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Definition Let $X$ and $Y$ be topological spaces. We say $f: X \rightarrow Y$ is Darboux provided that $F[C]$ is connected for every connected C $\subset$ X.

Problem Consider the Cantor set $C$ in $[0,1]$. if $(a,b)$ is an interval contiguous to $C$ we define $f(x)=\frac{2(x-a)}{b-a}-1$ for $x \in [a,b]$; otherwise put $f(x)=0$. Then $f$ is Darboux and is discontinuous at each point of $C$

My idea to prove f is not continuous, take $b \in [a,b]$ then $$lim_{b^{-} \rightarrow }f(x)=1 \neq 0=lim_{b^{+} \rightarrow }f(x)$$

And I have problems with prove Darboux function, if we denote $(a_{1},b_{1}), (a_{2},b_{2})$ with $b_{1} < a_{2}$ and $0 < a_{1}, b_{2}<1$ intervals contiguous to $C$, we can define the interval $[b_{1}, \frac{b_1+a_2}{2}]$ is a connected subset of $[0,1]$ then

$f[[b_{1}, \frac{b_1+a_2}{2}]]=\{1\} \cup \{0\}$ is not a connected set, is it?

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2 Answers 2

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The problem is the following. You can check that if $b_1$ is an the endpoint of an interval contiguous to $C$, then there are infinite intervals $(a_n,b_n)$ contiguous to $C$ converging to $b_1$ from the right. So, $[b_1,\frac{b_1+a_2}{2}]$ will contain some of these intervals, and its image will be the whole $[-1,1]$.

To prove that $f$ is Darboux, take and interval $I$ and let´s see that its image is continuous. If the interval is contained in some interval $[a_n,b_n]$ contiguous to the Cantor set, it is obvious. If not, it has to have a point of the Cantor set in its interior. And for every point of the Cantor set there is a convergent sequence of intervals contiguous to the Cantor set converging to it. So, the image will be $[-1,1]$, which is connected.

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  • $\begingroup$ And to prove the function is discontinuous, is it correct what I did? $\endgroup$
    – Luke
    Feb 1, 2021 at 23:29
  • $\begingroup$ You assumed the upper limit at $b$ was 0, but in fact the function hasn´t upper limit. But you can check that any nhood of $b$ has image $[-1,1]$, so the function can´t be continuous at $b$ (for example, $f^{-1}(0,2)$ doesn´t contain a nhood of $b$). $\endgroup$
    – Saúl RM
    Feb 1, 2021 at 23:48
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Denote $F$ the cantor set. If $C=(a,b)\subset[0,1]$ (or $[a,b],[a,b),(a,b]$) contains no cantor points, then $f$ is increasing and continuous on $(a,b)$, so $f((a,b))=(f(a),f(b))=(-1,1)$ (or $[-1,1],[-1,1),(-1,1]$), which is connected.

Cantor set has no isolated points, so any connected $C\subset[0,1]$ containing a cantor point $c\in F$ must contain another point $d\in F$ on the cantor set. Since the cantor set is nowhere dense there must exist a point $x\in(c,d)\setminus F$ outside of the Cantor Set. Since the Cantor Set is closed, $[0,1]\setminus F$ is open, and as $x\in[0,1]\setminus F$, there must exist $a,b\in[c,d]$ such that $x\in(a,b)\subset[0,1]\setminus F$, and both $a$ and $b$ are members of the cantor set. As we've already proved, $f([a,b])=[-1,1]$. Then, as $[a,b]\subset[c,d]\subset C$, $f([a,b])=[-1,1]\subset f(C)$. Also, by definition $f(C)\subset[-1,1]$, which completes the proof.

To prove $f$ is discontinuous on every $c\in [0,1]\setminus F$, its enough to use the previous. Let $\epsilon\in(0,\frac{1}{2})$. Then, for any $\delta\in\mathbb{R}^+$, $B(c,\delta)$ is connected; so as we proved before, $f(B(c,\delta))=[-1,1]$. In other words, whatever $f(c)$ is between $-1$ and $1$, there exists $x\in B(c,\delta)$ such that $|f(c)-f(x)|>\frac{1}{2}=\epsilon$.

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  • $\begingroup$ When we have $a$ and $b$, why can we affirm both are members of the cantor set? $\endgroup$
    – Luke
    Feb 1, 2021 at 23:49
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    $\begingroup$ At first you just know that there exists an open interval $(a,b)$ such that $x\in(a,b)\in[0,1]\setminus F$. But you can assume $a,b\in F$ by choosing the greatest open interval in $[0,1]\setminus F$ containing $x$. $\endgroup$
    – R.V.N.
    Feb 1, 2021 at 23:51
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    $\begingroup$ To be more rigorous, let $x\in(a',b')\in(c,d)$. Then define $a:=\sup\{t\in F : t\leq a'\}=\sup F\cap [0,a']$, $b:=\inf\{t\in F : b'\leq t\}=\inf F\cap[b',1]$, they both are in $F$ because $F\cap[0,a']$ and $F\cap [b',1]$ is compact (intersection of two compacts). $\endgroup$
    – R.V.N.
    Feb 1, 2021 at 23:56

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