# Boundary and closure

1. Dec 1, 2013

### Bipolarity

My first analysis/topology text defined the boundary of a set S as the set of all points whose neighborhoods had some point in the set S and some point outside the set S. It also defined the closure of a set S the union of S and its boundary.

Using this, we can prove that the closure of S is the smallest closed set containing S. We can also prove that the boundary of S is the intersection of the closure of S and the closure of the complement of S.

I was wondering, if we define the closure of S to be the smallest closed set containing S, and the boundary of S to be the intersection of the closure of S and the closure of the complement of S, will we have the machinery necessary to work backwards and prove the first two definitions, i.e. are the two definitions of boundary/closure equivalent?

BiP

2. Dec 1, 2013

### economicsnerd

Yes, they are! And your proof will show it!

Given a subset $S$ of your space $X$...
- Let $\partial S=\{x\in X: \enspace \text{ for any neighborhood } N \text{ of } x, \enspace N\cap S\neq\emptyset \text{ and } N\cap (X\setminus S)\neq\emptyset \}$.
- Let $\bar S = S \cup \partial S$.
- Let $\tilde S$ be the smallest closed set containing $S$. (You have to prove that there is such a set. But this is easy; it's just the intersection of every closed set containing $S$.)

Notice that I haven't used the words "closure" or "boundary" anywhere above.

It sounds like you know how to prove that $\bar S = \tilde S$ and $\partial S = \bar S \cap \overline{X\setminus S}$.

Having shown that, you can define:
- The closure of $S$ (denoted $cl(S)$) is either $\bar S$ or $\tilde S$, whichever definition you like. [We now know they're equivalent.]
- The boundary of $S$ is either $\partial S$ or $cl(S)\cap cl(X\setminus S)$, whichever definition you like. [We now know they're equivalent.]