# Connected subset of subset is connected in whole space.

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?

• I think it would be best to expand the definition of the subspace topology and relate to the topology of X. – BenB Mar 16 at 5:37

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.