Let $E\subset \mathbb{R}$ infinite such that for all $x<y$ in $E$ there is $z\in E$ such that $x<z<y.$ True or false? $\overline{E}$ has non-empty interior.
Attempt. I believe the answer is positive. We have to find $w\in \overline{E}$ such that $(w-\epsilon,w+\epsilon)\subset \overline{E}$ for some $\epsilon>0.$ But I don't see how to get there.
Thank you in advance.
This is a revisitation of an answer I gave last Christmas to basically the same question. I don't like to link it as duplicate anymore because the OP is (-3) and closed (therefore, there is no room for improvement). The apparent simplification of the argument is due to a learned observation by user Henno Brandsma.
Consider a set $C\subseteq \Bbb R$ such that:
$C$ is closed and the interior of $C$ is empty
for all $x,y\in C$ such that $x<y$, either $(x,y)\cap C=\emptyset$ or $\lvert(x,y)\cap C\rvert>\aleph_0$.
$\lvert C\rvert>\aleph_0$
Call $\mathcal U_C$ the family of connected components of $\Bbb R\setminus C$. It is self-evident that, if $x,y\in C$ and $x<y$, then $(x,y)\cap C=\emptyset$ if and only if $(x,y)\in\mathcal U_C$. Thus, call $$\mathcal Q=\{c\in C\,:\,c\text{ is an extremal point of some interval }I\in\mathcal U_C\}$$
Since $\mathcal U_C$ contains countably many intervals, each of which has at most two extremal points, $\lvert\mathcal Q\rvert=\aleph_0$. Therefore:
$\widehat C:=C\setminus\mathcal Q$ is uncountable by (3)
$\widehat C$ is nowhere dense by (1)
since we've eliminated all pairs of points $x<y$ such that $(x,y)\cap C=\emptyset$, for all $x,y\in \widehat C$ it holds $\lvert (x,y)\cap C\rvert>\aleph_0$; therefore $\left\lvert(x,y)\cap \widehat C\right\rvert=\lvert ((x,y)\cap C)\setminus\mathcal Q\rvert>\aleph_0\ge1$. Thus, $\widehat C$ is densely ordered a fortiori.
Such a $\widehat C$ is a counterexample to your claim.
An example of set that satisfies (1),(2),(3) is the Cantor set, and the answer I linked deals with that case. However, since for that instance $\mathcal Q\subseteq \Bbb Q$, the set $C'=C\setminus \Bbb Q$ works as well instead of the actual $\widehat C$.
One could also observe that the following holds:
Let $\mathbb P$ be an atomless $\sigma$-additive probability on $(\Bbb R,\text{Borel})$. Let $H$ be a closed subset of $\Bbb R$ such that $\mathbb P(H)>0$ and $H^\circ =\emptyset$. Then, there is a closed set $C\subseteq H$ satisfying (1), (2), (3) such that $\mathbb P(C)=\mathbb P (H)$. Specifically, the least closed subset of $H$ which has negligible complement has this property.
Proof: The existence and uniqueness of a maximal negligible open set is due to a topological fact: second-countable spaces are hereditarily Lindelöf and, therefore, the union of all the negligible open subsets of $H$ is actually a countable union. Now, let $C$ be the least closed subset of $H$ with negligible complement. Since $\Bbb P(C)=\Bbb P(H)>0$ and all points are $\Bbb P$-negligible, $\lvert C\rvert>\aleph_0$. Moreover, if $(x,y)\cap C\ne \emptyset$, then $\lvert (x,y)\cap C\rvert>\aleph_0$. In fact, $(x,y)\cap H=(((x,y)\cap H)\setminus C)\amalg ((x,y)\cap C)$. $(x,y)\cap H\setminus C$ is a Borel subset of $H\setminus C$, and therefore it is negligible. If $(x,y)\cap C$ were countable, it would be negligible as well. Therefore, $C\setminus (x,y)=C\setminus((x,y)\cap H)$ is a closed subset of $H$, with negligible complement, which is contained in $C$. By minimality, it must be $C$, therefore $C\cap (x,y)=\emptyset$. $\square$
The Cantor set is the instance where $\Bbb P$ is the Cantor probability on $[0,1]$. However, the lemma proves that there are nowhere dense, densely ordered subsets of the real line which have positive Lebesgue measure. Namely, consider $\Bbb P$ a probability such that $\Bbb P\ll \mathcal L\ll \Bbb P$ - for instance, $\Bbb P(A)=\frac1{\sqrt{2\pi}}\int_A e^{-x^2/2}\,dx$ - and take for $H$ this example from Wikipedia. Since $\widehat C$ is countably many points shy of being $C$, $\Bbb P\left(\widehat C\right)=\Bbb P(H)>0$ and thus $\mathcal L\left(\widehat C\right)>0$.
The answer is negative.
Consider a bijection $\varphi : \Bbb N \to \Bbb Q$.
Consider the function $f:\Bbb R \to \Bbb R$ where $f(r) = \sum \{2^{-n} \mid \varphi(n) < r\}$.
We have the property that:
Consider $x$ a limit point of $\operatorname{im}f$. Let $\{f(r_n)\}_{n=1}^\infty$ converge to $x$. Then, let $\{r_n\}_{n=1}^\infty$ converge to $r$. If $r$ is rational, then $x = f(r)$ or $x = f(r) + 2^{-\varphi^{-1}(r)}$, since $f$ is left-continuous and right-continuous at rationals. If $r$ is irrational, then $x = f(r)$ because $f$ is continuous at rationals. Therefore, $(f(\varphi(n)),f(\varphi(n))+2^{-n})$ is not in the closure of $E = \operatorname{im} f$.
Consider an interval in the closure of $E$. Its preimage under $f$ is open, which must contain a rational number, so the interval must not contain $(f(\varphi(n)),f(\varphi(n))+2^{-n})$, so is not connected, which gives us a contradiction.
