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Soppose $(X,T)$ is a topological space. Let $Y\subset S\subset X$. Is $Y$ connected in $S$ iff it is connected in $X$.

Possible Proof:

Since $Y$ is connected in S the only closed and open sets in $Y$ under the subspace topology of Y are $Y$ and $\emptyset$. Since the subspace topology is the same under X, Y is connected in X.

Is this proof valid?

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    $\begingroup$ I think it would be best to expand the definition of the subspace topology and relate to the topology of X. $\endgroup$ – BenB Mar 16 at 5:37
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You have the right idea, but it could use elucidation, unless you've got other results to which you can refer.

It seems like you're trying to show that, if $A$ is a clopen subset of $Y$ (as a subspace of $X$) then $A=\emptyset$ or $A=Y,$ and you want to use the fact that this is true when considering $Y$ as a subspace of $S.$ That's a nice idea! Here's how I might go about it.


Suppose that $A$ is a subset of $Y$, and that $A$ is clopen in $Y$ (as a subspace of $X$). We show that $A=\emptyset$ or $A=Y.$

Since $A$ is open in $Y$ as a subspace of $X,$ then there is some $U\in\mathcal T$ such that $A=U\cap Y.$ Since $A$ is closed in $Y$ as a subspace of $X,$ then there is some $V\in\mathcal T$ such that $Y\setminus A=V\cap Y.$

Now, since $A\subseteq Y\subseteq S,$ then $A$ is also a subset of $S,$ and so $$A=U\cap Y=U\cap(S\cap Y)=(U\cap S)\cap Y,$$ so $A$ is open in $Y$ as a subspace of $S.$ We similarly have $Y\setminus A=(V\cap S)\cap Y,$ so $A$ is closed in $Y$ as a subspace of $S.$ Since $Y$ is connected as a subspace of $S,$ then $A=\emptyset$ or $A=Y,$ as desired.

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