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Call a mapping of $X$ into $Y$ open if $f(V)$ is an open set in $Y$ whenever $V$ is an open set in $X$.

Prove that every continuous open mapping of $\mathbb{R}^1$ into $\mathbb{R}^1$ is monotonic.

Proof: Let $f$ is not monotonic function. Then exists $a<b<c$ such that $f(a)>f(b)$ and $f(c)>f(b)$. Hence $[a,c]$ is compact then continuous $f$ attains it's minimum at q, i.e. $\text{argmin}_{[a,c]}f=q.$ But $f(a)>f(b)\geqslant f(q)$ and $f(a)>f(b)\geqslant f(q)$ thus $\text{argmin} _{(a,c)}f=q.$

Let $(q-\delta,q+\delta)$ some neighborhood of $q$ and it's an open set in $\mathbb{R}^1$ and $f$ is an open mapping then $f((q-\delta,q+\delta))$ must be an open, but $f((q-\delta,q+\delta))$ is not open because the last set contains no neighborhood of $q$.

Is my proof true?

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  • $\begingroup$ The exercise reads "Prove that every continuous open mapping of $\mathbb R$ into $\mathbb R$ is monotonic" $\endgroup$ – user217285 Sep 20 '15 at 8:31
  • $\begingroup$ @Nitin, Ohh sorry please. I edited. Is proof true? $\endgroup$ – ZFR Sep 20 '15 at 8:32
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[edit] If you edit the proof, my answer becomes a little silly. Hence I have copied your original proof here: this is what I am commenting on.

Proof: Let $f$ is not monotonic function. Then exists $a<b<c$ such that $f(a)>f(b)$ and $f(c)>f(b)$. Hence $[a,c]$ is compact then continuous $f$ attains it's minimum at q, i.e. $\text{min}_{[a,c]}f=q.$ But $f(a)>f(b)\geqslant f(q)$ and $f(a)>f(b)\geqslant f(q)$ thus $\text{min} _{(a,c)}f=q.$

Let $(q-\delta,q+\delta)$ some neighborhood of $q$ and it's an open set in $\mathbb{R}^1$ and $f$ is an open mapping then $f((q-\delta,q+\delta))$ must be an open, but $f((q-\delta,q+\delta))$ is not open because the last set contains no neighborhood of $q$.

It is essentially right. Some points-

  • It is not true that you will get a minimum, since it could be a maximum. Of course this follows, but you didn't say so.
  • When you say $\min_{[a,c]}f = q$, what you actually want to say is $\text{argmin}_{[a,c]} f = q$
  • You repeated $f(a) > f(b) \geq f(q)$ twice; I assume one of them should be $f( c ) > f(b) \ge f(q)$.
  • You did not show that $f((q-δ,q+δ))$ is not open: what is true is that it does not have a neighbourhood around of $f(q)$, and also only for $δ$ small enough so that $(q-δ,q+δ)\subset [a,c]$.

Lemma. Say a function f is non-monotone if there exist $a<b<c$ such that either $f(b) > max(f(a),f(c )) $ or $f(b) < min(f(a),f(c ))$. Then f is non-monotone ⇔ it is not monotone increasing and it is not monotone decreasing.

Proof. "$\implies$" is clear. Suppose f is not non-monotone. Then for all triples $a<b<c$, $f(b)$ is in between the values $f(a)$ and $f(b)$.

Suppose $f(a)\le f(b)\le f(c)$ for one triple. We need to show that for any $x<y, f(x)\le f(y)$.

There are a few cases to consider, but in each case you just need to draw it out and show that if $f(x)>f(y)$ we get a 'V' shape in the graph. I will illustrate one-

$x<y \le a < b < c$. We must have $f(y)$ in between $f(x)$ and $f(b)$. But if $f(x)>f(y)>f(b)$, then $f(b)<min(f(y),f(c ))$, which contradicts the assumption that $f$ is not non-monotone. Thus $f(x)\le f(y) \le f(a)$ as needed.

The other cases are clear by analogy. We can also proceed similarly if we have a decreasing triple instead of an increasing one to show that $f$ is monotone decreasing. QED

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  • $\begingroup$ 1) Can you explain in detail the first point? 2)Yeah I mean argmin. 3) The second inequality must to be $f(c)>f(b)\geqslant f(q)$. 4) I proved that $f((q-\delta,q+\delta))$ because it's contains no neighbothood of $f(q)$. $\endgroup$ – ZFR Sep 20 '15 at 9:17
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    $\begingroup$ 1)$ f(x) = -x^2$ does not have a minimum. We are in agreement for 2) and 3). 4): No, first of all, your proof says 'q' not 'f(q)'. Secondly, you didn't choose a very small $δ$. Consider $f(x) := \frac{1}{1+x^2} + x/10000$. The range of $f$ is $\Bbb R$ but it is not monotonic, due to the local maximum at 0. Using your proof, if we choose $δ \gg 1$, then $f(-δ,+δ)$ will actually still be an open interval. $\endgroup$ – Calvin Khor Sep 20 '15 at 9:24
  • $\begingroup$ 1) I mean that $f$ has minimum in any compact. Right? $\endgroup$ – ZFR Sep 20 '15 at 9:30
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    $\begingroup$ Well we know one of the two situations must occur if $f$ is not monotonic. Prove the one you thought of first. Then if the second situation (f(b) is the highest) occurs, apply your first result to $g:=-f$. $\endgroup$ – Calvin Khor Sep 20 '15 at 9:45
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    $\begingroup$ $f$ is non-monotonic if it is neither monotone increasing or monotone decreasing. $\endgroup$ – Calvin Khor Sep 20 '15 at 9:53

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