Closed and Connected Subset of a Metric Space

Question

My English may not be perfect since I'm not a native speaker, so please do point out the grammar mistakes if there are any.

I've been reading Conway's "Functions of One Complex Variable", and encountered following exercise in the book (p. 17). $X$ here is a metric space and $d$ is the metric of it.

Show that if $F\subset X$ is closed and connected then for every pair of points $a,b$ in $F$ and each $\epsilon>0$ there are points $z_{0},z_{1},\dots,z_{n}$ in $F$ with $z_{0}=a, z_{n}=b$ and $d(z_{k-1},z_{k})<\epsilon$ for $1\leq k\leq n$. Is the hypothesis that $F$ be closed needed? If $F$ is a set which satisfies this property then $F$ is not necessarily connected, even if $F$ is closed. Give an example to illustrate this.

Now, I guess the right way to solve this problem is to first fix a point $a$ and consider the subset $S$ of $F$ definded by $$S=\{b\in X;\ for\ all\ \epsilon>0,\ there\ are\ points \ z_{0},z_{1},\dots,z_{n}\ in \ F\ with\ z_{0}=a, z_{n}=b\ and \ d(z_{k-1},z_{k})<\epsilon\ for \ 1\leq k\leq n\}$$and show that this set is nonempty, open and closed, but I'm currently stuck. I'm not sure where I should use the condition that $F$ is closed.

Moreover, the rest of the problem is also challenging, and I can't think of any possible way to get started.

Am I going in the right direction? Please give me a clear motivation on this problem and possibly the solution!! :D

Answer 1

Here is an long alternative to Stefan's succinct proof:

If $a,b \in F$ and $\epsilon>0$, let an $\epsilon$-path be a finite sequence of points satisfying the condition in the question.

Suppose $F$ is connected. Let $a \in F$ and $\epsilon>0$. Let $U_a(0) = B(a,\epsilon) \cap F$, and $U_a(n+1) = \cup_{x \in U_a(n)} B(x,\epsilon) \cap F$. Note that $U_a(n) \subset U_a(n+1) \subset F$, and each set is open relative to $F$. Let $U_a = \cup_{n} U_a(n)$, and again note that $U_a$ is open relative to $F$.

Note that if $b \in U_a$, then there is an $\epsilon$-path joining $a$ and $b$.

It is straightforward to see that if $a \neq b$, then either $U_a=U_b$ or $U_a \cap U_b = \emptyset $, and since $a \in U_a$ for all $a$, we see that $ { \cal P} = \{ U_a \}_{a \in F}$ forms a partition of $F$.

Suppose ${ \cal P}$ has more than one element, then let $V \in { \cal P}$, $W = \cup_{O \in { \cal P}, O \neq V } O$. Then $V,W$ are open (relative to $F$), disjoint, non-empty and $F=V \cup W$, which contradicts $F$ being connected. Hence ${ \cal P}$ consists of one element and hence any two points of $F$ can be connected by an $\epsilon$-path.

Counterexample: You need two closed sets that are not connected but are arbitrarily close to each other.

Look for an example in $\mathbb{R}^2$.

If either set was bounded, then one would be compact, and the hypotheses would imply that their union was connected.

Hence both sets need to be unbounded, and the points at which they get closer and closer must get further and further away.

Example:

Let $C_1 = \{ (x, {1 \over x} ) \}_{ x >0 }$, $C_2 = \{ (x,0) \}_{x \ge 0 }$ and $F = C_1 \cup C_2$.

Answer 2

Your approach seems correct. Your set contains the point $a$, so it's nonempty. For the openness pick any point and show that an open ball of radius $\varepsilon$ is contained in the set too. For the closedness pick a point in the closure and show there's a point from the set that is $\varepsilon$ close to it.

So your set is a nonempty clopen subset of $F$ and must be the whole $F$ because of connectedness. You didn't need that $F$ is closed in $X$ anywhere.

As for an example of a closed set which satisfies the property, but is not connected consider the graph of $f(x)=\dfrac1{x^2}$.

Answer 3

The set $S$ should depend on a particular $ϵ$, so $$S_ϵ=\{b\in F\mid \exists\{z_o,...,z_n\}\subseteq F,\ z_0=a,\ z_n=b,\ d(z_n,z_{m+1})<ϵ\}$$ You can show that $S_ϵ$ is open in $F$: If $b\in S_ϵ$, and $(z_0,...,z_n)$ is the sequence from $a$ to $b$ in $F$, then the ball $B_{ϵ-d(z_n,z_{n-1})}(b)$ is contained in $S_ϵ$.
$S_ϵ$ is also closed in $F$: If $x\in\overline S_ϵ\cap F$, then the $B_ϵ(x)$ intersects $S_ϵ$, and the sequence to some point in this intersection can be extended to $z_{n+1}=x$.
So $S_ϵ$ is open and closed in $F$, and since $F$ is connected, we have $F=S_ϵ$. But this works for any $ϵ$, so the claim follows.
The closedness of $F$ is not needed here.