Question

Is f(x)=1/x uniformly continuous then x is in (0,1)

Answer 1

are you asking if this statement is true?

Answer 2

Yes, but I need a proof accompanying the statement, of course.

Answer 3

(0,1) is an open interval. \[x \neq 0\]

Answer 4

well then its true =)

Answer 5

And why is that?

Answer 6

cuz f(x) is defined on the open interval (0,1)

Answer 7

Well, I know that if a function is continious then it will be uniformly continious in any CLOSED interval. I haven't heard anything about such thing being true for open intervals

Answer 8

The function is not uniformly continuous for any closed interval. If the interval included 0, it would be undefined.

Answer 9

I was talking about functions in general. Of course I meant the intervals in which the function is defined

Answer 10

I think that it's not uniformly continuous so I am trying to prove that there exists an epsilon>0 for all deltas>0 such that there exist x,y such that \[|f(x)-f(y)|\geq \epsilon \] if \[|x-y|<\delta\]
I started by choosin epsilon=1. Then:
\[|\frac{ 1 }{ x } - \frac{ 1 }{ y }| = \frac{ |x-y| }{ xy }\]
Now, I think, I need to choose such values of x and y expressed through delta such that the equality above would be greater or equal than one, but I am having trouble thinking of such values. Is my approach any good?