# Prove that connectedness is not an extrinsic property.

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!

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