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I need to prove the following:

Let $A$ and $B$ be subsets of a metric space $(X, d)$ show that

  1. $A^o \cup B^o \subset (A \cup B)^o$.
  2. $(A \cap B)^o = A^o \cap B^o$

Here is my attempt:

For 1)

Let $x\in (A^o \cup B^o)$. Then by definition of open set and union $x\in A$ such that the open ball $B(x, r_1) \subset A$ for some $r_1>0$ OR $x\in B$ such that the open ball $B(x, r_2) \subset B$ for some $r_2>0$.

Check whether $x\in (A\cup B)^o$ which is equivalent to checking $x\in A\cup B$ such that the ball $B(x, r) \subset A\cup B$ for some $r >0$. How do I conclude this?

For 2) Any ideas?

Asaf Karagila
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CodeKingPlusPlus
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1 Answers1

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You're nearly there. Note that if $z\in C$ such that $C\subseteq A$ or $C\subseteq B$, then $z\in A$ or $z\in B$, meaning $z\in A\cup B.$ Thus, in either case, we've got an open ball around $x$ contained in $A\cup B$

For (2), you can proceed similarly, showing that $(A\cap B)^o\subseteq A^o\cap B^o$ and $(A\cap B)^o\supseteq A^o\cap B^o$ through element-chasing. On the one hand, if $x\in (A\cap B)^o$, then there is an open ball about $x$ contained in $A\cap B$, which means.... On the other hand, if $x\in A^o\cap B^o$, then there is an open ball about $x$ contained in $A$ and another open ball about $x$ contained in $B$. One of those open balls must be contained in the other (why?), so we then have an open ball about $x$ contained in both $A$ and $B$, which means....

Cameron Buie
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