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A function $f: \mathbb{R} \to \mathbb{R}$ is said to have the intermediate-value property if for any $a$,$b$ and $\lambda \in [f(a),f(b)]$ there is a $x \in [a,b]$ such that $f(x)=\lambda$

A function $f$ is injective if $f(x)=f(y) \Rightarrow x=y$

Now it is the case that every injective function with the intermediate-value property is continuous. I can prove this using the following steps:

  1. An injective function with the intermediate-value property must be monotonic.
  2. A monotonic function possesses left- and right-handed limits at each point.
  3. For a function with the intermediate-value property the left- and right-handed limits at $x$, if they exist, equal $f(x)$.

I am not really happy with this proof. Particularly I don't like having to invoke the intermediate-value property twice.

Can there be a shorter or more elegant proof?

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    Even if you do invoke the IVP twice, this proof is quite clear. If you want to explore other avenues along the same lines, you may want to avoid proving that it's monotonic, just proving instead that the left- and right-handed limits exist at all points. (I'm not saying it's possible.)2010-10-19
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    I agree with lhf: the proof given in the question is short, clear and conceptual. I find absolutely nothing wrong with it.2010-10-19
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    A bounded monotonic and injective function need not be continuous. But, you can show that it is Riemann integrable and thus is continuous almost everywhere. So if you use IVP to show monotonicity and don't want to use it again, you are stuck.2010-10-19
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    @Moron: I later found a proof using sequences that does not show monotonicity first. Have put it below as one of the answers. Not sure if it is more elegant than the prove above.2010-10-19

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