Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The closure of $A$ can be equal to $\operatorname{int}(A)\displaystyle\cup\operatorname{bdry}(A)$. Another definition is that the closure is the set of limit points of $A$.

How are these 2 definitions equivalent?

share|improve this question
10  
whathaveyoutried.com –  lhf Jan 23 '12 at 10:28
1  
It's a pretty trivial proof. Just have the right out the definitions and do it. Like we don't know what you have done or even the definitions you are using. –  simplicity Jan 23 '12 at 10:47
1  
$bdry(A)$ is the set of limit points of $A$ that are also limit points of $X\backslash A$. Then you need to show that a boundry point is an element of $\bar{A}$ that cannot be an interior point. Hence from $bdry(A) = \bar{A}\backslash int{A}$, the result follows. –  Henry Shearman Jan 23 '12 at 10:51
1  
The closure of a set is not the set of limit points. For example $\mathbb Z$ has no limit points in $\mathbb R$ (each $n \in \mathbb Z$ has a neighbourhood, $\lbrace n\rbrace$, which contains no other element of $\mathbb Z$, so $\mathbb Z' = \emptyset$), but it is still closed. –  kahen Jan 23 '12 at 11:05
1  
For me, by definition, $Bdary(A)=clo(A)\setminus int(A)$ ... –  Student Jan 23 '12 at 11:20
show 4 more comments

3 Answers

up vote 1 down vote accepted

The closure $\overline{A}$ of $A \subset X$ is the set of so-called adherence points of $A$: all $x$ in $X$ such that every neighborhood of $x$ intersects $A$ (for me, a limit point of $A$ is an $x$ such that every neighborhood of $x$ intersects $A \setminus {x}$).

If $x$ is such an adherence point, so every neighborhood of $x$ intersects $A$, there are two mutually exclusive options: either some neighborhood of $x$ sits completely inside $A$, and then it is in the interior of $A$ by definition, or every neighborhood of $x$ intersects both $A$ and its complement, and this by definition makes $x$ a point of the boundary of $A$. And clearly both types of points (interior and boundary) are by definition adherence points, so we have the equality that $\overline{A} = bdry(A) \cup int(A)$, as claimed.

share|improve this answer
add comment

We call $a$ a limit point of $A$ if every neighbourhood of $a$ contains a point of $A$ different from $a$.

We call $a$ a point of closure of $A$ if every neighbourhood of $a$ contains at least one point of $A$.

Let $C$ denote the set of points of closure of $A$. Then you want to show that $\overline{A} = C = \operatorname{int}{(A)} \cup \operatorname{bdry}{(A)}$

$C \subset \operatorname{int}{(A)} \cup \operatorname{bdry}{(A)}$:

Let $a \in C$, i.e. let $a$ be a point of closure of $A$. By the definition of point of closure we know that every neighbourhood of $a$ contains a point $a^\prime \in A$. Assume that $a$ is neither in the boundary nor in the interior of $A$. Then there is an open set $O$ such that $a \in O \subset A^c$. But this would be a contradiction because $O \cap A = \emptyset$ hence there is no $a^\prime \in O$. Hence $a$ has to be either in $\operatorname{int}{(A)}$ or $\operatorname{bdry}{(A)}$.

$\operatorname{int}{(A)} \cup \operatorname{bdry}{(A)} \subset C$:

Let $a \in \operatorname{int}{(A)} \cup \operatorname{bdry}{(A)}$. Then either $a \in \operatorname{int}{(A)}$ or $a \in \operatorname{bdry}{(A)}$. If $a \in \operatorname{int}{(A)}$ then every neighbourhood of $a \in \operatorname{int}{(A)} \subset A$ certainly contains $a$ hence $a$ is a point of closure, i.e. $a \in C$. If $a$ is in the boundary of $A$ then by the definition of boundary, every neighbourhood of $a$ has non-empty intersection with both $A$ and $\mathbb{R} \setminus A$ hence again $a$ is in $C$.

As kahen pointed out in the comments, if you use the common definition of limit point you can well have sets that have points that aren't limit points in the usual sense but are in the closure of the set.

Hope this helps.

share|improve this answer
    
"sequentially closed" is not equivalent to "close"... –  Student Jan 23 '12 at 11:23
    
@Student: Since the answer doesn’t mention or use sequential closure, I fail to see the point of your comment. –  Brian M. Scott Jan 23 '12 at 11:28
2  
The OP won’t be able to prove the other direction: as kahen pointed out in the comments, it’s not true for the usual definition of limit point. To make it true, you need to say that $x$ is a limit point of $A$ iff every nbhd of $x$ meets $A$, not necessarily in a point different from $x$. –  Brian M. Scott Jan 23 '12 at 11:31
    
@BrianM.Scott Thanks Brian, I had not thought it through enough. I changed the definition of limit point. –  Matt N. Jan 23 '12 at 11:41
add comment

Generally speaking if $A$ is a subset of a topological space, and $x\in X$, exactly one of three things can occur:

  1. Every neighborhood $U$ of $x$ meets (intersects novoidly) $A$ and $A^c$ Such points form the boundary of $A$.

  2. $A$ is a neigbhorhood of $x$. In this case, we have $x\in A$. Such points form the interior of $A$

  3. $A^c$ is a neighboorhood of x. Such points form the interior of the complement of $A$.

The point $x$ is a limit point of $A$ if for every neighborhood $U$ of $x$, $$U\cap A - \{x\} \not= \emptyset.$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.