2
$\begingroup$

$A'$ denotes the set of limit points of $A$ and $\partial A$ is the boundary of $A$. $A^{o}$ denotes the set of interior points of $A$.

Show that $\overline{A} \subseteq A^{o} \cup \partial A$.

Suppose $x \in \overline{A} \implies x \in A \cup A'$.

Suppose $x \not\in A$ (What if $x\in A$), then $x\in A'$.

$\implies$ for all open sets $U$ containing $x$, $U$ has another point in $A$ not equal to $x$.

$\implies U \cap A \neq \varnothing$ and $U \cap (X\setminus A) \neq \varnothing$.

$\implies x \in \partial A$.

This proof is not complete; what if $x\in A$? Can someone help me out?

$\endgroup$
  • $\begingroup$ Use this: $x \in \overline{A}$ if and only if every nbhd of $x$ contains a point of $A$. Now, if $x$ is not in the interior, then... $\endgroup$ – Randall Oct 23 '18 at 3:05
  • $\begingroup$ then what follows? $\endgroup$ – Niang Moore Oct 23 '18 at 3:46
  • $\begingroup$ Limit points are making this an awkward proof. Why is it so recklessly important to teach that limit point thing. It is rarely needed in topology. $\endgroup$ – William Elliot Oct 23 '18 at 4:10
  • 1
    $\begingroup$ What definition of the boundary of A are you using? If you use a common definition, $\partial A$ = $\overline{A}$/$A^{o}$, the proof practically writes itself. $\endgroup$ – Steve B Oct 23 '18 at 5:02
  • $\begingroup$ This looks like a problem you have collected from / inspired by some source. According to recent discussions in Meta, we are looking forward to including sources for all applicable questions. Can you provide the source by editing the question?Refer-math.meta.stackexchange.com/questions/29290/… $\endgroup$ – tatan Oct 23 '18 at 6:16
1
$\begingroup$

Let $A^c$ denote the complement of $A.$

The definition of $\partial A$ is $\bar A \cap \overline {A^c}.$

If $p\in \bar A$ then....

(i) If there exists an open set $U$ such that $p\in U$ and $U\cap A^c=\emptyset$ then $U\subset A$ so $p\in U\subset A^o$ so $p\in A^o$.

(ii) If every open $U$ such that $p\in U$ has non-empty intersection with $A^c$ then $p\in \overline {A^c}$ so $p\in \bar A\cap \bar {A^c}=\partial A.$

In both cases we have $p\in A^o\cup \partial A.$

Therefore $\bar A\subset A^o\cup \partial A.$

BTW, since $$A^o\cup \partial A\subset A\cup \partial A=A\cup (\bar A \cap \overline {A^c})\subset$$ $$\subset A\cup \bar A=\bar A$$ we also have $A^o\cup \partial A \subset \bar A.$ So we have $\bar A=A^o\cup \partial A.$

All of this holds in every topological space.

$\endgroup$
  • $\begingroup$ This Q, or variant relatives of it, keep appearing on this site. The same can be said for many other Q's. Not surprising, as new students keep appearing also. Some day someone might take this site and produce a fully cross-indexed compendium of such Q's & A's. $\endgroup$ – DanielWainfleet Oct 25 '18 at 3:34
2
$\begingroup$

$\partial A = \bar A - A^o.$ Thus
$A^o \cup \partial A = \bar A$ because
$A^o \subseteq \bar A.$

In fact, because $A^o \subseteq A \subseteq \bar A,$
$A \cup \partial A = \bar A.$

$\endgroup$
1
$\begingroup$

If $x \in A \cup A'$ then suppose $x \notin A^\circ$, then for all open neighbourhoods $B$ (or balls around $x$ if you prefer) of $x$ we have that $B \nsubseteq A$, so for all such $B$: $B \cap (X\setminus A) \neq \emptyset$. Both $x \in A$ and $x \in A'$ imply that also for all such $B$: $B \cap A \neq \emptyset$. So combined this means that $x \in \partial A$: all its neighbourhoods/balls intersect $A$ and its complement.

This shows $\overline{A} \subseteq A^\circ \cup \partial A$.

$\endgroup$
  • $\begingroup$ what if x is not a boundary point of A, can you show it's the interior point of A $\endgroup$ – Niang Moore Oct 24 '18 at 1:43
  • $\begingroup$ @NiangMoore you can yes, but I found this way more convenient: not a boundary point gives two cases to consider and not in the interior is easier. $\endgroup$ – Henno Brandsma Oct 24 '18 at 4:30

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.