Proof of Topological Property

**jzellt** · February 1st 2010, 09:29 PM

Show that the boundary of any set D is itself a closed set.

Thanks in advance for any advice...

**Plato** · February 2nd 2010, 06:31 AM

Originally Posted by jzellt

Show that the boundary of any set D is itself a closed set.

Suppose that $\beta (A)$ is the boundary of $A$ .
If $x$ is a limit point of $\beta (A)$ and $\mathcal{O}$ is and open set such that $x\in\mathcal{O}$ implies $\left( {\exists y \in \mathcal{O}} \right)\left[ {y \in \beta (A)\backslash \{ x\} } \right]$ .
That means that because $y$ is a boundary point, $\mathcal{O}$ must contain a point of $A$ and a point not in $A$ .
How does this prove that $x$ is a boundary point of $A$ ? How does that show that $\beta (A)$ is closed?

**Drexel28** · February 3rd 2010, 01:23 PM

$\partial E=\bar{E}\cap\overline{E'}$ ....sooo...

**mabruka** · February 3rd 2010, 02:18 PM

Maybe state what definition of boundary you are starting from ?

Thread: Proof of Topological Property

LinkBack

Thread Tools

Display

Proof of Topological Property

Similar Math Help Forum Discussions

property of a topological group??

proof of property of gcd

dot product property proof help

Topological property

Proof in a topological space

Search Tags