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I'm trying to prove something here which isn't necessarily hard, but I believe it to be somewhat tricky. I've looked online for the proofs, but some of them don't seem 'strong' enough for me or that convincing. For example, they use the argument that since A is contained in $ \bar{B} $, then $ \bar{A} \subset \bar{B} $. That, or they use slightly altered definitions. These are the definitions that I'm using:

Definition #1: The closure of A is defined as the intersection of all closed sets containing A.

Definition #2: We say that a point x is a limit point of A if every neighborhood of x intersects A in some point other than x itself.

Theorem 1: $ \bar{A} = A \cup A' $, where A' = the set of all limit points of A.

Theorem 2: A point x $ \in \bar{A} $ iff every neighborhood of x intersects A.

Prove: If $ A \subset B,$ then $ \bar{A} \subset \bar{B} $

Proof: Let $ \bar{B} = \bigcap F $ where each F is a closed set containing B. By hypothesis, $ A \subset B $; hence, it follows that for each F $ \in \bar{B} $, $ A \subset F \subset \bar{B} $. Now that we have proven that $ A \subset \bar{B} $, we show A' is also contained in $\bar{B} $.

Let $ x \in A' $. By definition, every neighborhood of x intersects A at some point other than x itself. Since $ A \subset B $, every neighborhood of x also intersects B at some other point other than x itself. Then, $ x \in B \subset \bar{B} $.

Hence, $ A \cup A' \subset \bar{B}$. But, $ A \cup A' = \bar{A}$. Hence, $ \bar{A} \subset \bar{B}.$

Is this proof correct?

Be brutally honest, please. Critique as much as possible.

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    It looks fine to me. Your proof is correct. : )2012-03-17
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    Thank you! I learned from the information below that I could've written a simpler proof, but I just wanted to see if this was correct.2012-03-17
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    I thought so : )2012-03-17
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    I think there's a small error in the first paragraph of the proof. Instead of $F\in \overline{B}$, seems like you mean $F\subseteq \overline{B}$.2012-03-19
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    (Re previous comment): Perhaps I'm missing something, but if $F$ is one of the closed sets in the intersection $\cap F$, then $F\subset \overline{B}$ doesn't follow. And if $F$ is just an arbitrary subset of $\overline{B}$ then $A \subset F$ doesn't follow. I don't see how to fix this.2012-03-19

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