# Why does the definition of open set in Euclidean topology involve the open ball and not the closed ball?

The definition of Euclidean topological space $( \Bbb{R}^n,\tau)$ in most standard texts is $$\text{\Omega is open iff \forall p \in \Omega \exists r>0 s.t. D_r(p) \subset \Omega where D_r(q):= \{x : |x-q|<r \} }$$

What exactly would not work if the open ball $D_r(q)$ were defined with a $\leq$ sign? As in $$\text{\Omega is open iff \forall p \in \Omega \exists r>0 s.t. D_r(p) \subset \Omega where D_r(q):= \{x : |x-q|\leq r \} }$$ I have thought about this and haven't been able to reach any definite conclusion.

• The two definitions are equivalent. – 5xum Oct 13 '16 at 14:01
• You might possibly get a more interesting answer if you asked this in HSM. hsm.stackexchange.com – Spencer Feb 20 '17 at 13:30

This generates the same topology as the classical definition.

To see this, note that any closed ball $\overline{B_\epsilon(x)}$ of positive radius $\epsilon>0$ contains an open ball of positive radius, namely $B_{\epsilon\over 2}(x)$; and conversely as well ($B_\epsilon(x)$ contains $\overline{B_{\epsilon\over 2}(x)}$). So any set which is open in one sense, is open in the other.

Note that there's a related idea you might try: take the topology generated by closed balls of positive radius! Unfortunately, this does not give the usual topology, and in fact yields rather silly topology - the discrete topology, in which every set is open.

Why? Well, any singleton $\{x\}$ is the intersection of two closed balls of positive radius (picture two spheres touching at one point), and since a topology must be closed under arbitrary unions, this means that every set is open: if $S$ is a set of points, then $S=\bigcup_{x\in S}\{x\}$ is a union of open sets, hence open.

Note that if in your definition you drop the requirement that the radius $r$ be positive, then every singleton is closed (since the closed ball of radius $0$ around $x$ is just $\{x\}$); so this again yields the discrete topology. So it's important to restrict attention to positive radii.

If we write $B_{r}(p)=\{x \in \mathbb{R}^{n}|\ |x-p|<r\}$, and $\bar{B_{r}}(p)=\{x \in \mathbb{R}^{n}|\ |x-p|\le r\}$, then for any $p \in \mathbb{R}^{n}$ and $r \in \mathbb{R}$, we have the inclusions $$\bar{B}_{r/2}(p) \subset B_{r}(p) \subset \bar{B_{r}}(p)$$ So let $S$ be a set and $p \in S$. If there is an open $r$-ball inside $S$ containing $p$ then there is a closed $r/2$-ball inside $S$ containing $p$, and similarly if there is a closed $r$-ball in $S$ containing $p$ then the open $r$-ball also contains $p$ and is inside $S$. In other words, a set is open in the usual sense if and only if it is open in your slightly different sense.

However, note that closed balls are still not open! In particular, points on the boundary of the ball do not have the property you give (you might not have thought this, but it's worth warning in case you did!)