Separated sets proof

Question

For sets $X, Y$ and $Z$, if $X$ and $Y$ are separated, and $X$ and $Z$ are separated, then $X$ and $Y \cup Z$ are also separated.

I am reading up on some set theorems and I am looking for a complete proof of this one.

What I know:

If $X$ and $Y$ are separated then: $X \cap \overline{Y} = \overline{X} \cap Y = \emptyset$.

In the same way $X \cap \overline{Z} = \overline{X} \cap Z = \emptyset$.

It needs to be shown that $X \cap \overline{Y \cup Z} = \overline{X} \cap (Y \cup Z) = \emptyset$.

Any assistance with a proof is much appreciated.

@TomCollinge Could you elaborate? I have limited experience with sets and metric spaces... If you leave it as an answer, I can mark it as the accepted one. — SupremePickle, Mar 07 '21 at 19:56
Not really worth rating as an answer. Pure topology, not metric space involvement. The closure of the union is the union of the closures.You should complete "It needs to be shown ..." from that. — Tom Collinge, Mar 07 '21 at 19:59
@TomCollinge So would it suffice to say $X \cap \overline{Y \cup Z} = X \cap (\overline{Y} \cup \overline{Z}) = (X \cap \overline{Y}) \cup (X \cap \overline{Z}) = \emptyset$? — SupremePickle, Mar 07 '21 at 20:02
Sufficiency depends on context. If you already have the closure result established then yes, otherwise you might need to include proof of that. The first answer in the link gives a very clear and simple proof. — Tom Collinge, Mar 07 '21 at 20:05

score 1 · Answer 1 · answered Mar 07 '21 at 22:46

$X \cap \overline{(Y \cup Z)} = X \cap (\overline{Y} \cup \overline{Z}) = (X \cap \overline{Y}) \cup ( X \cap \overline{Z}) = \emptyset \cup \emptyset= \emptyset$

and

$\overline{X} \cap (Y \cup Z) = (\overline{X} \cap Y) \cup (\overline{X} \cap Y) = \emptyset \cup \emptyset= \emptyset$

does the job. We only really need that closure commutes with finite unions for the first, plus the distributive properties of union and intersections.

Separated sets proof

1 Answers1