Question about density and the boundary of open sets. If $X \subset R^2$ and $X$ is open. $S \subset bdry(X)$ and moreover, dense on it.
Is it necessary that any path on $ bdry(X)$ intersects $S $?
(constant path is not allowed)
 A: Assuming you mean $S$ is dense in $bdry (X)=\partial X$ (in LaTex \partial X), the answer is NO.
For $n\in \Bbb Z^+$ let $S(n)= \{m/n: m\in \Bbb Z\}\times \{1/n\}.$ Let $S=\cup_{n\in \Bbb Z^+}S(n).$ Let $X=(\Bbb R^2 \setminus S)\setminus (\Bbb R \times \{0\}).$
Then  $X$ is open and $\partial X =S\cup (\Bbb R\times \{0\}).$  Now $S$ is dense in $\partial X,$ but $\Bbb R\times \{0\}$ is an infinite path-connected subset of $\partial X$ which is disjoint from $S.$
Remark: If we require $X$ to be connected and open, the answer is still NO. Replace $X$ in the example above with $(\Bbb R\times \Bbb R^+)\setminus S.$
A more complicated example. Let $S=\{(x,\sin 1/x): 0<x\leq 1\}$ and let $X=\Bbb R^2\setminus \overline S.$ 
We have $\partial S=\Bbb R^2\setminus X=S\cup (\{0\}\times \{-1,1\} ).$
Now $S$ is dense in $\partial X$ but clearly there is a path $P$ in $\partial X$ from the point $(0,-1)$ to the point $(0,1)$ with $P\cap S=\emptyset.$
Remark: $ \overline S=\partial X$ is a "topologists' sine curve," which is connected but not path-connected. Any path in $\overline S$ is a subset of $S$ or a subset of $\{0\}\times \{(-1,1)\}.$
So even if $X$  is connected and open, and $\partial X$ is connected, the answer is still NO.
