# Prove the following functions is a Darboux function but discontinuous everywhere

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?

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.

• And to prove the function is discontinuous, is it correct what I did?
– Luke
Feb 1, 2021 at 23:29
• 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 $b$). Feb 1, 2021 at 23:48

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 $$|f(c)-f(x)|>\frac{1}{2}=\epsilon$$.

• When we have $a$ and $b$, why can we affirm both are members of the cantor set?
– Luke
Feb 1, 2021 at 23:49
• 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 $x$. Feb 1, 2021 at 23:51
• 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). Feb 1, 2021 at 23:56