# Establishing that a function is discontinous

1. Jan 6, 2014

### center o bass

According to the definition of continuity in topology a function f:X -> Y is continous if for every open set V in Y, the set $f(V)^{-1}$ is open in X. I have found this definition powerfull in dealing with proofs of a more general nature, but when presented with the trivially discontinous function from the reals to the reals

$$f(x) = 1 \ \text{for} \ x< 0, \ 0 \ \text{for} \ \ x \geq 0$$

I have problems to argue why this function is obviously not continuous from the above definition.
Which open set do I choose in it's codomain which is not open in it's domain?

2. Jan 6, 2014

### Office_Shredder

Staff Emeritus
The set (-1/2,1/2) (or any open set containing zero and not 1) has as its inverse image $[0,\infty)$ which is not open.

Basically as long as you include a small open set around a discontinuity you are good to go if the set is small enough.

3. Jan 7, 2014

### center o bass

Thanks! It's clear now. My confusion lied in trying to construct open sets in the image of f instead of the codomain.