Injective and Continuity of split functions

[kingstrick]

Homework Statement

Let I:=[0,1], let f: I→ℝ defined by f(x):= x when x is rational and 1-x when x is irrational. Show that f is injective on I and that f(fx) =x for all x in I. Show that f is continuous only at the point x =1/2

**I think i addressed all of these questions but I am unsure how to build all three elements into one formal proof.

The Attempt at a Solution

proof: Let I:=[0,1], f:I→ℝ defined by f(x):= {f(x) = x if x in Q, f(x) = 1-x if x is not in Q,} Now the lim f(x) = f(x0) as x →x0, lim x ≠ 1 as x →+1 (since there is an irrational going away from the limit), lim x ≠ 1 as x→-1 (since there is an irrational going away from the limit)

Injective:
There exists, a & b in [0,1], where a ≠ b, if a or b are rational then a ≠ 1 - a and b ≠ 1 - b. If a and b are irrational then 1-a ≠1-b → -a ≠-b → a ≠ b. Since a and b are in the interval [0,1], and a ≠ b then WLOG [a,b] must be an interval contained in [0,1].

show f(f(x) = x
If x is rational then f(f(x)) = (x) = x
If x is irrational then f(f(x)) = 1 - (1-x) = 1-1+x=x
Therefore f(f(x)) =x

show that f is continuous only at the point x =1/2
To be continuous at a point, the function must have a limit at that point and since this is a split function each of its equations must be able to exist at the same point. Therefore, 1-x = x, 1 = 2x, 1/2 = x is the only solution that meets these requirements.

 

[jgens]

proof: Let I:=[0,1], f:I→ℝ defined by f(x):= {f(x) = x if x in Q, f(x) = 1-x if x is not in Q,} Now the lim f(x) = f(x0) as x →x0, lim x ≠ 1 as x →+1 (since there is an irrational going away from the limit), lim x ≠ 1 as x→-1 (since there is an irrational going away from the limit)

What is the point of this argument?

Injective:
There exists, a & b in [0,1], where a ≠ b, if a or b are rational then a ≠ 1 - a and b ≠ 1 - b. If a and b are irrational then 1-a ≠1-b → -a ≠-b → a ≠ b. Since a and b are in the interval [0,1], and a ≠ b then WLOG [a,b] must be an interval contained in [0,1].

What you wrote here doesn't make any sense. Start with the assumption that f(x) = f(y) and show that x = y.

show f(f(x) = x
If x is rational then f(f(x)) = (x) = x
If x is irrational then f(f(x)) = 1 - (1-x) = 1-1+x=x
Therefore f(f(x)) =x

This is correct. You might want an argument showing that 1-x is irrational if x is irrational, but that depends on what your instructor wants for the course.

show that f is continuous only at the point x =1/2
To be continuous at a point, the function must have a limit at that point and since this is a split function each of its equations must be able to exist at the same point. Therefore, 1-x = x, 1 = 2x, 1/2 = x is the only solution that meets these requirements.

I am not familiar with the term "split function" but I assume you mean it to be the same thing as a piece-wise function. To prove that f is continuous at x = 2^-1 just use the ε-δ criterion for continuity at x = 2^-1. To prove that f is discontinuous everywhere else, just show that the ε-δ criterion fails everywhere else.

 

[kingstrick]

What is the point of this argument?

I don't really know. I thought I needed to show that the function wasn't continuous for some reason... I'll remove that part

What you wrote here doesn't make any sense. Start with the assumption that f(x) = f(y) and show that x = y.

I will try, but isn't an alternative way of showing injective to show that if x ≠ y then f(x)≠f(y). That's what I was going for.

 

[kingstrick]

I don't see how to prove that f(x) is not continuous anywhere else. i tried making that argument in the beginning of my proof but that obviously failed to make sense. I've seen the argument made using sequences against Dirichelt's function but that was arguing that the whole thing was discontinuous. Would I need to prove that it is not continuous above and below 1/2 or is there one broad argument that can be made for anything other than 1/2?