# I Open balls contained in other open balls

1. Dec 3, 2018

### Mr Davis 97

I am reading a proof that claims that $\overline{B_{\epsilon /2}(x)} \subseteq B_{\epsilon}(x)$, where the overline means closure, without really explaining why. It's clear to me that $B_{\epsilon /2}(x) \subseteq B_{\epsilon}(x)$, but I can't quite see why the former is the case without additional explication.

2. Dec 3, 2018

### Staff: Mentor

Can you describe the closure algebraically? Which points are in the closed ball?

3. Dec 3, 2018

### Mr Davis 97

I'm not sure I am able... I might say that the closure of an open ball is a closed ball with same radius, but I know that this is not necessarily true in general metric spaces. So I'm not really seeing how to characterize the closure of the open ball...

4. Dec 3, 2018

### Staff: Mentor

The open ball is $B_\varepsilon(x) =\{\,y\,|\,d(x,y) <\varepsilon\,\}$. The closure includes all limit points, so what do we get if we take the $y's$ to the limits?

5. Dec 3, 2018

### Mr Davis 97

I'm not seeing it... I'm tempted to say that $\overline{B_\epsilon(x)} = \{y\mid d(x,y)\le \epsilon \}$, but this is just the closed ball, so this can't be right. I don't know what other avenue to take

6. Dec 3, 2018

### Staff: Mentor

Yes, that's the closed ball. So where is your problem? $d(x,y) \leq \dfrac{\varepsilon}{2} < \varepsilon$

7. Dec 3, 2018

### Mr Davis 97

Let's denote the closed ball by $\overline{B}_\epsilon(x)$.

So is the idea that $\overline{B_{\epsilon/2}(x)} \subseteq \overline{B}_{\epsilon/2}(x) \subseteq B_\epsilon(x)$?

8. Dec 3, 2018

### Staff: Mentor

I'm not aware of a difference between $\overline{B_{\varepsilon/2}(x)}$ and $\overline{B}_{\varepsilon/2}(x)$. The idea is simply that everything closer or equal $\varepsilon/2$ is still strictly closer than $\varepsilon$.

9. Dec 3, 2018

### Mr Davis 97

The difference would be that the closure of an open ball is not necessarily the same as the closed ball with the same radius. This fact is mentioned in the question here https://math.stackexchange.com/ques...sure-of-an-open-ball-equal-to-the-closed-ball

10. Dec 3, 2018

### Staff: Mentor

The $\varepsilon-$notation doesn't make sense in the discrete metric, so I assumed, that you have radii of continuous magnitudes - the usual situation when we consider closed balls. The discrete metric is quite an exception.

11. Dec 3, 2018

### Staff: Mentor

For the discrete metric, we have as well the inclusion.

$\emptyset \; , \; \{\,x\,\}\; , \; X$ are the only sets to be considered here, because as soon as a second point is within a ball, the entire space $X$ is. And $x$ is within the balls, so we only consider $\{\,x\,\}$ and $X$.

The only way to get $B_\varepsilon(x) \subseteq \overline{B_{\varepsilon /2}(x)}$ is to have $B_\varepsilon(x) =\{\,x\,\}$ and $\overline{B_{\varepsilon /2}(x)}=X$. The first condition yields $\varepsilon < 1$, so $\varepsilon /2 < 1/2$ and $\overline{B_{\varepsilon /2}(x)} = \{\,x\,\}$ and they are equal again.

12. Dec 8, 2018

### WWGD

There is a result, possibly for some a definition, in the metric case , that if x is in Class(S) then d(x,S)=0.