# If A is a subset of B, then the closure of A is contained in the closure of B.

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. : ) – Rudy the Reindeer Mar 17 '12 at 8:35
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. – Daavid M. Mar 17 '12 at 8:58
I thought so : ) – Rudy the Reindeer Mar 17 '12 at 9:03
I think there's a small error in the first paragraph of the proof. Instead of $F\in \overline{B}$, seems like you mean $F\subseteq \overline{B}$. – Patrick Mar 19 '12 at 7:54
(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. – Patrick Mar 19 '12 at 8:02

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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Ah, okay. When I read the proofs with one or two lines previously, I didn't know why/how they justified it so swiftly. It just seemed very awkward to me. I guess this is because I learned about the 2nd definition immediately after the first. – Daavid M. Mar 17 '12 at 8:55
@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 $\overline{A} \subseteq F$. – arjafi Mar 17 '12 at 11:35

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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I see, I see. When I saw other proofs use the definition, I thought something else had to be incorporated to make it 'strong.' – Daavid M. Mar 17 '12 at 8:51
If I handed in this proof to a professor, would this proof be reasonable? It's long and somewhat complicated, but is it correct? I apologize if I seem abrasive in asking this, but I'm trying to write up new ways to prove/write new proofs using other definitions. – Daavid M. Mar 17 '12 at 8:52
Also, I thank you for taking the time out to explain the concept to me. – Daavid M. Mar 17 '12 at 8:54
@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. – Austin Mohr Mar 17 '12 at 20:33

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, $\overline{A}\subseteq\overline{B}.$

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