Definition of Limit for vector fields

Click For Summary
SUMMARY

Apostol defines the limit for vector fields as lim_{x \rightarrow a} f(x)=b, which is equivalent to lim_{||x-a|| \rightarrow 0} ||f(x)-b||=0. This definition emphasizes the importance of neighborhood conditions in topological vector spaces, where for every neighborhood V_b of b, there exists a neighborhood U_a of a such that f(x) \in V_b whenever x \in U_a. The discussion highlights the necessity of understanding the transition from scalar limits to vector limits, particularly in the context of metric spaces being topological vector spaces.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with topological vector spaces
  • Knowledge of metric spaces
  • Basic concepts of epsilon-delta definitions in analysis
NEXT STEPS
  • Study the epsilon-delta definition of limits in detail
  • Explore the properties of topological vector spaces
  • Learn about metric spaces and their relationship with vector fields
  • Investigate examples of limits in vector fields and their applications
USEFUL FOR

Students of mathematics, particularly those studying analysis and topology, as well as educators seeking to clarify the concept of limits in vector fields.

Kashmir
Messages
466
Reaction score
74
Apostol defines limit for vector fields as
> ##\quad \lim _{x \rightarrow a} f(x)=b \quad(\rm or\; f(x) \rightarrow b## as ##x \rightarrow a)##
means that :
##\lim _{\|x-a\| \rightarrow 0}\|f(x)-b\|=0##

Can't we say it's equivalent to ##\lim _{x \rightarrow a}(f(x)-b)=0##
 
Last edited by a moderator:
Physics news on Phys.org
Scalars? vectors ?

##\ ##
 
Kashmir said:
Apostol defines limit for vector fields as
> ##\quad \lim _{x \rightarrow a} f(x)=b \quad(\rm or\; f(x) \rightarrow b## as ##x \rightarrow a)##
means that :
##\lim _{\|x-a\| \rightarrow 0}\|f(x)-b\|=0##

Can't we say it's equivalent to ##\lim _{x \rightarrow a}(f(x)-b)=0##
You can, after you accept the definition. I think going from f(x)= b to f(x)-b is 0, is obvious, but going from lim(x->a) to lim (||x-a|| ->0) is not.
There are certainly functions where it's important that || f(x) - b || < epsilon for all the x with || x-a || < delta.
 
  • Like
Likes   Reactions: Kashmir
Kashmir said:
Apostol defines limit for vector fields as
> ##\quad \lim _{x \rightarrow a} f(x)=b \quad(\rm or\; f(x) \rightarrow b## as ##x \rightarrow a)##
means that :
##\lim _{\|x-a\| \rightarrow 0}\|f(x)-b\|=0##

Can't we say it's equivalent to ##\lim _{x \rightarrow a}(f(x)-b)=0##
Say ##\lim_{x \to a} f(x)=b## means, that for each neighborhood ##V_b## of ##b## there is a neighborhood ##U_a## of ##a## such that ##f(x)\in V_b## whenever ##x\in U_a\,.## If we further assume that we are dealing with topological vector spaces, then this is equivalent to ##\lim_{x-a \to 0}(f(x)-b)=0\,.## It also implies that every neighborhood contains an ##\varepsilon -##ball. Thus we can replace all neighborhoods with ##\varepsilon -##balls, i.e. the norm convergence. And metric spaces are topological vector spaces.
 
BvU said:
Scalars? vectors ?

##\ ##
I couldn't understand. Could you please explain what you mean?
 
willem2 said:
You can, after you accept the definition. I think going from f(x)= b to f(x)-b is 0, is obvious, but going from lim(x->a) to lim (||x-a|| ->0) is not.
There are certainly functions where it's important that || f(x) - b || < epsilon for all the x with || x-a || < delta.
"lim(x->a) to lim (||x-a|| ->0)" because if x->a the magnitude of difference vector ||x-a|| will tend to zero.
Is that ok?
 
fresh_42 said:
Say ##\lim_{x \to a} f(x)=b## means, that for each neighborhood ##V_b## of ##b## there is a neighborhood ##U_a## of ##a## such that ##f(x)\in V_b## whenever ##x\in U_a\,.## If we further assume that we are dealing with topological vector spaces, then this is equivalent to ##\lim_{x-a \to 0}(f(x)-b)=0\,.## It also implies that every neighborhood contains an ##\varepsilon -##ball. Thus we can replace all neighborhoods with ##\varepsilon -##balls, i.e. the norm convergence. And metric spaces are topological vector spaces.
Thank you for your reply. I can't understand it right now. It's beyond my learning. Hopefully I'll come back to it after studying some more. :)
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K