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Math Help - continuous

  1. #1
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    continuous

    Suppose  I is an interval in  \mathbb{R} and let  f: I \to \mathbb{R} be a continuous function. Prove that  f is injective  \Longleftrightarrow f is strictly monotonic.

     \Rightarrow direction: Let  a,b \in I . Suppose  a \neq b . Then  f(a) \neq f(b) . So  a<b or  a > b . And  f(a) < f(b) or  f(a) > f(b) . From here, would you use the intermediate value theorem?


     \Leftarrow direction: Suppose  f is strictly monotonic. Then  x \neq y \implies f(x) \neq f(y) .
    Last edited by manjohn12; January 21st 2009 at 03:06 AM.
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Suppose  I is an interval in  \mathbb{R} and let  f: I \to \mathbb{R} be a continuous function. Prove that  f is injective  \Longleftrightarrow f is strictly monotonic.

     \Rightarrow direction: Let  a,b \in I . Suppose  a \neq b . Then  f(a) \neq f(b) . So  a<b or  a > b . And  f(a) < f(b) or  f(a) > f(b) . From here, would you use the intermediate value theorem?


     \Leftarrow direction: Suppose  f is strictly monotonic. Then  x \neq y \implies f(x) \neq f(y) .
    If f is strictly monotonic it means if a\not = b, say a<b WLOG then f(a) < f(b) or f(a)>f(b). Thus, we see a\not = b \implies f(a)\not = f(b). This implies that f is one-to-one.

    Now we will show that if f is one-to-one then f is strictly monotonic. This part is more involved. The hint that will help you solve this is if: a<b<c then f(b) lies between f(a),f(c). We will prove the hint and you can try finishing the proof. If f(b) did not lie between f(a),f(c) it means f(b) > M=\text{max} \{ f(a),f(c)\}. Pick a real number r so that f(b) > r > M. Then by intermediate value theorem there is a<x_0<b \implies f(x_0) = r \text{ and }b<x_1<c \implies f(x_1) = r. But then f(x_0) = f(x_1) for x_0\not = x_1. A contradiction.
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