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

1. ## 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?

2. 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 $\displaystyle (\overline{A}) \cap (\overline{X \setminus A})$.

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.

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

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

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

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

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

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

4. 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 $\displaystyle (\overline{A}) \cap (\overline{X \setminus A})$.

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.

5. 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.

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