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

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