Proving an open ball is connected in a metric space X

[mahler1]

Homework Statement .

Let $B(a,ε) (ε>0)$ in a metric space $(X,d)$. Decide whether this subset of $(X,d)$ is connected or not.

The attempt at a solution.
Well, I know open intervals in the real line are connected. I suppose that an open ball in a given metric space can be imagined as an open interval of a more general metric space instead of the real line; at least, that's the way I see it. So, by this analogy, I think that any open ball in a given metric space is always connected. My problem is I don't know how to prove it. I've tried it by the absurd:

So, suppose we can disconnect $B(a,ε)$. Then there exist $U$ and $V$, nonempty open subsets of $(X,d)$ such that
i)$B(a,ε)= U \cup V$
ii)$U \cap V=\emptyset$

How can I come to an absurd? By all the things above, I know that there exists $x \in B(a,ε)$ : $x \in U$ and $x \not\in V$. On the other hand, $U$ is an open set, so for some $δ>0$, $B(x,δ) \subset U$. That's all I got up to now. How can I continue to arrive to an absurd?

 

[gopher_p]

Homework Statement .

Let $B(a,ε) (ε>0)$ in a metric space $(X,d)$. Decide whether this subset of $(X,d)$ is connected or not.

The attempt at a solution.
Well, I know open intervals in the real line are connected. I suppose that an open ball in a given metric space can be imagined as an open interval of a more general metric space instead of the real line; at least, that's the way I see it. So, by this analogy, I think that any open ball in a given metric space is always connected. My problem is I don't know how to prove it. I've tried it by the absurd:

So, suppose we can disconnect $B(a,ε)$. Then there exist $U$ and $V$, nonempty open subsets of $(X,d)$ such that
i)$B(a,ε)= U \cup V$
ii)$U \cap V=\emptyset$

How can I come to an absurd? By all the things above, I know that there exists $x \in B(a,ε)$ : $x \in U$ and $x \not\in V$. On the other hand, $U$ is an open set, so for some $δ>0$, $B(x,δ) \subset U$. That's all I got up to now. How can I continue to arrive to an absurd?

Is it possible that the statement that you are trying to prove, that every open ball in every metric space is connected, is not true?

 

[Dick]

Homework Statement .

Let $B(a,ε) (ε>0)$ in a metric space $(X,d)$. Decide whether this subset of $(X,d)$ is connected or not.

The attempt at a solution.
Well, I know open intervals in the real line are connected. I suppose that an open ball in a given metric space can be imagined as an open interval of a more general metric space instead of the real line; at least, that's the way I see it. So, by this analogy, I think that any open ball in a given metric space is always connected. My problem is I don't know how to prove it. I've tried it by the absurd:

So, suppose we can disconnect $B(a,ε)$. Then there exist $U$ and $V$, nonempty open subsets of $(X,d)$ such that
i)$B(a,ε)= U \cup V$
ii)$U \cap V=\emptyset$

How can I come to an absurd? By all the things above, I know that there exists $x \in B(a,ε)$ : $x \in U$ and $x \not\in V$. On the other hand, $U$ is an open set, so for some $δ>0$, $B(x,δ) \subset U$. That's all I got up to now. How can I continue to arrive to an absurd?

The real line is too simple an example to make judgements based on it. Can't you think of a metric space that has disconnected open balls?

 

[mahler1]

The real line is too simple an example to make judgements based on it. Can't you think of a metric space that has disconnected open balls?

I got really mixed up trying to generalize the interval case. The counterexample I could think of was: consider X an infinite metric space with the discrete δ-distance. Pick any x in X and consider of the ball centered at x of radius 2. Then I can disconnect this ball by two open sets U and V consisting of union of open balls of radius 1 centered at any other point of the space. Is this correct? Can you give me another example to completely destroy my previous wrong assumption?

 

[mahler1]

Is it possible that the statement that you are trying to prove, that every open ball in every metric space is connected, is not true?

Yes, it's wrong, I got confused.

 

[Dick]

I got really mixed up trying to generalize the interval case. The counterexample I could think of was: consider X an infinite metric space with the discrete δ-distance. Pick any x in X and consider of the ball centered at x of radius 2. Then I can disconnect this ball by two open sets U and V consisting of union of open balls of radius 1 centered at any other point of the space. Is this correct? Can you give me another example to completely destroy my previous wrong assumption?

Sure. Any subset of a metric space is a metric space, right? Take the subset of the reals X=(-infinity,-1/2]U[1/2,infinity) and think about the open ball around 0 of radius 1. There's also plenty of examples where X is connected and still has disconnected open balls. Can you think of one?

 

[Office_Shredder]

Dick, your example doesn't work because 0 isn't a point in the metric space to have an open ball around, but the open ball around -1/2 of radius 2 does work.

I second the think of a connected metric space which has disconnected open balls call. There are simple subsets of R² that work.

 

[Dick]

Dick, your example doesn't work because 0 isn't a point in the metric space to have an open ball around, but the open ball around -1/2 of radius 2 does work.

I second the think of a connected metric space which has disconnected open balls call. There are simple subsets of R² that work.

Good catch. Thank you!