Undergrad Understanding metric space definition through concrete examples

[Ricster55]

Right now, I am studying Advanced Calculus (proof based), and it is hard thinking through some of the definitions without first thinking about it concretely (meaning how to visualize it better geometrically, if that makes any sense?). I need help with this definition.

Definition
Let X be a metric space. A set G ⊂ X is open if for every a ∈ G there exists r > 0 such that Br(a) ⊂ G. A subset F ⊂ X is closed if F^C = X - F is open.

How do I try to "visualize" this definition, through say, a diagram or a set example?

 

[member 587159]

It just means that for every point in the set, you can find a small open ball that is contained entirely in the set.

If you think about this in general $\mathbb{R}^n$ space (where the intuition comes from), and consider a bounded set, this means that nothing of a boundary is included in the set itself, because if there would be a point on the boundary, any ball with center that point will intersect the complement of the set, so the set isn't open.

Therefore, I like to think about openness as if the set has no boundary.

For example, $(0,1)$ is open in the reals with the usual metric, because it does not contain its 'boundary points' and $[0,1)$ is not open, because any ball with center 0 will contain a point smaller than $0$.

Notice that this intuition starts to break down in general metric spaces. Take for example any set with the discrete metric.

 

[fresh_42]

I'd first take the real line as example, so the open sets are open intervals. The $B_r(a)$ are required to be open: $B_r(a)=\{x\in X\,: \,||x-a|| < r\}$ which is usually written $U_r(a)$. The $B_r(a)$ are commonly reserved for closed balls.

After that you could do the same in the plane.

 

[Ricster55]

It just means that for every point in the set, you can find a small open ball that is contained entirely in the set.

If you think about this in general $\mathbb{R}^n$ space (where the intuition comes from), and consider a bounded set, this means that nothing of a boundary is included in the set itself, because if there would be a point on the boundary, any ball with center that point will intersect the complement of the set, so the set isn't open.

Therefore, I like to think about openness as if the set has no boundary.

For example, $(0,1)$ is open in the reals with the usual metric, because it does not contain its 'boundary points' and $[0,1)$ is not open, because any ball with center 0 will contain a point smaller than $0$.

Notice that this intuition starts to break down in general metric spaces. Take for example any set with the discrete metric.

aah, I think I get it now. Thanks for the reply