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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.

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  • $\begingroup$ According to my copy, 2.20 asks if closure and interior of a connected set is connected. (Which edition are you using?) $\endgroup$
    – user99914
    Commented Jul 6, 2015 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
    Commented Jul 6, 2015 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
    Commented Jul 6, 2015 at 16:49

2 Answers 2

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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".

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  • $\begingroup$ 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
    Commented Jul 6, 2015 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 $\endgroup$ Commented Jul 9, 2015 at 15:54
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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.

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