Statement: $A\subset B \subset X$ with subset $B$ equipped with subspace topology then "$A$ is connected in $B$ if and only if $A$ is connected in $X$"
I am not able to find counter example of above statement! Is it true? I know that if word connected is replaced by compact then it is true!! But, for connected is it true? Please help me..
On
Yes, it is true. That's because the subspace topology induced on $A$ by the topology from $B$ is the same as the subspace topology on $A$ induced by the topology on $X$. And this is trua because, if $A^*\subset A$, then asserting that $A^*$ is open with respect to the subspace topology induced on $A$ by the topology from $B$ means that $A^*=B\cap Z$ where $Z$ is an open subset of $B$. And asserting that $A^*$ is open with respect to the subspace topology induced on $A$ by the topology from $X$ means that $A^*=B\cap W$ where $W$ is an open subset of $X$. But these assertions are equivalent: in order to pass from the second one to the first one, just take $Z=B\cap W$. And in order to pass from the first one to the second one, take an open subset $W$ of $X$ such that $Z=B\cap W.$
It holds as a consequence of the fact that the topology $A$ inherits as a subspace of $B$ is the same as the topology $A$ inherits as a subspace of $X$.
Namely, if $U \subseteq A$ is open relatively to $B$, then there exists $V \subseteq B$ such that $U = A \cap V$. $V$ is open in $B$ so there exists $W \subseteq X$ open in $X$ such that $V = W \cap B$.
We have $$U = A \cap V = A \cap (W \cap B) = A \cap W$$
so $U$ is open relatively to $X$. The converse is similar.