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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?

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  • $\begingroup$ For your first definition, do you want $A\cup B\supseteq Y$? If not, then $A$ and $B$ would have to be subsets of $Y$ and the two definitions are equivalent. $\endgroup$ – Joe Johnson 126 Oct 14 '13 at 13:16
  • $\begingroup$ A connected subset, considered as equipped with the subspace topology, is a connected (sub)space. So there's essentially no difference. $\endgroup$ – user43208 Oct 14 '13 at 13:17
  • $\begingroup$ @User43208: yes, it is true there is no difference. But Jj asks a specific question about the reason for this non-difference. $\endgroup$ – GEdgar Oct 14 '13 at 13:21
  • $\begingroup$ Yes, I know @GEdgar. I'm going to let someone else answer this question. You care to do the honors? :-) $\endgroup$ – user43208 Oct 14 '13 at 13:23
  • $\begingroup$ The interesting question would be how that relates to: There are open subsets $A$ and $B$ of $X$ such that $A\cap Y\ne\emptyset$, $B\cap Y\ne\emptyset$ and $A\cup B\supset Y$. $\endgroup$ – Carsten S Oct 14 '13 at 14:18
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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:

$\left\{ A,B\right\} $ is a separation of $Y$ in $X$ if and only if $\left\{ A,B\right\} $ is a separation of $Y$ in $Y$ .

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.

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