The Real And Complex Number Systems

c lim n
c
lim n
fx n
f lim n
x n by continuity of f at 0
f0
From the preceding, we have proved that f is constant.
4.14 Let f be continuous at the point a a 1 , a 2 ,...,a n R n . Keep a 2 , a 3 ,...,a n fixed
and define a new function g of one real variable by the equation
gx fx, a 2 ,...,a n .
Prove that g is continuous at the point x a 1 . (This is sometimes stated as follows: A
continuous function of n variables is continuous in each variable separately.)
Proof: Given 0, there exists a 0 such that as y Ba; D, whereD is a
domain of f, wehave
|fy fa| . *
So, as |x a 1 | , which implies |x, a 2 ,...,a n a 1 , a 2 ,...,a n | , wehave
|gx ga 1 | |fx, a 2 ,...,a n fa 1 , a 2 ,...,a n | .
Hence, we have proved g is continuous at x a 1
Remark: Here is an important example like the exercise, we write it as follows. Let
j : R n R n ,and j : x 1 , x 2 ,...,x n 0,.,x j ,..,0. Then j is continuous on R n for
1 j n. Note that j is called a projection. Note that a projection P is sometimes
defined as P 2 P.
Proof: Given any point a R n ,and given 0, and choose , then as
x Ba; , wehave
| j x j a| |x j a j | x a for each 1 j n
Hence, we prove that j x is continuous on R n for 1 j n.
4.15 Show by an example that the converse of statement in Exercise 4.14 is not true in
general.
Proof: Let
fx, y
x y if x 0ory 0
1otherwise.
Define g 1 x fx,0 and g 2 y f0, y, then we have
limg 1 x 0 g 1 0
x0
and
limg 2 y 0 g 2 0.
y0
So, g 1 x and g 2 y are continuous at 0. However, f is not continuous at 0, 0 since
limfx, x 1 0 f0, 0.
x0
Remark: 1. For continuity, if f is continuous at x a, then it is NOT necessary for us
to have
lim xa
fx fa