# In arcwise connected and completely normal space, two mutually separated sets can be enclosed by two disjoint neighbourhoods

How do I prove or disprove above claim ? These are the definitions of Completely normal space, arcwise connected space and mutually separated sets.

Normal and Completely Normal spaces

Arcwise Connected space

Mutually Separated: Two sets A and B of M are said to be mutually separated if they are disjoint and open in their union.

• Are the negative numbers and the non-negative numbers mutually separated? Commented Oct 11, 2018 at 9:57
• No. Their union is R. They are certainly disjoint in R with usual topology but not open. Commented Oct 11, 2018 at 11:22
• But you wrote: “... we’re open in their union.” What precisely means “mutually separate”? Commented Oct 11, 2018 at 14:46
• If you take two sets A and B from parent space M such that they are disjoint, if they are open in union of A and B as subspace topology then they are mutually separated. Commented Oct 11, 2018 at 21:01
• In your example, union of A and B is R(space of real numbers). A and B are disjoint but set of non negative real numbers is not open since 0 is not interior point of B in R Commented Oct 11, 2018 at 21:03

The definition of being completely normal is that two completely separated subsets of $$X$$ can be separated by disjoint neighbourhoods, where your stated definition that $$A$$ and $$B$$ are mutually separated iff they are open and disjoint in their union is equivalent to the more standard one that $$A$$ and $$B$$ are totally separated, I.e. $$A \cap \overline{B} = \emptyset = \overline{A} \cap B$$ (that Wikipedia also uses). A small moment's thought will convince you of this.
• @user579315 if $A$ and $B$ are mutually separated then there is an open set $O$ such that $O \cap (A \cup B)=A$ and this implies that no point of $A$ is a limit point of B$, so$A\cap \overline{B}=\emptyset\$. Etc. Commented Oct 12, 2018 at 15:33