# Continuity of a function

1. Nov 20, 2008

### kbfrob

1. The problem statement, all variables and given/known data
Suppose f: [0,1] -> [0,1] is such that f attains each of its values exactly twice
Show that f cannot be continuous

3. The attempt at a solution
I assumed that f is continuous and tried to break it up into cases and show that there must be a value that is obtained 3 times.
since f is defined on an interval it has a sup and an inf (each attained twice by hypothesis). my cases are the order in which the are attained, i.e. 1) max,min,max,min or 2) max,max,min,min
case 1 is easy, but i can't figure out how to do case 2
is this the right approach or is there an easier way?

Last edited by a moderator: Nov 20, 2008
2. Nov 20, 2008

### Vid

Are you sure you phrased the question right? Your sentence ends abruptly and the hypothesis you used in the proof doesn't match the hypothesis you gave.

3. Nov 20, 2008

### kbfrob

my bad. poor mistake for me to make

4. Nov 20, 2008

### Dick

kfrob, stick to your guns and don't just admit you are wrong when you aren't. If you have either (min,max,min,max) or (max,min,max,min) if min<y<max how many times is f(x)=y if f is continuous? At least. Use the IVT. If it's (max,max) what does that mean?

5. Nov 20, 2008

### kbfrob

i got that part, but i can't figure out how to show that f attains a given value 3 time if you have (max,max,min,min) or (min,min,max,max). drawing it i can see that is true, but any heuristic argument involves me saying that f has a "relative min" between the maximums and a "relative max" between the minimums. IVT is about as far as we've gotten so i can't make any argument about f'. is there a way to get around this?

6. Nov 20, 2008

### Dick

If you have two different consecutive maxes, doesn't there have to be a min in between? Otherwise the function must be constant in between? Which means it achieves that value an infinite number of times?

7. Nov 20, 2008

### kbfrob

i finally got it. thank you