# Continuity, given topologies.

1. May 16, 2011

### Slats18

Is the function f: R -> R, x -> x^2 continuous when the domain and codomain are given the Half interval topology? (Or Lower Limit topology).

I'm not sure where to go with this. On inspection, I know that the intervals are open sets, so preservance of open sets in preimages are defined for x > 0. But what if there is a set [x^2,x^2+r) that is in the negative part of the real line, there is no pre-image for this set. Is there something I'm missing, or just not realizing (most likely the second one)?

2. May 16, 2011

### Hurkyl

Staff Emeritus
Every set has a pre-image.

You don't seem to have missed any elements of the pre-image....

Proof?

3. May 16, 2011

### Slats18

It was only by inspection, assuming that any sets in the negative real line for this particular function don't have pre-images in the real line. If that assumption is wrong, then I've got nothing to go on to prove it's not continuous, so it must be, but that's a very weak justification.

4. May 16, 2011

### micromass

Staff Emeritus
Hi Slats18

Can you tell me what $$f^{-1}([x,x+r[)$$ is?

5. May 27, 2011

### Slats18

Sorry for the really late reply, been busy with other topological concerns, namely product topologies haha.

I'm completely blanking on this at the moment, no matter how interesting topology is, it just doesn't stick. Would it be [sqrt(x),sqrt(x) + r) ?

6. May 27, 2011

### micromass

Staff Emeritus
Not exactly. You'll need to figure out what f-1(x) and f-1(x+r) are (there are multiple values). Then you need to figure out what happens to the points between x and x+r...

7. May 29, 2011

### Slats18

On further, concentrated inspection, given [x,x+r) the pre-image of this is
( -(sqrt(x+r)),-(sqrt(x)) ]U[ sqrt(x),sqrt(x+r) )
which isn't open as -sqrt(x) is an element of the pre-image, but there is no r > 0 such that [-sqrt(x),r) is an element of the pre-image as well.

8. May 29, 2011

### micromass

Staff Emeritus
Looks right to me!!

9. May 29, 2011

### Slats18

Could it be also said, not neccessarily proven, that because the mapping is from R to R, the pre-image is not defined for certain R and hence, not continuous?
Ex: Take the interval [-1,0). The preimage of this is obviously in the complex plane, hence not in R.

10. May 29, 2011

### micromass

Staff Emeritus
The pre-image is always defined. The pre-image of the set you mention is empty:

$$f^{-1}([-1,0[)=\emptyset$$

11. May 29, 2011

### Slats18

Ohh, duh, of course haha. My mistake, I'm doing topology and complex analysis so sometimes the two subjects mix haha.