$X$ topological space and $S\subseteq X$ and $\overline{S}=X$. Show that if $S$ is connected, then $X$ is also $X$ topological space and $S\subseteq X$ and $\overline{S}=X$. Show that if $S$ is connected, then $X$ is also.
First of all, what is the intuition behind this? For example, what is a subset $S$ such that $\overline{S} = X$?. I can't even think of such thing... Is it a space that is almost $X$? And then when we take the closure and it's $X$. Well, what argument should be used to argue that $X$ will also be connected?
 A: Suppose $X$ isn't connected. Then there are closed sets $V, W\subsetneq X$ with $V\cup W=X, V\cap W=\emptyset$. Then we have relatively closed sets $V\cap S\subseteq S, W\cap S\subseteq S$ with $(W\cap S)\cap (V\cap S)=\emptyset$ and we also have $W\cap S\subsetneq S$ and $V\cap S\subsetneq S$, then if e.g $W\cap S= S$ was correct, we had $\bar{S}\subseteq W\subsetneq X$.
So all in all we see $S$ isn't connected, a contradiction.
A: Assume that there is a continuous surjection $\varphi:X\rightarrow\{0,1\}$ and we let without loss of generality that some $s\in S$ be such that $\varphi(s)=0$. As $S$ is connected, we have $\varphi[S]=\{0\}$. Then for any $x\in X-S$, we could find some net $\{x_{\delta}\}\subseteq S$ such that $\varphi(x)=\lim_{\delta}\varphi(x_{\delta})=\lim_{\delta}0=0$, it violates the condition that some $x\in X$ is such that $\varphi(x)=1$.
A: Assume X is not connected. Then X=Y U Z where Y= U{B(y)|y is in Y and S} and Z=U {B(z)| z is in Z and S} are non-empty. Then S=Y' U Z' where Y'=U{B(y) intersect S| y is in Y and S} and Z'= U {B(z) intersect S | z is in Z and S}. These are open sets in S.
