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Connection between Monotone and One-to-one Functions

  1. Oct 4, 2009 #1
    Hello,

    I was curious about the following point. I know that if a function is monotone, then it is one to one (meaning for x1 != x2, then f(x1) != f(x2) ).

    But what about the converse? I can't seem to think of a counter-example.
     
  2. jcsd
  3. Oct 4, 2009 #2
    How about [tex]f : \mathbb{R} \to \mathbb{R}[/tex] given by

    [tex]f(x) = \left\{ \begin{array}{ll}
    1/x & x \neq 0 \\
    0 & x = 0
    \end{array} \right.[/tex]

    It's injective, but not monotone (because it's not continuous ;) ).
     
  4. Oct 4, 2009 #3
    Thanks!

    So, is that to say, that any injective, continuous function is strictly monotone?
     
  5. Oct 4, 2009 #4
    Correct.
     
  6. Jun 15, 2011 #5
    If f is bijective and monotone with both its range and domain being a closed and bounded interval...then what can we say about the continuity of f ?
     
  7. Jun 15, 2011 #6

    micromass

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    This thread is 2 years old. I guess the OP already found it by now.
     
  8. Jun 15, 2011 #7
    Hello, Micromass

    Thanks for your reply.

    If f is injective and continuous then it is strictly monotone...that is clear

    My question is : If f is montone then is f continuous ?
     
  9. Jun 15, 2011 #8

    micromass

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    No, there are a lot of counterexamples, for example

    [tex]f:\mathbb{R}\rightarrow\mathbb{R}:x\rightarrow \left\{\begin{array}{c} x~\text{if}~x\leq 0\\ x+1~\text{if}~x>0\\ \end{array}\right.[/tex]

    This is injective and strictly increasing, but not continuous. If f is required to be surjective however, then it will be continuous.

    Hint: next time you may get a faster reply if you just start a new topic about it! :smile:
     
  10. Jun 15, 2011 #9
    Thanks for your help with an example. I appreciate it.

    Thanks for yesterday's reply too.

    Regards,
    Bhatia
     
  11. Jun 15, 2011 #10

    disregardthat

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    A function which is injective and continuous need not be monotonic. E.g. f : R-{0} --> R defined by f(x) = 1/x is continuous and injective, but not monotonic.

    If however f : A --> B is injective and continuous, A and B are totally ordered topological spaces and A is connected, then f will be monotonic. The key here is that R itself in the standard topology is connected, and so continuous injective functions f : R --> R will be monotonic.
     
  12. Jun 15, 2011 #11
    Thanks for the insight. I get what you mean now.
     
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