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If $X$ is a topological space and $Y$ is a subspace of $X$, then is it true that any subset $Z$ of $Y$ which is connected in the topology of $X$, is also connected in the subspace topology of $Y$?

I believe it is not true.

Explanation: Consider $X$ to be the topological space $\mathbb{R}$ equipped with the cofinite topology. Let $Y=\{1,2,3\}$ and $Z=\{1,2\}$. Then the subspace topology on $Y$ is the discrete topology, as $\{1\}=\{2,3\}^c\cap Y$, and so on. Therefore, $Z$ is not connected in the subspace topology of $Y$. However $Z$ is connected in the topology of $X$, as any two open set in $X$ have non-empty intersection.

However, is my query true for metric spaces? If so, what property of a metric space ensures such a nicety?

Thanks in advance!

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I think that you misunderstood the definition of connectedness: A subset of a topological space $Z\subseteq X$ is disconnected if it can be divided into two nonenmpty disjoint open in $Z$ sets $A_1,A_2$.

Recall that a set $A_i\subseteq Z$ is open in $Z$ if there is an open in $X$ set $B_i$ such that $A_i=B_i\cap Z$.

It is not required for the definition of connectedness that also the sets $B_i$ are disjoint.

It is of course a valid question (though not directly related with the definition of connectedness) whether a space $X$ has the property that in such cases you can always choose even $B_i$ to be disjoint.

If you only consider closed subsets $Z$ it happens that this is the case if and only if the space satisfies the $T_4$ separation axiom.

For arbitrary subsets $Z$, it holds if (and probably only if) the space satisfies even the $T_5$ separation axiom.

Metric spaces satisfy all $T_i$ separation axioms.

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