# Given a metric space (X,d), the set X is open in X. HELP!

MechanicalEngr
I must be overlooking something! Given a metric space (E,d), the improper subset E is open in E. How? Here is my understanding:

1) We call a set S(subset of E) open iff for all x(element of S) there exist epsilon such that an open ball of radi epsilon centered about s is wholly contained in S.

So, how can E always be open. Here is my counterexample:

Take E to be a subset of R^2 described by...

E={(x,y):0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 }

If now analyze E as a subset of itself, the set E is clearly not open as it contains its boundary. What am I missing here?

Thanks,

Topology Newbie

## Answers and Replies

The set E is not open in R^2, but it is open in E with the subspace topology. The balls of that topology might not be circles. For example a ball at (1/2,0) with radius 1/4 is really only the upper half of a circle.

##\partial E## is relative to the larger space you are taking ##E## to be a subset of. If you are looking at ##E\subseteq E##, with the topology on ##E## being the subspace topology inherited from ##\mathbb{R}^{2}##, then ##\partial E = E\setminus (\text{int}E \cup (E\setminus \bar{E})) = \varnothing ## so it has empty topological boundary. Keep in mind that whether or not sets are closed and/or open in a space depends entirely on the topology endowed on that space. If ##A\subseteq X##, where ##X## is a topological space, there is a natural topology induced on ##A## called the subspace topology (alluded to above) given by ##\mathcal{T}_{A} = \left \{ U\subseteq A:U = A\cap V \text{ for some V open in X} \right \}##.

MechanicalEngr
Wow. i think I see it now! Is this correct:

If one examines the definition of the open ball, it must be in the context of a particular set... or else it is not sensible. The precise definition of an open ball of radi r centered at p(element of E) in the set E would be the set.....

{x(ELEMENT OF E): d(p,x) < r}

So, in my metric space (E,d), if I take E to be an improper subset of itself, it must be that at any x(element of E) an open ball does exist of some radi that is wholly contained in the set E..... this is because of the precise definition of the open ball IN THE CONTEXT OF THE SET E, precisely the part of the set description... all x WHICH ARE ELEMENTS OF E for which the distance between p and x is less than r.

Do I have the right understanding now?

Thanks.

Yes, that's correct. The open balls of a metric space $(X,d)$ are defined to be $B_\epsilon(x) = \{y \in X : d(x, y) < \epsilon\}$ and thus if you check your definition of openness, you will find that your set $E$ is open in $(E,d)$, but closed in $(\mathbb R^2,d)$.