# Thread: Basic Topology

1. ## Basic Topology

1)Show that a set E is open iff E intersection boundary of E = empty set

2)Show that a set E is closed iff boundary is contained in E

2. Originally Posted by hebby
1)Show that a set E is open iff E intersection boundary of E = empty set

2)Show that a set E is closed iff boundary is contained in E
Don't you dare take a peek until you've tried them yourself.
Spoiler:

1) If $\displaystyle E$ is open then every point of $\displaystyle E$ is an interior point of $\displaystyle E$. Clearly then no point of $\displaystyle E$ may be a boundary point of $\displaystyle E$ so that it follows that $\displaystyle E\cap\partial E=\varnothing$. Conversely, suppose that that $\displaystyle E\cap\partial E=\varnothing$, then no point of $\displaystyle E$ is a boundary point of $\displaystyle E$. Thus, for every $\displaystyle \xi\in E$ we must have that there exists some neighborhood of $\displaystyle E$, which we'll call $\displaystyle N_{\delta}(\xi)$ such that either $\displaystyle N_{\delta}(\xi)\subseteq E$ or $\displaystyle N_{\delta}(\xi)\subseteq E'$. But, since $\displaystyle \xi\in E$ the latter is impossible and the conclusion follows.

2) If a set $\displaystyle E$ is closed then it contains all it's limit points. So let $\displaystyle \xi\in\partial E$. If $\displaystyle \xi$ is in $\displaystyle E$ we are done so assume that $\displaystyle \xi\notin E$. Since $\displaystyle \xi\in\partial E$ we must have that every neighborhood $\displaystyle N_{\delta}(\xi)$ contains points of both $\displaystyle E$ and $\displaystyle E'$, and since $\displaystyle \xi\notin E$ we must conclude that the points of $\displaystyle E$ are distinct from $\displaystyle \xi$. But, this implies that $\displaystyle \xi$ is a limit point of $\displaystyle E$ and the conclusion follows. Conversely, suppose that $\displaystyle \partial E\subseteq E$. Let $\displaystyle \xi$ be a limit point of $\displaystyle E$. If $\displaystyle \xi\in E$ we are done, so assume that $\displaystyle \xi\notin E$. Then every neighborhood of $\displaystyle \xi$ contains points of $\displaystyle E$ but since every neighborhood of $\displaystyle \xi$ also contains $\displaystyle \xi$ it follows that $\displaystyle \xi\in \partial E\implies \xi\in E$. We may therefore conclude that $\displaystyle E$ contains all it's limit points. The conclusion follows.