Sequential Compactness and Connectedness

Question

Let $K\subset \mathbb{R}^n$ be sequentially compact. Prove that $K$ is not connected if and only if there are nonempty disjoint subsets $A$ and $B$ of $K$, with $A\cup B = K$ and a positive number $\epsilon$ such that $d(u,v)> \epsilon$, $ \forall u\in A,v\in B$. Give an example that does not hold if $K$ is assumed to be only bounded or only closed.

$\underline{\Rightarrow}$

Suppose that $K$ is not connected.

Then by definition, there exist two open subsets of $\mathbb{R}^n$, $U$ and $V$, such that $$U\cap K = A \neq \emptyset,\ \ V\cap K = B \neq \emptyset,\ \ A\cap B = \emptyset, \ \ A\cup B = K.$$

Because $K$ is sequentially compact, it is bounded. Therefore, for any two points, $u$ and $v$ in $K$, $\alpha \leq d(u,v) \leq \beta$, where $\alpha$ and $\beta$ are nonnegative real numbers.

Let $\epsilon$ be a lower bound of $K$, but not the greatest lower bound. Then $\epsilon < d(u,v)\ \forall u,v\in K$, including points where $u \in A$ and $v \in B$.

I feel fairly confident about this section of the proof since I used all material given.

$\underline{\Leftarrow}$

Suppose that there are nonempty disjoint subsets $A$ and $B$ of $K$ whose union is $K$.

Then we can choose two open subsets $U$ and $V$ of $\mathbb{R}^n$ whose intersections with $K$ are $A$ and $B$ respectively.

Then by definition, $K$ is not connected.

I feel less confident about this section of the proof because I did not use the distance or sequential compactness assumptions. I am not sure how they are relevant here, though.

If $\pmb{K}$ is only closed and not bounded

Then $K$ is not guaranteed to have bounded distances between two numbers, and the second claim is not guaranteed to be satisfied.

I honestly get mixed up about closed sets and bounded sets, so I am not sure how to give a decent example for why both qualities are required.

Answer 1

Counter examples.

Let A = { (x,y) : 1 <= xy }, B = { (x,y) : -1 <= xy }.
A and B are closed.
inf{ d(a,b) : a in A, b in B } = 0.

Let A = (-1,0), B = (0,1).
A and B are bounded.
inf{ d(a,b) : a in A, b in B } = 0.

Exercise. Show if A is compact, then A is bounded.

Answer 2

If $K$ is not connected it means that there exists a continuous $f: K \to \{0,1\}$ that is not constant; this corresponds to a partition of $K$ into two clopen subsets.

Now apply the fact that $f$ must be uniformly continuous as $K$ is (sequentially) compact to $\varepsilon = \frac{1}{2}$ and we get the required $\delta$.