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Hi I just want to find out if the following is a acceptable proof for the proposition:

"Consider metric space $(X,d)$. If $A \subset X$ is connected and $A \subset B \subset \bar{A}$ then $B$ is connected."

Proof: Assume that $A \subset X$ is connected and $B$ is not connected. There exists disjoint open sets (in $(B,d)$) $V_{1}$ and $V_{2}$ such that $B = V_{1} \cup V_{2}$. Since $A \subset B$, it follows that $A \cap V_{1}$ and $A \cap V_{2}$ are open in $(A,d)$. Also then $A = (A \cap V_{1}) \cup (A \cap V_{2})$. Thus since $A $ is assumed connected, either $A \cap V_{1} = \emptyset$, $A \cap V_{2} = \emptyset$, $A \cap V_{1} = A$ or $A \cap V_{2} = A$.

W.L.O.G assume that $A \cap V_{1} = \emptyset$, then there exists an $x \in B \subset \bar{A}$ and $\epsilon > 0$ such that $B(x,\epsilon) \cap A = \emptyset$. This contradicts $x \in \bar{A}$. Assume now W.L.O.G that $A \cap V_{1} = A$, then $A \subset V_{1}$ and $A \cap V_{2} = \emptyset$. If we take $x \in V_{2} \subset B \subset \bar{A}$ then again we find $\epsilon > 0$ such that $B(x,\epsilon) \cap A =\emptyset$. This contradicts $x \in \bar{A}$. Therefore there cannot exist disjoint open sets $V_{1}$ and $V_{2}$ in $B$ such that $B = V_{1} \cup V_{2}$. Hence, $(B,d)$ cannot be disconnected and is therefore connected.

$\square$

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    $\begingroup$ What's $(B,d)$ and $(A,d)$? $\endgroup$ – MyUserIsThis Jun 17 '15 at 14:56
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    $\begingroup$ This holds for all toplogical spaces. $\endgroup$ – DanielWainfleet Jun 7 '17 at 20:00
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Proof: Assume that $A⊂X$ is connected and $B$ is not connected. There exists disjoint open sets (in $(B,d)$) $V_1$ and $V_2$ such that $B=V_1∪V_2 \dots$

I would add the word "non-empty" since you want to have a contradiction to the assumption that both sets are inhabited, and you later use that you can find an element in the "other" $V_i$.

... Thus since $A$ is assumed connected, either $A∩V_1=∅, A∩V_2=∅, A∩V_1=A$ or $A∩V_2=A$.

Here it looks as if you distinguish four cases. However, $A\cap V_1=\emptyset$ is equivalent to $A\cap V_2=A$ (and $A\subseteq V_2$). So you have only two cases, and of these it suffices to consider the case $A\cap V_1=\emptyset$, since the situation is completely symmetric in $V_1$ and $V_2$ (That's what is expressed by the W.L.O.G.).

Otherwise, the proof is fine. You take an $x\in V_2$ and conclude by the openness of $V_2$ that the relative neighborhood $B(\varepsilon,x)\cap B$ is contained in $V_2$ and thus disjoint from $A$, contradicting $x\in\overline A$. At least that's what it looks like, as you didn't include all the details.

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It's not harder to prove that this is true for any connected space:

Let $A\subset X$ be connected and suppose $A \subset B \subset \bar{A}$. If $U,V \neq \emptyset $ form a separation of $B$, then we have that $ U \cap \bar{V} = \bar{U} \cap V = \emptyset $.

Furthermore, either $A\subset U$ or $A\subset V$:

For if not then we have $U\cap A\neq \emptyset$ and $V\cap A\neq \emptyset$ open in $A$ and disjoint and so form a separation of $A$, which is impossible.

So, wlog $A\subset U$. Then $\bar {A} \subset \bar {U} $ and now since $V\cap \bar{ U }=\emptyset $, we have that $\bar {A}\cap V=\emptyset$, and therefore that $$ V \overset{(V \subset B)}{=} B\cap V=\emptyset\;, $$ a contradiction.

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  • $\begingroup$ In the given context, it might be more helpful to indicate how $U\cap \bar V = \bar U \cap V$ comes about; i.e. that this holds because: 1. Open sets contain no boundary points, and 2. $\partial V \overset{always}= \partial (X \setminus V) \overset{(X\setminus V = U)}{=} \partial{U}$ $\endgroup$ – polynomial_donut Jun 7 '17 at 18:52

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