Let $A$ and $B$ be subsets of a metric space $(M,d)$. If $A\subseteq B$, then $A^{\circ} \subseteq B^{\circ}$.

Proof : Assume that $a\in A^{\circ}$. Then there exists a $r>0$ such that $K(a,r)\subseteq A$. Since $A\subseteq B$ it shows that $K(a,r)\subseteq B$, then $a\in B^{\circ}$.

I wondered if it's possible to write it differently, but I am not sure if it's correct, \begin{align} A^{\circ}&=\left \{x\in M \mid \exists r>0: K(x,r)\subseteq A \right \}\\ &\subseteq \left \{x\in M \mid \exists r>0: K(x,r)\subseteq B \right \}\qquad (\textrm{since } A\subseteq B)\\ &=B^{\circ}. \end{align}

  • 3
    $\begingroup$ That's not a different way! That's the same proof, just written differently. $\endgroup$ – David C. Ullrich Jul 21 '15 at 18:44
  • $\begingroup$ @DavidC.Ullrich Yes, that's what I meant. I've edited it. $\endgroup$ – UnknownW Jul 21 '15 at 18:45
  • $\begingroup$ Your first proof is more rigorously correct. $\endgroup$ – Omnomnomnom Jul 21 '15 at 18:46
  • $\begingroup$ This is actually the same proof. In the first proof you stop before concluding that $A^° \subset B^°$ because $\forall x(x \in A^° \Longrightarrow x \in B^°)$, and in the second, you don't justify why $\{x \ | \ \exists r > 0: K(x,r) \subset A\} \subset \{x \ | \ \exists r > 0: K(x,r) \subset B\}$. $\endgroup$ – nombre Jul 21 '15 at 18:48
  • $\begingroup$ @nombre I thought it's better left it unmentioned since $A$ is assumed to be a subset of $B$, or am I mistaken? $\endgroup$ – UnknownW Jul 21 '15 at 18:53

That is correct, but it's not really 'another way'. What you've done is taken the first proof and expressed it in set-theoretic language. This is often a useful thing to do, though the first proof has the advantage that everything is explained in words rather than with symbols.

  • $\begingroup$ There are some proofs explained in words that are not easy to understand. That's why I mostly prefer something that is explained with symbols. $\endgroup$ – UnknownW Jul 21 '15 at 18:57

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