# Why is the boundary of a set in the closure of the set?

The definition of the closure of a set $S$ is the intersection of all the closed subsets of the topological space $X$ which contain $S$ as a subset.

The definition of boundary of $S$ is $\text{Comp}\{\text{Int}(S) \cup \text{Ext}(S)\}$.

Since the boundary of $S$ does not contain $\text{Int}(S)$ as a subset, it is not a closed set that contains $S$, so the boundary of $S$ is not in the closure of $S$, but apparently it is. Why is this?

• What is $\operatorname{Comp}$? – Asaf Karagila Apr 1 '15 at 12:24
• $\text{Comp}(S)$ is the complement of the set $S$ – mr eyeglasses Apr 1 '15 at 12:27

Now let $p\in\partial S$. Then $p\not\in\mathrm{Ext} S$, and so any open neighborhood of $p$ intersects $S$. It follows that $p$ lies in any closed set which contains $S$, i.e. $p$ is in the closure.
• How do you know that any open neighborhood of $p$ intersects $S$? – mr eyeglasses Apr 1 '15 at 12:03
• @ᴇʏᴇs otherwise, by definition we have $p\in\mathrm{Ext}S$. – Amitai Yuval Apr 1 '15 at 12:10
• Sorry, which definition are you using? Our definition of $\text{Ext}S$ is $\text{Int}(\text{Comp}S)$, and our definition of neighborhood of $p$ is a subset of the topological space that contains an open set such that $p$ is contained in the open set – mr eyeglasses Apr 1 '15 at 12:12
• @ᴇʏᴇs ... and the definition of open neighbourhood of $p$ should be: an open set containing $p$. – Hagen von Eitzen Apr 1 '15 at 12:36