連續性－証明

a. The statement is false.Consider the following counter example.
Define f:[0,2]→[0,1] such that f(x)=1 if 0≤x≤1andf(x)=0 if 1<x≤2.
Define g:[0,2]→[0,1] such that g(x)=0 if 0≤x≤1and g(x)=1 if 1<x≤2.
Then fg:[0,2]→[0,1] and fg(x)=f(x)g(x)=0 for all x in [0,2]
Each of the graphs of f and g consists of two horizontal line segments broken at x=1.
The graph of fg is a horizontal line segment.
Therefore fg is continuous at x=1 but f and g are not.


b. The statement is false. Consider the following counter example.
Define f:[0,2]→[0,1] such that f(x)=1 if 0≤x≤1andf(x)=0 if 1<x≤2.
Define g:[0,2]→[0,1] such that g(x)=0 if 0≤x≤1and g(x)=1 if 1<x≤2.
Then f+g:[0,2]→[0,1] and (f+g)(x)=f(x)+g(x)=1 for all x in [0,2]

Each of the graphs of f and g consists of two horizontal line segments broken at x=1.
The graph of f+g is a horizontal line segment
Therefore f and g are both discontinuous at x=1 but f+g is not.


Statements a and b are conditional statements in the form p→q.
The negation of p→q is [p and (~q)].
So we may disprove p→q by finding p and q such that p is true but q is false.

我仲想補問多少少... ( 請別介意啊! )
我想知証明連續, 如果我想証明在一定區間內函數連續, 該如何証明?

2012-07-25 22:17:47 補充：
What can I do if I don't know whether it is ture or fulse?