4
$\begingroup$

I am working on an exercise in baby Rudin (Ex 2.20 in particular) and as part of that I am trying to show that any neighborhood in a metric space is connected. I've seen several differing definitions of neighborhoods and connectedness on this site while searching for a solution so I want to be clear about what definition Rudin uses at this point.

In metric space $X$ with metric $d$ a neighborhood of $x \in X$ of radius $r \in \mathbb{R}$ and $r > 0$ is $N_r(x) = \{y \in X \,|\, d(y,x) < r\}$.

The closure of a set $E \subseteq X$ is $\overline{E} = E \cup E'$, where $E'$ is the set of limit points of $E$.

The interior of $E \subseteq X$ is the set of interior points of $E$ and is denoted here by $E^\circ$.

Two subsets of $X$, $A$ and $B$, are separated if $A \cap \overline{B} = \emptyset$ and $\overline{A} \cap B = \emptyset$.

A set $E \subseteq X$ is connected if it is not the union of two separated nonempty sets, i.e. if $\lnot \exists A \exists B (E=A \cup B \land A \neq \emptyset \land B \neq \emptyset \land A \cap \overline{B} = \emptyset \land \overline{A} \cap B = \emptyset)$

This last definition is logically equivalent to $\forall A \forall B ([E=A \cup B \land A \neq \emptyset \land B \neq \emptyset] \to [A \cap \overline{B} \neq \emptyset \lor \overline{A} \cap B \neq \emptyset]) $

So if $E$ is some neighborhood in $X$ then consider any $A$ and $B$ where $E = A \cup B$, $A \neq \emptyset$, and $B \neq \emptyset$. We must then show that $A \cap \overline{B} \neq \emptyset \lor \overline{A} \cap B \neq \emptyset$. First if $A \cap B \neq \emptyset$ then clearly they are not separated so we need only worry about the case where $A \cap B = \emptyset$, i.e. they are disjoint. I think I can show this if both $A$ and $B$ are not open because if $A$ is not open then $\breve{A} = A - A^\circ \neq \emptyset$ and I can show that any $x \in \breve{A}$ is a limit point of $B$ thus $A \cup \overline{B} \neq \emptyset$, using the facts that $E=A \cup B$ and $E$, being a neighborhood, is open.

However I am having trouble showing that both $A$ and $B$ cannot both be open despite thinking about it for days. It was shown in a previous exercise that two disjoint open sets are separated but obviously we can't use this since we are trying to show that the arbitrary $A$ and $B$ are not separated. I am thinking that a proof by contradiction is the way to go here and to show that if they are both open then $\exists x \in E (x \not\in A \cup B)$ so that $E \neq A \cup B$ but I cannot pin down how to show this.

Anybody have any insights that might help me out here?

NOTE: It has been brought to my attention below that not every $N_r(x)$ in every metric space is connected, e.g. $X=\mathbb{Z}$ for $r > 1$. So really I am talking specifically about $\mathbb{R}^2$, though I suspect it is true in any $\mathbb{R}^n$. Of course this implies that we'd probably need to use the fact that we are in $\mathbb{R}^n$ in the proof.

$\endgroup$
  • $\begingroup$ According to my copy, 2.20 asks if closure and interior of a connected set is connected. (Which edition are you using?) $\endgroup$ – user99914 Jul 6 '15 at 15:44
  • $\begingroup$ Yes, mine is the same. This is regarding the interior part of the question, to which the answer is no. Consider the set he has for the solution here. I am trying to first show that the set he constructs is in fact connected. Showing that its interior is not will be pretty easy. $\endgroup$ – kyp4 Jul 6 '15 at 16:25
  • $\begingroup$ The rational numbers, $\Bbb Q$, are a perfectly good metric space, and the only connected subsets are the singletons. $\endgroup$ – Lubin Jul 6 '15 at 16:49
2
$\begingroup$

No, not all neighbourhoods are connected. For example, in $\mathbb R$ with the usual metric, $(-1,1) \cup (2,3)$ is a disconnected neighbourhood of $0$. A neighbourhood of $x$ is not $N_r(x)$, it's any set that contains $N_r(x)$.

But it's also not true that $N_r(x)$ is connected. For example, in $\mathbb Z$ with the usual metric, $N_r(x)$ is disconnected if $r > 1$.

It's also not true that in a connected metric space, $N_r(x)$ is connected. A counterexample is the "topologist's sine curve".

$\endgroup$
  • $\begingroup$ I am using Rudin's definition of neighborhood, which unless I am mistaken about his definition at the top of page 32, is just $N_r(x)$ as defined above and does not include any points outside of $N_r(x)$. You are right about $\mathbb{Z}$ though, good point. In particular I am trying to show this for any $N_r(x)$ in $\mathbb{R}^2$, though I suspect any such neighborhood (as defined above) is connected in any $\mathbb{R}^n$. $\endgroup$ – kyp4 Jul 6 '15 at 16:29
  • $\begingroup$ I was very surprised that baby Rudin used this non-standard definition of "neighborhood", which everybody else would call "open ball". When you get to grown-up Rudin ("Real and Complex Analysis") you'll see "A neighborhood of a point $x$ is, by definition, an open set that contains $x$". But even that is not the usual definition: a neighborhood of $x$ is a set that contains an open set that contains $x$; it doesn't have to be open itself. en.wikipedia.org/wiki/Neighbourhood_%28mathematics%29 $\endgroup$ – Robert Israel Jul 9 '15 at 15:54
0
$\begingroup$

It would seem that a good strategy for this is to first prove that any convex set in $\mathbb{R}^n$ is connected. The proof of this is similar to that presented here.

Then show that any $N_r(x)$ in $\mathbb{R}^n$ is convex, from which it follows that $N_r(x)$ is connected. Since $N_r(x)$ is an open ball, this proof is rather trivial and is actually presented at the bottom of page 31 in Rudin.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.