Thread: function and continuity

When Im trying to do this question , I got stuck

Prove that f:A-->R^m is continuous ata if and only if each f^i is continuous at a for i=1,2,...,m

This is my attempt:
We have to prove 2 sides(=> and <=)

(=>) Let a belongs to A then f(a) is in R^m
Since f is continuous at a, we have:
Given e>0, I can find d>0 such that 0<|x-a|< d implies |f(x)-f(a)|<e
But each f^i is a scalar field component of f(x) = {(f^1(x),...,f^m(x)}

My question is can I put f(x) = {(f^1(x),...,f^m(x)} into |f(x)-f(a)|to yield the results?
Is there any better ideas to do this question

Thank you so much

Originally Posted by knguyen2005

When Im trying to do this question , I got stuck

Prove that f:A-->R^m is continuous ata if and only if each f^i is continuous at a for i=1,2,...,m

This is my attempt:
We have to prove 2 sides(=> and <=)

(=>) Let a belongs to A then f(a) is in R^m
Since f is continuous at a, we have:
Given e>0, I can find d>0 such that 0<|x-a|< d implies |f(x)-f(a)|<e
But each f^i is a scalar field component of f(x) = {(f^1(x),...,f^m(x)}

My question is can I put f(x) = {(f^1(x),...,f^m(x)} into |f(x)-f(a)|to yield the results?
Is there any better ideas to do this question

Thank you so much

Yes, that is what you have to do!
In . |f(x)- f(a)|=\sqrt{(f^1(x)- f^1(a))^2+ ... (f^m(x)- f^m(a))^2}. You should be able to prove that , perhaps by induction.