# how to prove the boundary of a set is closed?

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• November 17th 2009, 06:19 PM
p00ndawg
how to prove the boundary of a set is closed?
i know it is the definition of a boundary, but im supposed to prove it somehow and I really dont have a clue for this one.

also, ive been looking for proofs using google for this, but most if not all use the closure of the set to prove it.

however, in my class we havent gone over the closure of sets, so im assuming I have to prove the bd is closed without using the closure.

is this possible?
• November 17th 2009, 07:29 PM
aliceinwonderland
Quote:

Originally Posted by p00ndawg
i know it is the definition of a boundary, but im supposed to prove it somehow and I really dont have a clue for this one.

also, ive been looking for proofs using google for this, but most if not all use the closure of the set to prove it.

however, in my class we havent gone over the closure of sets, so im assuming I have to prove the bd is closed without using the closure.

is this possible?

Let A be a subset of a metric (or topology) space X. A point x in the metric (or topology) space X is a boundary point of A provided that x belongs to $(\overline{A}) \cap (\overline{X \setminus A})$.

An intersection of closed set is closed, so bdA is closed.

EDIT: plz ignore this post. I didnt notice that using a closure is not allowed in your question.
• November 17th 2009, 07:33 PM
Drexel28
Quote:

Originally Posted by p00ndawg
i know it is the definition of a boundary, but im supposed to prove it somehow and I really dont have a clue for this one.

also, ive been looking for proofs using google for this, but most if not all use the closure of the set to prove it.

however, in my class we havent gone over the closure of sets, so im assuming I have to prove the bd is closed without using the closure.

is this possible?

I assume for my convenience that you are referring to subsets of the reals.

Problem: Let $S\subset\mathbb{R}$ and denote it's boundary by $\partial S$. Prove that $\partial S$ is closed.

Proof: Let $x\in\mathbb{R}-\partial S$. There are two cases

Case 1- $x\in S$. Since $x\in S$ clearly every neighborhood of $x$ will contain a point of $S$. Therefore since $x\in\mathbb{R}-\partial S$ we know there exists some neighborhood $N_{\delta}(x)$ such that $N_{\delta}(x)\cap \left(\mathbb{R}-S\right)=\varnothing$. But clearly for any $x'\in N_{\delta}(x)$ there exists some neighborhood of $N_{\delta'}(x)'$ such that $N_{\delta'}(x')\cap \left(\mathbb{R}-S\right)=\varnothing\implies x'\in\mathbb{R}-\partial S$. Therefore $x\in\left(\mathbb{R}-\partial S\right)^{\circ}$

Case 2- $x\in\mathbb{R}-S$. Similarly, since $x\notin S$ every neighborhood of $S$ will contain a point of $\mathbb{R}-S$. Thus since $x\in\mathbb{R}-S$ we know there exists some $N_{\delta}(x)$ such that $N_{\delta}(x)\cap S=\varnothing$. Let $x'\in N_{\delta}(x)$. Clearly there exists some $N_{\delta'}(x')$ such that $N_{\delta'}(x')\cap S=\varnothing\implies x'\in\mathbb{R}-\partial S$. Therefore, once again $x\in\left(\mathbb{R}-\partial S\right)^{\circ}$

Therefore every element of $\mathbb{R}-\partial S$ is an interior point, thus $\mathbb{R}-\partial S$ is open and consequently $\partial S$ is closed. $\blacksquare$
Alternatively.

Proof: Let $x\in \left[\partial S\right]'$. Then for every $\delta>0$ there exists some $x'\in N_{\delta}(x)$ such that $x'\in \partial S$. Now since $x'\in\partial S$ we know that every neighborhood contains points of both $S$ and $\mathbb{R}-S$, but since $N_{\delta}(x)$ is open I may choose a $\delta'>0$ such that $N_{\delta'}(x')\subset N_{\delta}(x)$. But this clearly implies that there exists points of $S$ and $\mathbb{R}-S$ in $N_{\delta}(x)$. Therefore $x\in \partial S$ and the conclusion follows.
• November 17th 2009, 07:38 PM
aliceinwonderland
Quote:

Originally Posted by aliceinwonderland
Let A be a subset of a metric (or topology) space X. A point x in the metric (or topology) space X is a boundary point of A provided that x belongs to $(\overline{A}) \cap (\overline{X \setminus A})$.

An intersection of closed set is closed, so bdA is closed.

EDIT: plz ignore this post. I didnt notice that using a closure is not allowed in your question.

If you don't use "closure" at all, just use the definition. Every open set containing x in the above post contains a point of A and a point of X\A. Use the definition of limit points and closed sets, I think you can derive the conclusion.
• November 17th 2009, 07:49 PM
p00ndawg
thanks guys i get it.

I kind of knew what I wanted to show I just didnt know how to phrase it in math terms.