Open set (equivalent definitions?)

[center o bass]

I've seen open sets $S$ of a bigger set $X$ being defined as

1) for every $x\in S$ one can find an open disk $D(x,\epsilon)$ centered at $x$ of radius $\epsilon$ such that $D$ is entirely contained in $S$. Where

$$D(x,\epsilon)= \left\{y \in X: d(x,y) < \epsilon\right\}$$
and $d$ is a metric.

2) An open set is a set that can be written as a union of open disks.

Are these two definitions equivalent in general? Or does it require $X$ to be Hausdorff. If they are in general equivalent, can you outline a proof?

 

[center o bass]

I found a proof here:
http://people.hofstra.edu/stefan_waner/diff_geom/openballs.html

 

[mklejeune]

The two definitions are equivalent if the topological space in question is metrizable.

I recommend proving for yourself that every metric space satisfies the Hausdorff axiom.

 

[WWGD]

More generally, the open subsets of a topological space $X$ are , or can be, any collection of subsets of $X$ that are closed under unions and closed under finite intersection, and the collection includes the whole space $X$ and the empty set. You then have a sub -collection of the collection of open sets that is called a basis, so that for every element $x$ is an open set $U$ , there is a basis element $B$ with $x$ contained in $B$, and $B \subset U$.

 

[blue_raver22]

These two definitions are equivalent in general. The first definition states that for every point x in the set S, there exists an open disk D(x,ε) that is entirely contained in S. This means that every point in S has a small neighborhood within S that is also open. The second definition states that an open set is a union of open disks. This means that every point in S is contained in at least one open disk, and since the union of open sets is also open, S is an open set.

To prove their equivalence, we can start with the first definition and show that it implies the second definition. Let S be an open set defined by the first definition. For any point x in S, we can find an open disk D(x,ε) that is entirely contained in S. Since every point in S has such a neighborhood, we can take the union of all these open disks to cover S. This union is also open since it is a union of open sets. Therefore, S can be written as a union of open disks, satisfying the second definition.

To show the reverse implication, we can start with the second definition and show that it implies the first definition. Let S be an open set defined by the second definition. This means that for every point x in S, there exists an open disk D(x,ε) that contains x. Since S is a union of open disks, every point in S is contained in at least one open disk. Therefore, for every point x in S, we can find an open disk D(x,ε) that contains x and is also entirely contained in S. This satisfies the first definition, proving the equivalence of the two definitions.

This equivalence holds in general, regardless of whether X is a Hausdorff space or not. This is because the definitions only involve the properties of open sets and open disks, which are defined in terms of the metric d. The Hausdorff property is not necessary for this equivalence.