Just to be clear, I will define Cantor's ternary set and its ternary property.
Let $A_0:=[0,1]$
Let $A_1:=[0,1]\backslash\left(\frac{1}{3},\frac{2}{3}\right)=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3},1\right]$
Let $A_2:=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9},3\right]\cup\left[\frac{2}{3},\frac{7}{9}\right]\cup\left[\frac{8}{9},1\right]$
So, $A_n$ is union of $2^n$ closed intervals, each of which is contained in [0,1],for some $n \in \mathbb{N},n>2$
$\implies A_n$ is closed, $\forall n \in \mathbb{N}(\because$ each $A_n$ is finite union of closed sets(closed intervals)$)$
Define $C:=\cap_{n \in \mathbb{N}} A_n=$ The Cantor's ternary set. $\implies C \subset [0,1]$
Since, arbitrary intersection of closed sets is closed, $C$ is closed.
Ternary property:
$\alpha \in C \iff \forall u \in \mathbb{N}(\alpha=0._3a_1a_2...a_na_{n+1}...$ such that $a_u\neq 1 \implies a_u=0$ or $a_u=2)$
$\therefore |C|=|\{0,2\}^{\mathbb{N}}|=2^{\aleph_0}$
Proceeding to my query:
No interval as a subset exists in $C$
(If not, assume $F\subset C$ to be an interval. Since $F$ is not free from $Y$(where $Y\in\{\mathbb{Q},Ir, Al, Tr\})$, there exists some $s \in F\subset [0,1]$ such that $s=0._3a_1a_2...a_na_{n+1}...$, and at least one $a_i=3\implies s\notin C\implies F \not \subset C$
Here $Ir:=$ The set of all irrationals, $Al:=$ The set of all real algebraic numbers, $Tr:=$ The set of all real transcendental numbers.)
From this, we get no open interval is a subset of $C$
Also, based on how $C$ is defined, there is quite a large gap between any two elements of $C$.
So, $C$ is a non-dense in $W(\forall W \not \subset C)$, uncountable subset of $\mathbb{R}$, where $C=\cap_{x \in X}\{x\}$, where $\{x\}\cap \{y\}=\varnothing,\forall x\neq y \in X:=\{x\ |\ x=0._3a_1a_2...a_na_{n+1}.. \text{ and } a_i\neq1,\forall i \in \mathbb{N}\}$
By a dense set in $G$, I mean its closure is $G$
Am I correct in my interpretation so far?
Also, in case of $\mathbb{R}$, if two numbers have nothing in between them, then they have to be equal. Comparing this with Cantor's set, 'gap can be seen between' any two reals, yet $C$ turns out to be uncountable.
Are there other sets $T$ like $C$ which are non-dense in all subsets of $\mathbb{R}$ except for any subset of $T$ and uncountable, that are totally different from $C$, in terms of their definition(nothing of the sort of removing middle third, or a process similar to it: removing middle fifth, middle eleventh, etc...)
"$C$ is not dense in any set that is not a subset of $C$" is equivalent to "$C$ is closed". So you're asking about uncountable closed sets. Maybe you also want uncountable closed sets that contain no intervals.
Now it's not quite clear what you mean by "totally different in terms of their definition". In principle, any closed subset of $\mathbb R$ can be defined by removing intervals. The complement of a closed set is an open set, and in the real line that means a union of at most countably many disjoint open intervals (this because each open interval contains a rational number, and there are only countably many rational numbers). So start with the real line, remove all maximal open intervals disjoint from your set, and what you're left with is your set.