metric space the two definitions are equivalent.

you've misunderstood what i meant.

mathman states that algebraic closure and set theoretic closure are two different things. i claim they're not that different. you claim what's completely true is my statement. hence contradiction.

 

[sutupidmath]

A set C of the topological space X, is said to be closed if its complement, X-C, is open. (While a set is said to be open if it is an element of the topology T on X).

This is how Munkres defines it. A property/characterisitc of closed sets, as previously said, is that they contain all their limit points. Alternatively, the closure of a closed set is that set itself.

 

[ice109]

A set C of the topological space X, is said to be closed if its complement, X-C, is open. (While a set is said to be open if it is an element of the topology T on X).

This is how Munkres defines it. A property/characterisitc of closed sets, as previously said, is that they contain all their limit points. Alternatively, the closure of a closed set is that set itself.

that's only true in metric spaces. in general topological spaces that doesn't necessarily make sense . hence the topological definition using complements.

 

[sutupidmath]

that's only true in metric spaces. in general topological spaces that doesn't necessarily make sense . hence the topological definition using complements.

I am not that sure what are you referring to here? But, if you are saying that closed sets do not contain their limit points in a general topological space, i will have to disagree with you...but again this might depend on how you define the limit point. In Munkres, he defines it this way: x is said to be a limit point of A, if every neighnorhood of x intersects A in a point other than x. With this definition in mind, then: a set is closed iff it contains all its limit points.

 

[snipez90]

More importantly, if a topological space T satisfies the first axiom of countability (i.e. each point of T has a countable neighborhood base), then the limit point characterization of closed sets is equivalent to the closed under limit operation definition (closed iff every sequence in the space converges to a point in the space). Obviously getting rid of the metric doesn't mean that sequences are just useless, but the point ice109 was making regarding the closure of the limit operation still holds in a topological space if we impose additional conditions.

Anyways I'm pretty sure over half of these posts are completely useless to the OP, oh well.

 

[mathman]

that's not completely true since in a metric space a set is closed iff it's closed under "taking limits" i.e. closed with respect to convergent sequenes.

"closed" in your statement is a particular example of the topological context. It doesn't mean that my assertions are not correct.