# Let $(X,d)$ be metric space, let $A$ be open, then $\partial(A\cup A')\subseteq A'$.

Is this statement true? I tried to prove by myself, but not sure if it is error-free. We define $$b_r(x)$$ be the open ball of radius $$r$$ around $$x$$ and define $$b^*_r(x):=\{y\in X: d(x,y) be the punched open ball of radius $$r$$ around $$x$$. We define the closure of $$A$$ to be $$\overline{A}$$, the complement to be $$A^c$$, and the boundary of $$A$$ to be $$\partial A$$.

Remark: Thanks for a comment to point out my error in the original statement. In particular, all I need is $$\partial(A\cup A')\subseteq A'$$, whereas before I claimed that $$\partial(A\cup A')=A'$$, which failed miserably when we take $$A=X$$.

Proof:

Let $$x\in \partial(A\cup A')$$, then $$\forall r>0, b_r(x)\cap \overline{A}\neq \emptyset\wedge b_r(x)\cap (\overline{A})^c\neq \emptyset$$

In particular, we have $$\forall r>0, b_r(x)\cap \overline{A}\neq \emptyset$$

Now, suppose $$\exists r_0>0$$ so that $$b_{r_0}^*(x)\cap \overline{A}=\emptyset$$. This would imply $$x$$ is the only point intersect $$\overline{A}$$ in $$b_{r_0}(x)$$. However, recall that $$B:=\overline{A}=A\cup A'$$ so $$x\in A$$ or $$x\in A'$$. It is impossible for $$x\in A$$ as that would imply $$\exists r_1>0, b_{r_1}(x)\subseteq A\subseteq B$$ and so $$\emptyset\neq b_{\min\{r_0,r_1\}}(x)\cap B\subseteq b_{r_0}(x)\cap B$$. Thus $$x\in A'$$, but then there exists a sequence $$\{a_n\}$$ in $$A$$ so that $$\lim a_n\to x$$, i.e. $$\exists k\in \mathbb{N}, d(a_k,x) so that $$a_k\in A, a_k\neq x$$ and so $$\emptyset\neq b^*_{r_0}(x)\cap U\subseteq b^*_{r_0}(x)\cap B$$. Either way, we would have contradiction. Thus, we have $$\forall r>0, b_r^*(x)\cap B\neq \emptyset$$

We also note this is the same as $$\forall r>0, b_r^*(x)\cap (A\cup A')=(b_{r}^*(x)\cap A)\cup (b_{r}^*(x)\cap A')\neq \emptyset$$

Let $$x\neq y_n\in A$$ be so that $$y_n\in (b_{1/n}^*(x)\cap A)\cup (b_{1/n}^*(x)\cap A')$$ If for all $$n\in \mathbb{N}$$, $$y_n\in (b_{1/n}(x)^*\cap A)$$, then $$\{y_n\}$$ is a sequence in $$A$$ converges to $$x$$ and $$y_i\neq x$$ for all $$i\in \mathbb{N}$$ so that $$x\in A'$$. If for all $$n\in \mathbb{N}$$, $$y_n\in b_{1/n}(x)^*\cap A'$$, then $$\{y_n\}$$ is a sequence in $$A'$$ converges to $$x$$ such that $$y_n\neq x$$, so that $$x\in (A')'\subseteq A'$$. Finally, suppose we have infinite many $$y_i\in A$$ and infinite many $$y_j\in A'$$, then since this is clearly a Cauchy sequence (the distance between $$y_n$$ and $$x$$ is getting smaller as $$n\to \infty$$, so is the distance between $$y_i$$ and $$y_j$$ by triangle inequality), and the subsequence $$\{y_{n_i}\}_{i=1}^\infty$$ where $$y_{n_i}\in A$$ converges to $$x$$, we have the entire sequence converges to $$x$$, i.e. $$x\in A'$$.

• @Bungo hmmm, wait, just realized my proof only showed one direction, and even if we take $A=X$ we still have $\partial(X\cup X')=\emptyset\subseteq X'=X$, right? – Hyacinth Sep 15 at 3:30
• @Bungo Thanks, that's a way better argument than mine. – Hyacinth Sep 15 at 3:43
• @Bungo It's not true that "if $A$ is open, every point of $A$ is a limit point of $A$" in general metric spaces: let $X=[0,1]\cup \{2\}$ (induced metric from the reals) and $A=\{2\}$ or $X=\mathbb{Z}$ in the discrete metric and $A=X$. Metric spaces can have isolated points... – Henno Brandsma Sep 15 at 5:39
In any space $$\overline{A} = A \cup A'$$ and for an open $$A$$, $$\partial A = \overline{A}\setminus A = (A \cup A')\setminus A \subseteq A'$$