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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.

311411
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Isaac Bullock
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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.)

311411
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  • 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? – Isaac Bullock May 24 '21 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 $p \in \overline{B}$. – 311411 May 24 '21 at 23:04
  • This is perfect, thank you! – Isaac Bullock May 24 '21 at 23:05
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    You're welcome. One of the reasons that your definition is a nice one is that this proof becomes a quickie. – 311411 May 24 '21 at 23:06
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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.

Vizag
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  • 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. – Isaac Bullock May 24 '21 at 22:55
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    Then how has 'closure' been defined for you? @IsaacBullock – Graham Kemp May 24 '21 at 22:57
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    @GrahamKemp It's explicitly defined above as the set of all closure points. – Isaac Bullock May 24 '21 at 22:57
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    @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 '21 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...) – Graham Kemp May 24 '21 at 23:38