1
$\begingroup$

The following question is from Fred H. Croom's book "Principles of Topology"

Let $A, B$ be subsets of a metric space with $A \subset B$. Prove that:

  1. int $A$ $\subset$ int $B$;
  2. $\bar{A} \subset \bar{B}$;
  3. $A^{'}\subset B^{'}$.

The following are my attempts at proving the above statements.

Part 1. By definition, int$A$ is the set of all interior points of $A$. Let $x$ be such an interior point. Thus $A$ is a neighborhood of $x$ provided there is an open set $O$ which contains $x$ and is contained in $A$. Since $x \in O \subset A$, then $x \in O \subset B.$ Therefore int$A$ $\subset$ int$B$.

Part 2. Let $x\in \bar{A}$. Then every open set containing $x$ contains a point of $A$ by definition of the closure of a set $A$. Since $A \subset B$, then every open set containing $x$ contains a point of $B$. Thus $x\in\bar{B}$. Therefore $\bar{A} \subset \bar{B}$

Part 3. Let $x\in A^{'}$ Thus $x$ is a limit point $A$. This means every open set containing $x$ contains points of $A$ distinct from $x$. Since $A\subset B$, we say every point in $A$ is in $B$. Thus $x$ is a limit point of $B$. Therefore $A^{'}\subset B^{'}$.

Am I on the right track? Any suggestions or feedback?


I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.

$\endgroup$
  • 1
    $\begingroup$ It's fine for me. For the interior, an alternative proof would be to say that, since $\operatorname{int} A$ is the largest open set contained in $A$, and $A\subset B$, it is an open set contained in $B$, hence in the largest open set contained in $B$. $\endgroup$ – Bernard Feb 17 '15 at 19:15
1
$\begingroup$

All proofs seem correct and clear to me, well done!

$\endgroup$
  • 1
    $\begingroup$ I'm sorry, but a for a limit point $x$, the neighbourhoogs must contain a point of $A$ distinct from $x$. An isolated point of $A$ is not a limit point. $\endgroup$ – Bernard Feb 17 '15 at 19:26
  • $\begingroup$ @Bernard You are right! Thank you for pointing that out. My apologies to OP. $\endgroup$ – Uncountable Feb 17 '15 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.