# absolute value continuity

• Jan 6th 2012, 03:08 PM
grandunification
absolute value continuity
It is very easy to prove that abs(x) is continuous everywhere using epsilon-delta techniques. Yet, let U = (0,1). This is open in [0, infinity). Yet, f inverse (U) can be said to equal (0,1) U {-1/2} right? This set, call it V is certainly not open in (- infinity, infinity).

Thanks!
• Jan 6th 2012, 03:19 PM
Plato
Re: absolute value continuity
Quote:

Originally Posted by grandunification
It is very easy to prove that abs(x) is continuous everywhere using epsilon-delta techniques. Yet, let U = (0,1). This is open in [0, infinity). Yet, f inverse (U) can be said to equal (0,1) U {-1/2} right? This set, call it V is certainly not open in (- infinity, infinity).

What?
\$\displaystyle f^{-1}(U)=(-1,0)\cup(0,1)\$ which is open.
• Jan 6th 2012, 03:38 PM
grandunification
Re: absolute value continuity
Quote:

Originally Posted by Plato
What?
\$\displaystyle f^{-1}(U)=(-1,0)\cup(0,1)\$ which is open.

The preimages aren't unique. One preimage would be the union of (0,1) with the singleton set {-1/2}. f((0,1)) = (0,1) and f({-1/2}) = 1/2 which is an element of (0/1).
• Jan 6th 2012, 03:45 PM
grandunification
Re: absolute value continuity
Disregard that comment above. I was not really using the proper definition of a preimage. I have to consider ALL elements which map onto the image. Thank-you.