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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 $b^*_r(x):=\{y\in X: d(x,y)<r\wedge x\neq y\}$ 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. $\exists k\in \mathbb{N}, d(a_k,x)<r_0$ 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. $x\in A'$.

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  • $\begingroup$ @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? $\endgroup$ – Hyacinth Sep 15 at 3:30
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    $\begingroup$ @Bungo Thanks, that's a way better argument than mine. $\endgroup$ – Hyacinth Sep 15 at 3:43
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    $\begingroup$ Then I think this question is closed, thanks a lot. $\endgroup$ – Hyacinth Sep 15 at 3:56
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    $\begingroup$ @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... $\endgroup$ – Henno Brandsma Sep 15 at 5:39
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In any space $\overline{A} = A \cup A'$ and for an open $A$, $$\partial A = \overline{A}\setminus A = (A \cup A')\setminus A \subseteq A'$$

Don't overcomplicate things...

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