Let $(X,\tau)$ be a topological space.
Let $(Y,\tau_Y)$ be a subspace of $X$.
Let's say "$Y$ is a connected subset of $X$ iff 'there does not exist nonempty subsets $A,B$ of $X$ such that $\overline{A} \cap B = \emptyset, \overline{B} \cap A= \emptyset, A\cup B= Y$'".
(Closures are taken with respect to $\tau$)
Let's say $Y$ is a connected subspace of $X$ iff 'there does not exist nonempty subset $A,B$ of $Y$ such that $\overline{A} \cap B =\emptyset, \overline{B}\cap A=\emptyset, A\cup B =Y$'".
(Closures are taken with respect to $\tau_Y$)
It's trivial to see that every connected subspace is a connected subset. But how do i prove the converse? Or is it false?
Best Answer
If $A,B\subseteq X$ where $X$ is a topological space then the pair $\left\{ A,B\right\} $ is a separation if both sets are not empty and $\bar{A}\cap B=\emptyset=A\cap\bar{B}$.
If $\left\{ A,B\right\} $ is a separation, and $Y=A\cup B$ then $\left\{ A,B\right\} $ is a separation of $Y$ in $X$.
Then we have:
This affirmes that the definitions are equivalent.
In the following proof $\bar{A}^{Y}$ denotes the closure of $A$ as a subset of $Y$ and we have $\bar{A}^{Y}=Y\cap\bar{A}$ .
Proof: Let $\left\{ A,B\right\} $ be a separation of $Y$ in $X$. Then $\bar{A}^{Y}\cap B=Y\cap\bar{A}\cap B\subseteq\bar{A}\cap B=\emptyset$ and likewise $A\cap\bar{B}^{Y}=\emptyset$ showing that $\left\{ A,B\right\} $ be a separation of $Y$ in $Y$. Conversely let $\left\{ A,B\right\} $ be a separation of $Y$ in $Y$. Denoting the complement of $Y$ in $X$ by $Y^{c}$ we find: $\bar{A}\cap B=\left(Y\cap\bar{A}\cap B\right)\cup\left(Y^{c}\cap\bar{A}\cap B\right)=\left(\bar{A}^{Y}\cap B\right)\cup\emptyset=\emptyset$ and likewise $A\cap\bar{B}=\emptyset$.
Note: if $Y\subset Z\subset X$ then the subspace topology on $Y$ inherited from $X$ agrees with the subspace topology on $Y$ inherited from $Z$. So if $\left\{ A,B\right\} $ is a separation of $Y$ in $Z$ then above it has been shown that it is a separation of $Y$ in $X$ as well, and vice versa. So terms like 'in $X$', 'in $Z$' or 'in $Y$' become redundant here and can be omitted.