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It is intuitive and it seems very obvious that if a function $f : X \rightarrow Y$ is continuous on whole $X$ and it's injective, then it must be monotonic, but I can't come up with any neat proof for that. Could you maybe help me?

Hagrid
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    What are $X$ and $Y$? Parts of $\mathbb{R}$? – Seirios Jan 15 '13 at 18:22
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    Depends on what kind of topological spaces $X$ and $Y$ are. To speak of monotonic, they should be ordered. Maybe you even want just subsets of $\mathbb R$? Then $X=(-3,-2)\cup(0,1)$, $Y=\mathbb R$, $f(x)=x^3-9x$ is a counterexample. – Hagen von Eitzen Jan 15 '13 at 18:25
  • show that there exist a continuous map$f^{-1}:Y \to X $, $f(a) \neq f(b) \implies a \neq b \implies a < b $ or $ b < a$, Let $a = f^{-1} (A) $ and $ b = f^{-1} (B)$. $ a < b \implies f^{-1}(A) < f^{-1}(B) \implies A < B \implies f(a) < f(b)$ hence monotonic. – S L Jan 15 '13 at 18:42
  • I'm so sorry. I only need X, Y to be connected subsets of R. because I proved that inverse of a continuous and injective function $f:(a, b) \rightarrow R$ is also continuous and I think I should include that proof there. – Hagrid Jan 15 '13 at 18:52
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    @HagenvonEitzen: I learned a function is monotonic if the preimage of a connected set is connected, so I don't think you need ordering at all, and under this definition, you would have to include either $-3$ or $0$ in your example so that the image is connected. Then any open interval around $0$ would be disconnected, showing the function isn't monotone. – Clayton Jan 15 '13 at 19:02
  • The familiarest example is $f(x)=1/x$ on the real line punctured by the removal of $0$. – Lubin Jan 15 '13 at 19:19

2 Answers2

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I assumed that $X$ and $Y$ are parts of $\mathbb{R}$. If $X$ is not connected, Hagen von Eitzen gave a counterexample.

HINT: otherwise, you can use the intermediate value theorem to prove the statement.

FD_bfa
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Seirios
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  • Could you tell me exactly how do I use the intermediate value theorem here? – Hagrid Jan 15 '13 at 18:53
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    @Hagrid: If $f$ is not monotonic, there exists $x<y,z<t \in X$ such that $f(x) \geq f(y)$ and $f(z) \leq f(t)$. You should make a drawing. – Seirios Jan 15 '13 at 19:07
  • Ok, it's all clear now. We need to apply the fact that if $f'(x)<0$, then $f$ is decreasing and if $f'(x)>0$, then $f$ is increasing. Is that ok? – Hagrid Jan 15 '13 at 19:24
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    @Hagrid: $f$ is not supposed to be differentiable and it is not necessary. You only have to distinguish the possible cases; for example, if $y<z$ and $f(y) \geq f(z)$, then you can apply the intermediate value theorem to find $a \in [y,z[$ and $b \in [z,t]$ such that $f(a)=f(b)$, ant that it is impossible because $f$ is injective. – Seirios Jan 15 '13 at 21:40
  • "You can use the intermediate theorem to prove it" is a terrible answer. It is just a hint. – ashpool Mar 21 '24 at 01:30
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I am assuming that $X$ is an interval in $\mathbb{R}$, and $Y=\mathbb{R}$.

Lemma: Let $f:X\to Y$ be a function with the following property: For any $x,y,z\in X$ such that $x<y<z$, $f(y)$ is strictly between $f(x)$ and $f(y)$. Then $f$ is strictly monotonic.

Proof: $f$ should injective. For, if $x,y\in X$ and $x<y$, then there exists $z\in X$ such that $x<z<y$, and either $f(x)<f(z)<f(y)$ or $f(x)>f(z)>f(y)$, i.e., $f(x)\neq f(y)$.

Next, let $x\in X$. There exists $y\in X$ such that $x\neq y$. First suppose that $x<y$. Since $f$ is injective, either $f(x)<f(y)$ or $f(x)>f(y)$. We will show that $f$ is strictly increasing in the former case, and strictly decreasing in the latter case. Suppose that $f(x)<f(y)$. Let $z\in X$. Note the following:

  • If $x<z<y$, then necessarily $f(x)<f(z)<f(y)$.
  • If $x<y<z$, then necessarily $f(x)<f(y)<f(z)$.
  • If $z<x<y$, then necessarily $f(z)<f(x)<f(y)$.

It follows that if $x<z$, then $f(x)<f(z)$, and if $z<x$, then $f(z)<f(x)$. This shows that $f$ is strictly increasing.

Now suppose that $f(x)>f(y)$. Let $z\in X$. Note the following:

  • If $x<z<y$, then necessarily $f(x)>f(z)>f(y)$.
  • If $x<y<z$, then necessarily $f(x)>f(y)>f(z)$.
  • If $z<x<y$, then necessarily $f(z)>f(x)>f(y)$.

It follows that if $x<z$, then $f(x)>f(z)$, and if $z<x$, then $f(z)>f(x)$. This shows that $f$ is strictly decreasing.

Next, suppose that $y<x$. Since $f$ is injective, either $f(y)<f(x)$ or $f(y)>f(x)$. We will show that $f$ is strictly increasing in the former case, and strictly decreasing in the latter case. Suppose that $f(y)<f(x)$. Let $z\in X$. Note the following:

  • If $y<x<z$, then necessarily $f(y)<f(x)<f(z)$.
  • If $y<z<x$, then necessarily $f(y)<f(z)<f(x)$.
  • If $z<y<x$, then necessarily $f(z)<f(y)<f(x)$.

It follows that if $x<z$, then $f(x)<f(z)$, and if $z<x$, then $f(z)<f(x)$. This shows that $f$ is strictly increasing.

Now suppose that $f(y)>f(x)$. Let $z\in X$. Note the following:

  • If $y<x<z$, then necessarily $f(y)>f(x)>f(z)$.
  • If $y<z<x$, then necessarily $f(y)>f(z)>f(x)$.
  • If $z<y<x$, then necessarily $f(z)>f(y)>f(x)$.

It follows that if $x<z$, then $f(x)>f(z)$, and if $z<x$, then $f(z)>f(x)$. This shows that $f$ is strictly decreasing.

Now let $f:X\to Y$ be a continuous and injective function. Suppose that $f$ is not strictly monotonic. Then by the lemma above, there exist $a,b,c\in X$ such that $a<b<c$, and $f(a)>f(b)<f(c)$ or $f(a)<f(b)>f(c)$, either of which contradicts the intermediate value theorem.

ashpool
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