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kingstrick
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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.