Vtyjkyuk

I would like to show the following:

Let $(X,d)$ be a metric space.
If a set and its complement both have empty interior, then the boundary of each of them is $X$.

Could you give me a hint?

See the boundary of any set, say $A$ in $(X, d)$ is equal to $cl(A) setminus int(A)$. So if $A$ has an empty interior then $partial A =cl(A)$ (here, $partial A :=$Boundary of $A$).

Now let $A$ be a given set in $X$ such that $int(A)=emptyset=int(Xsetminus A)$.

To prove: $A$ and $X setminus A$ both has $X$ as their boundary. i.e., $A$ and $X setminus A$ are dense in $X$ (Why?)

Proof Suppose $U$ be a non empty open set in $X$. Now since $int(A) =emptyset$, $U not subset A$ (Recall that interior is the largest open set inside a set and in this case interior of $A$ is $emptyset$) . This means $U cap X setminus A ne emptyset$. Since $U$ was an arbitrary non empty open set in $X$ so it means $cl(Xsetminus A)=X$ but see $cl(Xsetminus A)=partial (Xsetminus A)$.

Similarly you can prove that $partial A=X$.

This completes the proof.

Recall that if $S$ has no interior, then $partial S = barS setminus S^° = barS$. So let's show that $barS = overlineX setminus S = X$ if $S$ and its complement have no interior. Since

$$
barS = bigcap_S subseteq F\ textclosedF
$$

We ought to see that the only closed set containing $S$ is $X$. In effect, take $F$ closed that contains $S$. Now, $X setminus F$ is open and is contained in $X setminus S$ which has empty interior. Thus, $X setminus F$ is empty, and so $F = X$.

Denote the set by $A$. Let $xin X$ and let $B$ be an arbitrary ball around $x$. If $xin A$, then $Bnotsubset A$ because otherwise $xinoperatornameint(A)$. Hence, $B$ contains $xin A$ and a point from $Xbackslash A$. As $B$ was arbitrary, $xinpartial A$. If $xnotin A$, the above reasoning applies with the roles of $A$ and $Xbackslash A$ exchanged.

Other approach: By definition of the boundary, $partial A = partial(Xbackslash A)$. For any set $S$ we have $overline S = partial Scupoperatornameint(S)$. So, $Asubsetoverline A = partial A$ and, similarly, $Xbackslash Asubsetpartial(Xbackslash A)$. So, $X = Acup(Xbackslash A)subsetpartial A$.

In any topological space $X,$ for any $Asubset X$ we have $overline A=Int(A)cup partial (A).$ And of course with $A^c=Xbackslash A$ we have $overline A^c=Int(A^c)cup partial (A^c).$

And $partial (A)=overline A cap overline A^c=partial (A^c)$ by definition of $partial.$

So if $Int(A)=emptyset=Int (A^c)$ then $overline A=partial (A)=partial (A^c)=overline A^c.$ Hence $$X=Acup A^csubset overline Acup overline A^c=$$ $$=partial (A)cup partial (A^c)=partial (A)=partial (A^c)subset X.$$