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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\subset \overline{B} $, then $ \overline{A} \subset \overline{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: $ \overline{A} = A \cup A' $, where $A'$ = the set of all limit points of $A$.

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

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

Proof: Let $ \overline{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 \overline{B} $, $ A \subset F \subset \overline{B} $. Now that we have proven that $ A \subset \overline{B} $, we show $A'$ is also contained in $\overline{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 \overline{B} $.

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

Is this proof correct?

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

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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

5 Answers 5

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I think it's much simpler than that. By definition #1, the closure of A is a subset of any closed set containing A; and the closure of B is certainly a closed set containing A (because it contains B, which contains A). QED.

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Using Definition #1 makes it quite easy. For each $A \subseteq X$, let $\mathcal{C}_A = \{ F \subseteq X : F\text{ is closed and }A \subseteq F \}$. Then by Definition #1 it follows that $\overline{A} = \bigcap \mathcal{C}_A$.

Note that if $A \subseteq B$, then $\mathcal{C}_B \subseteq \mathcal{C}_A$, and therefore $\bigcap \mathcal{C}_A \subseteq \bigcap \mathcal{C}_B$.

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    @DaavidM.: Yeah, this is a somewhat unnatural way of looking at the closure. The notion of limit points make the closure something that can be visualised, whereas the formal definition seems a bit odd. David Wallace's proof above is very nice, and the same idea is used quite often: If $A \subseteq F$ and $F$ is closed, then $\overline{A} \subseteq F$.2012-03-17
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I think it's simplest to see from the first definition.

Let $\mathcal{A}$ be the collection of closed sets containing $A$ and $\mathcal{B}$ the collection of closed sets containing $B$. Since $A \subset B$, we know $\mathcal{B} \subset \mathcal{A}$, and so $\bigcap \mathcal{A} \subset \bigcap \mathcal{B}$ (i.e. $\overline{A} \subset \overline{B}$).

Loosely speaking, adding more sets to an intersection can only make it smaller.

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    @DaavidM. Your proof is a good one given the approach you've taken (decomposing the closure into $A$ and $A^\prime$). I just wanted to show that there is a more direct approach right from the definition using a little knowledge about intersections. (In fact, David Wallace's proof is even more direct.) Your primary concern in writing proofs should be to convey your idea to your intended audience as clearly as possible. Don't try to incorporate theorems solely for the sake of gravitas.2012-03-17
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You say that some of the proofs you have looked use the argument "that since $A$ is contained in $\overline{B}$, then $\overline{A}\subseteq\overline{B}$" and that they don't seem strong enough for you but this follows directly from definition #1. Any closed subset containing $B$ contains $A$ and consequently $A\subseteq \overline{B}$. Since $\overline{B}$ is closed, $\overline{A}\subseteq\overline{B}.$