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How to prove that strictly monotonic continuous function carries open intervals to open intervals? if $f:\mathbb{R}\to \mathbb{R}$ continuous and monotonic, we need to to Prove that If $X \subset$ R is an open interval then $f(X)$ is an open interval.

Clearly $f(X)$ is an interval(as continuous functions takes connected sets to connected sets and connected sets in $\mathbb{R}$ is an interval ) we have to prove that it is open. i am not able to use the monotonicity to prove an arbitrary point is an interior point.

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  • $\begingroup$ If $f(X)$ contains an endpoint $a$, consider an open subinterval of $X$ containing $f^{-1}(a)$. $\endgroup$ – David Mitra Mar 16 '14 at 10:50
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I give the proof for the maximum, and the minimum is likewise. suppose the domain is(a,b), and the range is $X$.

i.e. $$x\in(a,b)$$.

without loss of generality, we suppose the function is strictly monotonic increasing

Let M denote the supremum of the range of f(x), $$f(x)\le M$$ for $x\in (a,b)$.

Now, we prove M cannot be reached. we demonstrate a proof by contradiction.

Suppose not, then then there is a $x_0\in (a,b)$, such that $$ f(x_0)=M$$

since $x_0<b$, we can find a $x'$ such that, $$x_0<x'<b$$. The function is strictly increasing(as supposed above).

then $$M=f(x_0)<f(x')$$

so this means that $M$ is not the supremum of f(x), which is a contradiction.So the supremum M cannot be reached for $x\in (a,b)$. we finish our proof.

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Hopefully my answer is simpler than the ones already there (in particular I only use the mean value theorem, not Bolzano-Weierstrass).

Let $X = (a,b)$ be an open interval. Let $f(x) \in f(X)$, for $a < x < b$. Choose $u,v$ such that $a < u < x < v < b$ (note that $u,v \in X$). then by monotonicity, $f(u) < f(x) < f(v)$. By the MVT, the whole of $(f(u), f(v))$ is included in $f(X)$, and $f(x) \in (f(u), f(v))$, and therefore $f(x)$ is an interior point of $f(X)$.

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As you have noticed $f(X)$ is an interval. To prove that it is open notice that given any $x\in X$ there exist $\varepsilon>0$ such that $[x-\varepsilon,x+\varepsilon]\subset X$. Since $f$ is continuous $f(X)$ contains $[f(x-\varepsilon),f(x+\varepsilon)]$ and since $f$ is strictly monotone this is a neighbourhood of $f(x)$ (e.g. if $f$ is strictly increasing $f(x-\varepsilon) < f(x) < f(x+\varepsilon)$).

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Consider the open interval $(a,b)$. First note that $[a,b]$ is compact and connected, so $f[ [a,b]]$ is also compact and connected, and so must be a closed bounded interval (closed and bounded from compact and interval from connectedness), say $f[[a,b]] = [c,d]$. Now monotonicity implies that in fact $f[(a,b)] = (c,d)$ as well (the minimum goes to the minimum and the same for the maximum).

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