0
$\begingroup$

Is the notion of separation and separated sets interchangeable? I was looking at the following problem: let $A,B$ be separated subsets of a space $X$. If $C$ is a connected subset of $A \cup B$, then either $C \subset A$ or $C \subset B$.

The definition of separated sets given to me is that $A,B$ are separated sets if $A,B$ are disjoint nonempty and $A \cap \overline{B}=\overline{A} \cap B=\varnothing$.

To solve the problem in question I think it would be easier to assume $A,B$ are disjoint nonempty open sets whose union is $X$. Then if $C$ is not entirely contained in $A$ or $B$, $A \cap C$ and $B \cap C$ are a pair of disjoint nonempty open sets with union $C$. This contradicts the assumption that $C$ is connected.

So it seems easier to use the nonempty open sets to prove this, not two separated sets. My question is, is it valid to interchange the notion of disjoint nonempty open sets constituting a "separation" of a space with separable sets in a space? If not how would this proof go using the definition of two separated sets?

$\endgroup$
2
  • 1
    $\begingroup$ Unless you are able to prove that the condition $A\cap\bar B=\bar A\cap B=\emptyset$ implies $A,B$ are open (which is by the way not true, say $A=[1,2],B=[3,4]\subseteq\Bbb R$), I don't think it is valid. $\endgroup$ May 11 '21 at 5:39
  • 1
    $\begingroup$ $A$ and $B$ are open in $A \cup B$, not always in $X$. So the argument can be adapted. $\endgroup$ May 11 '21 at 6:26
3
$\begingroup$

First of all, separable (for a space $X$ or a subspace $A$) means that it has a countable dense subset, and has nothing to do with connectedness properties. So keep the term separable out of it, please.

$A$ and $B$ are separated in $X$ iff $\overline{A} \cap B = \emptyset = A \cap \overline{B}$ (i..e no point of $A$ is "close to" $B$ and vice versa). For disjoint sets $A$ and $B$ this means the same as saying that $A$ and $B$ are closed-and-open (clopen) in $C=A \cup B$ (in the subspace topology); this is easy to see. A space being disconnected is defined as being able to write $X$ as the union of two non-empty separated sets; and connectedness is defined as "not disconnected".

So if $S$ is connected and $S \subseteq A \cup B$, $S$ cannot contain points from $A$ and $B$ both, because that would indeed mean that $S \cap A$ and $S \cap B$ are relatively clopen non-empty subsets of $S$ partioning $S$, contradicting connectedness of $S$. So indeed $S \subseteq A$ or $S \subseteq B$ holds.

$\endgroup$
3
  • $\begingroup$ Henno, would the following approach be ok? Since $A,B$ are open in $A \cup B$ and $C \subset A \cup B$. We know that $A \cap C$ and $B \cap C$ are open in $C$. But then if both of these intersections are nonempty, $A \cap C$ and $B \cap C$ would be a separation of $C$, contradicting the connectedness of $C$? $\endgroup$
    – ernesto
    May 12 '21 at 2:00
  • $\begingroup$ @PatrickR separated trivially implies disjoint, yes. $\endgroup$ Jul 10 '21 at 13:03
  • 1
    $\begingroup$ @PatrickR It's more accurate to say that two disjoint sets are separated if they're both clopen in their union, or some such formulation, yes. I added this proviso in the body too. $\endgroup$ Jul 11 '21 at 7:36
1
$\begingroup$

No, it is not legitimate: it does not follow from the hypotheses that you were given. For example, let $X=\Bbb R$ with the usual topology, $A=[0,1)$ and $B=(1,2]$; then $A$ and $B$ are separated sets in $X$, but they are not open, and their union is not $X$. It is nevertheless true that if $C$ is a connected subset of $A\cup B=[0,1)\cup(1,2]$, then $C\subseteq A$ or $C\subseteq B$.

Suppose that $C\subseteq A\cup B$, and let $C_A=C\cap A$ and $C_B=C\cap B$; clearly $C_A\cap C_B=\varnothing$, and $C_A\cup C_B=C$. Let $U=X\setminus\operatorname{cl}A$ and $V=X\setminus\operatorname{cl}B$; $U$ and $V$ are open sets in $X$.

  • Verify that $C_A=V\cap C$ and $C_B=U\cap C$, so that $C_A$ and $C_B$ are open in the subspace $C$.

Since $C$ is connected, it follows that (at least) one of $C_A$ and $C_B$ must be empty and hence that $C\subseteq A$ or $C\subseteq B$.

$\endgroup$
6
  • $\begingroup$ Would $A=X- \text{cl} \ B$ and $B=C- \text{cl} \ A$ be the reason why for the verification? $\endgroup$
    – ernesto
    May 11 '21 at 6:37
  • 1
    $\begingroup$ @ernesto: Not quite, because $A\cup B$ need not be all of $X$. But $$V\cap C=(X\setminus\operatorname{cl}B)\cap C=(X\cap C)\setminus\operatorname{cl}B=C\setminus\operatorname{cl}B=C_A\,,$$ and a similar calculation takes care of $C_B$. $\endgroup$ May 11 '21 at 18:19
  • $\begingroup$ I mean $B=X-\text{cl} A$ and would my statement above this comment hold for all separable sets, that are disjoint, nonempty, whose union is $X$? $\endgroup$
    – ernesto
    May 11 '21 at 18:19
  • $\begingroup$ @ernesto: But $B$ in general isn’t $X\setminus\operatorname{cl}A$. If $A\cup B=X$, though, you’re right: $A$ and $B$ are then disjoint open sets whose union is $X$. $\endgroup$ May 11 '21 at 18:20
  • 1
    $\begingroup$ @ernesto: That’s correct. However, one can let $Y=A\cup B$ and conclude that $A$ and $B$ are relatively open sets in the subspace $Y$; this shows immediately that $Y$ is not connected. $\endgroup$ May 11 '21 at 18:26

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.