If a set and its complement both have empty interior, then the boundary of each of them is $X$

Question

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?

Answer 1

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 $int(A)=\emptyset=int(X\setminus A)$.

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 $cl(X\setminus A)=X$ but see $cl(X\setminus A)=\partial (X\setminus A)$.

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

This completes the proof.

Answer 2

Recall that if $S$ has no interior, then $\partial S = \bar{S} \setminus S^° = \bar{S}$. So let's show that $\bar{S} = \overline{X \setminus S} = X$ if $S$ and its complement have no interior. Since

$$ \bar{S} = \bigcap_{S \subseteq F\\ \text{closed}}F $$

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 $F = X$.

Answer 3

Denote the set by $A$. Let $x\in X$ and let $B$ be an arbitrary ball around $x$. If $x\in A$, then $B\not\subset A$ because otherwise $x\in\operatorname{int}(A)$. Hence, $B$ contains $x\in A$ and a point from $X\backslash A$. As $B$ was arbitrary, $x\in\partial A$. If $x\notin A$, the above reasoning applies with the roles of $A$ and $X\backslash A$ exchanged.

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Other approach: By definition of the boundary, $\partial A = \partial(X\backslash A)$. For any set $S$ we have $\overline S = \partial S\cup\operatorname{int}(S)$. So, $A\subset\overline A = \partial A$ and, similarly, $X\backslash A\subset\partial(X\backslash A)$. So, $X = A\cup(X\backslash A)\subset\partial A$.

Answer 4

In any topological space $X,$ for any $A\subset X$ we have $\overline A=Int(A)\cup \partial (A).$ And of course with $A^c=X\backslash A$ 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 $$X=A\cup A^c\subset \overline A\cup \overline {A^c}=$$ $$=\partial (A)\cup \partial (A^c)=\partial (A)=\partial (A^c)\subset X.$$