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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.

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  • $\begingroup$ Are the negative numbers and the non-negative numbers mutually separated? $\endgroup$
    – Michael Hoppe
    Commented Oct 11, 2018 at 9:57
  • $\begingroup$ No. Their union is R. They are certainly disjoint in R with usual topology but not open. $\endgroup$
    – Ganesh Gani
    Commented Oct 11, 2018 at 11:22
  • $\begingroup$ But you wrote: “... we’re open in their union.” What precisely means “mutually separate”? $\endgroup$
    – Michael Hoppe
    Commented Oct 11, 2018 at 14:46
  • $\begingroup$ 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. $\endgroup$
    – Ganesh Gani
    Commented Oct 11, 2018 at 21:01
  • $\begingroup$ 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 $\endgroup$
    – Ganesh Gani
    Commented Oct 11, 2018 at 21:03

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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.

So the arcwise connected bit is a red herring (i.e. totally unrelated), and I suppose this must be a detail in a larger proof that you were stuck on (?), but I'm afraid it's just a restatement of a definition..

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  • $\begingroup$ Arcwise connectedness cannot be dispensed since you have examples where above claim becomes false without arcwise connectedness. Take trivial topology on two element set. $\endgroup$
    – Ganesh Gani
    Commented Oct 12, 2018 at 3:26
  • $\begingroup$ Moreover I have taken sets to be "mutually separated" which is different from "completely separated". $\endgroup$
    – Ganesh Gani
    Commented Oct 12, 2018 at 3:32
  • $\begingroup$ @user579315 the trivial topology is completely normal and satisfies your statement. $\endgroup$
    – Henno Brandsma
    Commented Oct 12, 2018 at 3:45
  • $\begingroup$ For A and B when mutually separated, they are completely separated in their union, but why would they be completely separated in parent space. For instance, limit point of A may be in B with respect to parent space but not in the subspace AUB $\endgroup$
    – Ganesh Gani
    Commented Oct 12, 2018 at 10:53
  • $\begingroup$ @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. $\endgroup$
    – Henno Brandsma
    Commented Oct 12, 2018 at 15:33

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