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Def'n of Limit Point? and limit.

by calvino
Tags: defn, limit, point
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calvino
#1
Oct22-05, 10:49 AM
P: 108
I know that there are different definitions for a limit point .

"A number such that for all , there exists a member of the set different from such that .

The topological definition of limit point of is that is a point such that every open set around it contains at least one point of different from ."-MATHWORLD

Are they all equivalent, when defining "the limit of f"? Or, this may help too, does my definition of a limit sound correct?...(bold-faced variables are vectors)

Let f: U->R^n
Let a be an element of the reals such that for all delta' >0 there exists an x in U, different than a, such that ||x-a||<delta'.
We say that lim f(x)=L as x->a, if for every epsilon>0 there exists a delta''>0 so that if ||x-a||<delta'' then ||f(x)-L||<epsilon.

Also, Is it right that I used delta' and delta"?
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calvino
#2
Oct22-05, 10:53 AM
P: 108
Is it necessary that I write U to be an open set?
DeadWolfe
#3
Oct22-05, 11:11 AM
P: 461
Limit point in this sense is not the same thing at all as the limit of a function as it aproaches a point.

They are two different things.

calvino
#4
Oct22-05, 11:20 AM
P: 108
Def'n of Limit Point? and limit.

i know that, but you DO need to define a limit point in order to define the limit. It would really help if you could please look at my definition of a limit.
DeadWolfe
#5
Oct22-05, 11:35 AM
P: 461
To be honest, I'm not sure I understand the question.
hypermorphism
#6
Oct22-05, 12:48 PM
P: 506
Quote Quote by calvino
Let f: U->R^n
Let a be an element of the reals such that for all delta' >0 there exists an x in U, different than a, such that ||x-a||<delta'.
We say that lim f(x)=L as x->a, if for every epsilon>0 there exists a delta''>0 so that if ||x-a||<delta'' then ||f(x)-L||<epsilon.
In set form, the definition then reads that for every open set U about a in R, there exists an open set V in Rn about L that contains f(U), the image of U under f.
If D is the domain of f, then we see that L is a limit point of f(D).
HallsofIvy
#7
Oct22-05, 12:59 PM
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Except for one small point, since x and a are real numbers you mean |x-a|, not ||x-a||, Your definition of limit is correct. You really have 2 definitions:

"Let a be an element of the reals such that for all delta' >0 there exists an x in U, different than a, such that ||x-a||<delta'."
is saying that a is a limit point of the set (of real numbers) U.

"We say that lim f(x)=L as x->a, if for every epsilon>0 there exists a delta''>0 so that if ||x-a||<delta'' then ||f(x)-L||<epsilon."
is the definition of limit of the function (which only exists a a limit point of U).

No, U does not have to be an open set. Although in that case a would be member of U.

If you keep ||x-a|| then U can be a subset of any R[sup]m[/tex].

In fact, if you use a ||x-a|| to represent a general metric (distance) function, the definitions are correct for a function between any two metric spaces.

If you replace "||x-a||< delta" with "there exist an element of U in every open set containing a". Then your definitions work in any topological space.


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