# Show that $A \subset B \implies \overline{A} \subset \overline{B}$

NOTE: I know that a question asking for help to prove this same property already exists, but I would like an answer specifically based on the definition(s) and / or remark below, please.

Definition 1: A point $$x \in \mathbb{R}$$ is a point of closure of a set $$E \subset \mathbb{R}$$ if
$$\quad \forall \ \delta>0,\; \ \exists \ y \in E \ \;$$ s.t. $$\ |x-y| < \delta$$.
Equivalently, $$x$$ is a point of closure of $$E$$ if every open interval containing $$x$$ also contains a point of $$E.$$
We call the set of all points of E the closure of $$E$$ and denote it by $$\overline{E}.$$

Remark: Every point in $$E$$ belongs to its closure. Particularly, $$E \subset \overline{E}$$.

Definition 2: $$E$$ is closed if $$E=\overline{E}$$.

Question: Show that $$A \subset B \implies \overline{A} \subset \overline{B}$$.

Attempt: $$A \subset B \implies A \subset B \subset \overline{B} \implies A \subset \overline{B}$$. If $$A$$ is closed then $$A= \overline{A} \implies \overline{A} \subset \overline{B}.$$

So I can do this when $$A$$ is closed but I'm not certain on how to use either of the definitions / remark to show that it holds when $$A$$ is open.

Let $$p$$ belong to the closure of $$A$$. Then every interval $$I$$ including $$p$$ has some $$a \in A \cap I.$$ But then $$a \in B \cap I.$$ That makes $$p$$ a closure point for $$B$$, by your definitions. (We did not use definition 2 here.)

• And since $p$ is a closure point of $B$, it follows that all closure points of $A$ are closure points of $B$ and hence the result? May 24, 2021 at 23:02
• Yeah, that's right. To show $S$ is a subset of $T$ (for any sets), we can use the "pick a point" method. So we've showed that $p \in \overline{A}$ implies $p \in \overline{B}$. May 24, 2021 at 23:04
• This is perfect, thank you! May 24, 2021 at 23:05
• You're welcome. One of the reasons that your definition is a nice one is that this proof becomes a quickie. May 24, 2021 at 23:06

Let $$x \in \overline{A}$$. Then $$x \in A \cup A'$$ where $$A'$$ is the set of limit points of A. If $$x\in A \subset B$$, then $$x \in B \subset \overline B$$. So let $$x \in A'$$. Then for any $$\epsilon$$ nbd of $$x$$, there exists $$y \neq x$$ such that $$y \in A$$. But since $$A \subset B$$, $$y \in B$$. Hence, $$x \in B' \subset \overline B$$. This completes the proof.

• Hi, @Vizag thank you for your help but as I said I would like an explanation based strictly on the information above... I don't know "officially" know what limit points are. I, of course, have searched it up and it seems they later come up in the module but as we haven't learnt it about them yet, I cannot use them in my proof. Hope that makes sense. May 24, 2021 at 22:55
• Then how has 'closure' been defined for you? @IsaacBullock May 24, 2021 at 22:57
• @GrahamKemp It's explicitly defined above as the set of all closure points. May 24, 2021 at 22:57
• @GrahamKemp I think Isaac provided that in the question. Limit points are not the only way to provide a definition of the closure.
– B2K
May 24, 2021 at 22:59
• You can adjust Vizags logic to show every point of closure for $A$ is also a point of closure for $B$. (Any point $x$ is a point of closure of $A$ if every open interval containing $x$ also contains at least a point in $A$; and since $A\subset B$, therefore...) May 24, 2021 at 23:38