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Let $(S, d)$ be a metric space and $E \subset S$. Let $T$ be another subset of $S$ that contains $E$. Then $(T, d)$ is clearly a metric space. Show that $E$ is connected in $S$ iff $E$ is connected in $T$.

My attempt:

$\Rightarrow$ Suppose that $E$ is connected in $S$, then there is no nontrivial decomposition $E_1, \ E_2 \subset S$ satisfies the following

$\overline{E_1} \cap E_2=\emptyset$ and $\overline{E_2} \cap E_1=\emptyset$ ... (*)

Therefore, there is no nontrivial decomposition in $T$ satisfies (*). As a result, $E$ is connected in $T$.

$\Leftarrow$ I will prove its contrapositive. Suppose that $E$ is disconnected in $S$. This means that there is nontrivial decomposition $E_1, \ E_2 \subset S$ satisfies (*).

Claim: $E$ is disconnected in $T$ and its nontrivial decomposition in $T$ is $E_1 \cap T$ and $E_2 \cap T$.

  1. $E_1 \cap T\neq \emptyset$:

Suppose by contradiction that $E_1 \cap T=\emptyset$, then $E_1 \subset T$. But we know that $E=E_1\cup E_2 \subset E$, so $E_1 \subset T$, which is a contradiction. Hence $E_1 \cap T\neq \emptyset$. Similarly, $E_2 \cap T\neq \emptyset$.

  1. $(E_1\cap T)\cap (E_2\cap T)=(E_1\cap E_2)\cap T=\emptyset$.

  2. $(E_1\cap T)\cup(E_2\cap T)=(E_1\cup E_2)\cap T=E \cap T=E$, since $E\subset T$.

This means $E_1 \cap T$ and $E_2 \cap T$ comprise a nontrivial decomposition of $E$.

  1. $\overline{E}_T$ denote the closure of $E$ in $T$. We have $\overline{E}_T=\overline{E}\cap T$, so

$(\overline{E_1\cap T})_T\cap (E_2 \cap T)=(\overline{E_1}\cap T)\cap (E_2 \cap T)=(\overline{E_1} \cap E_2)\cap T=\emptyset \cap T=\emptyset$

Similarly, $(\overline{E_2\cap T})_T\cap (E_1 \cap T)=\emptyset$

Therefore, $E$ is disconnected in $T$.

As a result, $E$ is connected in $S$ iff $E$ is connected in $T$.

Is my answer correct? Any help would be appreciated. Thanks!

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Your proof is fine, but the $(\Longrightarrow)$ direction can be cleaned up a bit.

  1. $E_1 \cap T\neq \emptyset$:

You don't need a proof by contradiction here. Note that $E_1 \subset E \subset T$, and so $E_1 \cap T = E_1 \neq \emptyset$.

  1. $(E_1\cap T)\cap (E_2\cap T)=(E_1\cap E_2)\cap T=\emptyset$.

This is implied by (4), so it's not really needed. The rest of the proof looks fine.

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