# Inverse Image of an Open Set

1. Aug 22, 2004

### Ed Quanta

Why is the inverse image of the open set {t|t<1/2},
{t|t>=1/2 or t<1/4}?

the t>+1/2 sort of makes sense to me, but can't seem to grasp how t<1/4 is an inverse image

2. Aug 22, 2004

### HallsofIvy

What do you mean by inverse image?? A set has an image or inverse image under some function. What function are you talking about?

(My first week in grad school, I was called upon to do a proof in class about "inverse image" of sets. I completely embarrased myself by assuming that, since the word "inverse" was used, f must have and inverse function!)

3. Aug 22, 2004

### mathwonk

one thing seems clear, the function will not likely be continuous, since an open set has a non open inverse image (unless the domain is a little special).

4. Aug 22, 2004

### Ed Quanta

http://www.csh.rit.edu/~pat/math/papers/topology/topology.pdf

At the end of section 3.2 on Multipilication of Paths, they speak of the inverse image of the open set I mentioned earlier.

I suppose the function is the step function y=2t when 0<=t<1/2
and y=0 when t>=1/2

Thanks for the replies

5. Aug 22, 2004

### HallsofIvy

Yes, and it specifically refers to the inverse image under a given function. In particular, the function of the problem you are referring to (on page 5 of your reference) is :
&tau;&sigma;(t)= 2t if 0<= x< 1/2
0 if x>= 1/2
and notes that this is not a "path" because it is not continuous (as mathwonk pointed out from the given solution). (&tau;(t) and &sigma;(t) were defined separately.)

We really need to know that before we can answer your question!

Now, what is the inverse image of {t|t< 1/2}?
That is, what are the values of t such that &tau;&sigma(x)< 1/2?

Well, certainly 0< 1/2 so all t> =1/2 qualifies.
In order that 2t< 1/2, we must have t<1/4. That is certainly less than 1/2 so it fits the formula.
f(t)< 1/2 as long as t is either < 1/4 or >= 1/2 as claimed.

6. Aug 22, 2004

### Ed Quanta

Ah, I see. Thanks for clearing up.