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 $x \in [a,b]$ 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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    $\begingroup$ 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.) $\endgroup$ – lhf Oct 19 '10 at 12:13
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    $\begingroup$ I agree with lhf: the proof given in the question is short, clear and conceptual. I find absolutely nothing wrong with it. $\endgroup$ – Pete L. Clark Oct 19 '10 at 14:33
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    $\begingroup$ 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. $\endgroup$ – Aryabhata Oct 19 '10 at 18:43
  • $\begingroup$ @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. $\endgroup$ – Jyotirmoy Bhattacharya Oct 19 '10 at 23:24

[I thought of another proof that uses the IVP and injectiveness once. Putting it as a community wiki answer.]

Assume on the contrary that $f$ is not continuous at $x$. Then there is a sequence $x_n$ converging to $x$ such that $f(x_n)$ does not converge to $f(x)$. Then there is $\epsilon>0$ and a subsequence $x_{n_k}$ such that $f(x_{n_k}) \notin (f(x)-\epsilon,f(x)+\epsilon)$.

There must either be a further subsequence $x_{n_j}$ such that $f(x_{n_j}) \leq f(x)-\epsilon$ or a subsequence $x_{n_q}$ such that $f(x_{n_q}) \geq f(x)+\epsilon$ (or both). Assume without loss of generality the former.

Since $f(x_{n_j}) \leq f(x)-\epsilon < f(x)$, by the IVP for every $j$ there is a $y_j$ such that $x_{n_j}\leq y_j < x$ and $f(y_j)=f(x)-\epsilon$.

Because $f$ is injective, all the $y_j$ must be the same, say $y$. Because $x_{n_j}$ converges to $x$, $y=x$ by the sandwich theorem. But $f(y)\neq f(x)$. Hence a contradiction.


I don't know if this is shorter or elegant, and in particular it does use the hypotheses more than once, but here's something.

Given your 1., assume WLOG that $f$ is increasing. Then for all $a\lt b$, $f(a,b)\subseteq (f(a),f(b))$, and thus $(a,b)\subseteq f^{-1}(f(a),f(b))$. Since each element of $(f(a),f(b))$ has a preimage point in $(a,b)$ and $f$ is injective, $f^{-1}(f(a),f(b))\subseteq (a,b)$. Combining, $f^{-1}(f(a),f(b))=(a,b)$. I'll show that this implies that the inverse image of each open interval is open, which implies continuity.

Let $x\lt y$ and suppose that $t$ is in $f^{-1}(x,y)$. Let $b>t$. Since $f^{-1}(f(t),f(b))=(t,b)$ and $f(t)\lt y$, there is a $b'$ with $t\lt b' \lt b$ and $f(b')\lt y$. Thus $f^{-1}(f(t),f(b'))=(t,b')$ is contained in $f^{-1}(x,y)$. Applying a similar argument on the other side, there is an $a'\lt t$ such that $(a',t)$ is contained in $f^{-1}(x,y)$. Together this implies that $t$ is in the interior of $f^{-1}(x,y)$.


As you note, f is injective and has the intermediate value property => f is monotonic. We may assume f is increasing.

Now for any x and any small* ε > 0, we have by the IVP

  • c in [x-1, x] such that f(c) = f(x)-ε and

  • d in [x, x+1] such that f(d)=f(x)+ε.

Then we choose δ=min(x-c,d-x). Since f is increasing, any point within δ of x maps to a point within ε of f(x); thus f is continuous.

*By "small" ε, I mean ε smaller than both f(x)-f(x-1) and f(x+1)-f(x), so that the condition of the IVP is satisfied.

I also invoke the IVP more than once. Perhaps the other Jonas will supply a proof with just one use of the IVP!


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