1
$\begingroup$

Let $(X,\tau)$ be a topological space. We say that a subset $S\subset X$ is connected iff there does not exist a set $A\subset X$ that is both open and closed in the induced topology. I need to prove the following:

A subset $S \subset X$ is connected if and only if cannot be written as a union of two disjoint subsets $H,K$, such that $S=H \cup K$ with $\overline H \cap K= H \cap \overline K=\emptyset$

$(\implies)$ Suppose that there exist $H,K\subset S $ such that $S=H \cup K$ with $\overline H \cap K= H \cap \overline K=\emptyset$. Then $S\setminus \overline H$ is open because $\overline H$ is closed by definition. Since we know that $\overline H \cap K=\emptyset$ and $S=H \cup K$, clearly $S\setminus \overline H \subset K$. I would like to show that $\overline H$ is also open, but I don't know how to do it; I thought about showing that its complement $S\setminus \overline H$ is closed using that $H \cap \overline K$ is empty but I am not able to do it.

$(\impliedby)$ Suppose $S$ is not connected. Then there exists a set $H\subset S$ such that $H$ is both open and closed. Take $K=S\setminus H$ and we are done. In fact, $S=H\cup (S\setminus H)$ by definition. On the other hand, $\overline H=H$ since $H$ is closed, so $\overline H \cap K=\emptyset$. Since $H$ is open, $K$ is closed, hence $H \cap \overline K=\emptyset$.

How can I show the first implication to be true? It seems simple, but I feel like I am missing something.

$\endgroup$
  • $\begingroup$ Be careful about which topology you are referring to: the topology on $X$ or the induced topology on $S$. There are instances in both implications where this is unclear. $\endgroup$ – Aweygan Aug 8 '16 at 16:38
  • $\begingroup$ I am using the induced topology in $S$. Could you please point out in which instances the exposition is not clear so I can edit my post? Thank you. $\endgroup$ – Lonidard Aug 8 '16 at 16:39
  • $\begingroup$ I changed the question title to something more eloquent, if you don't mind. $\endgroup$ – Alex Provost Aug 8 '16 at 17:15
0
$\begingroup$

I'll use the equivalent definition of connectedness: $A$ is disconnected iff there exist two closed disjoint subsets whose union is $A$. I'll also use subscripts to denote the space in which a closure is taken.

If $S$ is disconnected by $H$ and $K$, then $$\begin{align}H \cap \overline{K}_X &= (H \cap S) \cap \overline{K}_X \\&= H \cap (S \cap \overline{K}_X) \\ &= H \cap \overline K_S = \emptyset, \end{align}$$ and similarly for $\overline{H}_X \cap K$.

Conversely, if $H,K$ satisfy the criterion, then $$\begin{align}\overline H_S &= S \cap \overline{H}_X \\&=(H \cup K) \cap \overline H_X \\&= (H \cap \overline H_X) \cup (K \cap \overline H_X) = H \end{align}$$ and hence $H$ is closed in $S$. Similarly, $K$ is closed in $S$.

$\endgroup$

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.