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1) Stuck with this problem, can you please help? Show that if $X$ is connected and for a connected subspace $A$ of $X$ we have $X \setminus A = U \cup V$ where $U,V$ are open in $X \setminus A$ and disjoint, then the sets $A \cup U$ and $A \cup V$ are connected.

2) The following result appears in Encyclopaedia of mathematics by Hazewinkel page $375$:

Any compact subset of the Sorgenfrey line is countable.

I tried to prove this without any luck. Why is this true?

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Maybe this link will give ideas. – André Nicolas Jul 13 '11 at 19:37
up vote 2 down vote accepted

(1) Let $U_0$ and $V_0$ be open sets in $X$ such that $U = U_0 \setminus A$ and $V = V_0 \setminus A$, and suppose that $A \cup U$ is not connected; then there are non-empty sets $H,G \subseteq A \cup U$ such that $A \cup U = G \cup H$, $G \cap H = \emptyset$, and $H$ and $G$ are closed and open in the subspace topology on $A \cup U$.

Suppose that $G \cap A \ne \emptyset \ne H \cap A$. Since $G$ and $H$ are open in $A \cup U$, there are open sets $W_G,W_H$ in $X$ such that $G = W_G \cap (A \cup U)$ and $H = W_H \cap (A \cup U)$. Then $G \cap A $ = $W_G \cap (A \cup U) \cap A = W_G \cap A$, since $U \cap A = \emptyset$, and similarly $H \cap A = W_H \cap A$. Thus, $G \cap A$ and $H \cap A$ are relatively open, non-empty subsets of $A$, and since $A \subseteq G \cup H$, their union is $A$; this contradicts the assumption that $A$ is connected. We may therefore assume that $A \subseteq G$, say, so that $H \cap A = \emptyset$, $W_H \cap A = \emptyset$, and hence $H = W_H \cap U$.

Now $W_H \cap U \subseteq W_H \cap U_0 \subseteq W_H \cap (U \cup A) = (W_H \cap U) \cup (W_H \cap A) = W_H \cap U$, so $W_H \cap U = W_H \cap U_0$, and hence $H = W_H \cap U$ is open in $X$. I claim that $H$ is also closed in $X$, contradicting the assumption that $X$ is connected.

Clearly $X \setminus H = G \cup V = \left(W_G \cap (A \cup U) \right) \cup V = \left(W_G \cap (X \setminus V) \right) \cup V = W_G \cup V$. But $A \subseteq G \subseteq W_G$, so $V_0 \setminus V \subseteq A \subseteq W_G$, and $X \setminus H = W_G \cup V = W_G \cup V_0$, an open subset of $X$, as claimed.

(2) Let $K$ be an uncountable subet of the Sorgenfrey line. Suppose that $x \in K$ and there is a sequence $\langle x_n:n \in \omega \rangle$ in $K$ such that $x_n < x_{n+1}$ for all $n \in \omega$ and $x = \sup\{x_n:n \in \omega \}$ (i.e., the sequence converges monotonically up to $x$ in the Euclidean topology). Let $V_L = (\leftarrow,x_0)$, $V_R = [x, \to)$, and for $n \in \omega$ let $V_n = [x_n,x_{n+1})$; then $\mathcal V = \{V_L,V_R\} \cup \{V_n:n \in \omega\}$ is an open cover of $K$ with no finite subcover. Thus, it suffices to show that $K$ must contain such a point $x$ and sequence $\langle x_n:n \in \omega \rangle$.

If not, for each $x \in K$ there must be a rational number $r(x)<x$ such that $(r(x),x) \cap K = \emptyset$. But $\mathbb Q$ is countable, so there must be distinct $x,y \in K$ such that $r(x) = r(y)$. (In fact the function $r$ must be constant on an uncountable subset of $K$, but we need only two points with the same $r$ value.) Without loss of generality $x<y$. But then $r(y) = r(x) < x < y$, and $x \in (r(y),y) \cap K$, contradicting the choice of $r(y)$. Thus, the desired point and sequence must exist, and $K$ cannot be compact (or even countably compact, since $\mathcal V$ is countable).

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thanks, impressive. – user10 Jul 15 '11 at 16:35

I will assume sets of the form $[a,b)$ are open.

Let $C$ be a compact subset of the Sorgenfrey line $E$. As the topology on $\mathbb{R}$ is weaker than the one on $E$, $C$ is also compact in the standard topology of $\mathbb{R}$. In particular it is closed and bounded there. Now, if $A$ is any non-empty subset of $C$, its supremum $\sup(A)$ is not only in $C$ but already in $A$: otherwise the sets $\{x: x < a \}$ for $a \in A$ and $\{ x: x \ge \sup(A) \}$ form an open cover of $C$ without a finite subcover, contradicting its compactness. So compact subsets of the Sorgenfrey line are well-ordered under $>$, the reverse standard order on $\mathbb{R}$.

It is well-known such subsets are at most countable, e.g. using that the order topology is second countable.

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I'm not sure how to answer a question with two questions in it, so here's an answer to the first question. The argument seems more convoluted than it should be, so please let me know if there's some simplification you see.

We'll show $A\cup U$ is connected, the other case is symmetric. Suppose that $C \subseteq A \cup U$ is clopen. Since $A$ is connected, by replacing $C$ with $(A \cup U)\setminus C$ if necessary we may assume $C \cap A = \emptyset$. Since $U$ is open in $X\setminus A$, there is an open set $U_1 \subset X$ disjoint from $V$ such that $U = U_1 \setminus A$. Since $C$ is open in $A \cup U$, there is an open set $U_2 \subseteq X$ disjoint from $A$ such that $C = U_2 \cap (A \cup U)$. Thus, $C = U_1 \cap U_2$, so $C$ was actually open in $X$ all along.

It suffices then to show that $C$ is closed in $X$ (since $X$ is connected, this forces $C$ to be empty granting the connectedness of $A \cup U$). An argument analogous to the one above works to show that $X \setminus C$ is the union of two open subsets of $X$.

I was going to answer the second question in a second answer, but then I realized that made absolutely no sense. The point to this question is that any uncountable subset of the reals contains a countably infinite, strictly increasing sequence. If not, then you'd have a reversed copy of $\omega_1$ (the first uncountable ordinal) sitting inside the reals, which contradicts separability (squeeze a rational between two successive points).

Once you have your increasing sequence $(x_n)_n$, you can build a counterexample to compactness by looking the cover $\{(-\infty, x_n) : n \in \mathbb{N}\} \cup \{[\lim_n x_n, +\infty)\}$ (if the limit exists -- if the sequence blows up to infinity you don't need to worry about that last part).

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