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Let $I \subset R$ be an interval. Let $f : I \to R$ be a continuous function.

Assume that $I := [a, b]$. Assume that for all $c, d \in [a, b]$ such that $c < d$, there exists $e \in [c,d]$ such that $f(e) = f(a)$ or $f(e) = f(b)$. Prove that $f$ is a constant.

Consider this statement: For all $c \in [a,b], f(c) \in \{f(a),f(b)\}$

I figured that proving this statement would allow me to prove the function to be constant, but I'm unable to do so. Any thoughts?

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  • $\begingroup$ Are (or were) you unable to prove the result from your statement, your statement from the given information or both? $\endgroup$ – PJTraill Nov 22 '18 at 21:53
  • $\begingroup$ The latter actually. @PJTraill $\endgroup$ – Ahmad Lamaa Nov 23 '18 at 1:43
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We will proceed by contradiction:

I assume you want to prove that the function is constant given the statement

$c \in [a, b] \Rightarrow f(c) \in \{f(a), f(b) \}$

if that's the case, then we get a continuous function to a discrete space. If the function takes more than one value, the image of our function will be$ {f(a), f(b)}$ which is disconnected. But continous functions preserve connectedness.

Contradiction Q.E.D.

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  • $\begingroup$ I suppose here $f(c)\in[f(a), f(b)]$. Otherwise the function is discontinuous assuming $f(a)\neq f(b)$. $\endgroup$ – Yadati Kiran Nov 22 '18 at 15:52
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    $\begingroup$ Yes, assuming $f(a) \neq (fb) $ would mean it is discontinous, which is precisely the contradiction we're using that in fact $f(a)$ must be equal to $f(b)$ and the function is in fact constant. [think about the fact that if in the end both of them were to not be the same, the function would not be constant anyway ] $\endgroup$ – Aleks J Nov 22 '18 at 15:55
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Let $c$ in $(a,b)$ and let $\epsilon$ be a positive real number such that $[c-\epsilon,c+\epsilon]\subseteq[a,b]$. Define a sequence of intervals $(I_n)_{n\in\mathbb{N}}$ by the formula $I_n=[c-\frac{\epsilon}{n},c+\frac{\epsilon}{n}]$. By hypothesis, there exists a sequence of points $(c_n)_{n\in\mathbb{N}}$ such that $c_n\in I_n$ and $f(c_n)\in\{f(a),f(b)\}$. Then $c_n$ tends to $c$ and, by continuity of $f$, we have that $f(c)\in\{f(a),f(b)\}$. Thus, by continuity, $f$ is constant because it takes values in a discrete space.

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  • $\begingroup$ @PJTraill Yes, I corrected it. $\endgroup$ – Dante Grevino Nov 22 '18 at 23:41
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The set $D:=f^{-1}(f(a))\cup f^{-1}(f(b))$ is closed by continuity of $f$ and dense in $I$ by the special property. Clearly $D$ does not intersect the open set $I\setminus D$. By defnition of dense, this means that $I\setminus D$ is empty. Hence $I=D$. This makes $I$ the union of the two non-empty closed sets $f^{-1}(f(a))$, $f^{-1}(f(b))$. As $I$ is connected, these sets must overlap, which means that $f(a)=f(b)$ and ultimately that $f^{-1}(f(a))=I$, i.e., $f$ is constant.

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You need to use intermediate value property of continuous functions.

Suppose $f(a) =f(b) $. If $f$ is not constant in $[a, b] $ then there is some value of $f$ which differs from $f(a) =f(b) $ and let's suppose this value $f(c) > f(a) $ (the case $f(c) <f(a) $ is handled in a similar manner). By continuity there is an interval of type $[c-\delta, c+\delta] $ where all values of $f$ are greater than $f(a) $. And this contradicts the given hypotheses. Thus $f$ must be a constant function.

The case when $f(a) \neq f(b) $ leads us to a contradiction as shown next. Let's us assume $f(a) <f(b) $ and choose $k$ such that $f(a) <k<f(b) $. By intermediate value property there is a $c\in(a, b) $ for which $f(c) =k$. And by continuity there is an interval of type $[c-\delta, c+\delta] $ where all values of $f$ lie strictly between $f(a)$ and $f(b) $. This contradiction shows that we must have $f(a) =f(b) $ and as shown earlier the function is constant.

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