Topology: Show that $\partial A=\emptyset$

Question

Problem: Let $X$ be a topological space and let $A\subset X$ such that $A$ is both open and closed. Show that $\partial A = \emptyset$

My proof: Let $A=∅$ since it is known that the empty set is a subset of every set and therefore $∅⊂X$.

By definition, $A$ is closed iff $A=Cl(A)$ and $A$ is open iff $Int(A)=A$. Since $A=∅$ and $Cl(A)=Cl(∅)=∅=A$, then $A$ is closed. And since $Int(A)=Int(∅)=∅=A$, then $A$ is open, therefore $A$ is both open and closed or said to be clopen.

Since by definition $b(A)=∂A=Cl(A)∩A^∁$ and the complement of $A$ is $A^∁=X$\ $A =X$\ $∅$, then $b(A)=∂A=Cl(A)∩A^∁=∅∩X=∅$

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Did I prove it correct?

Answer 1

You cannot just assume $A = \varnothing$, because that is not the only open and closed subset of $X$. Another example is $X$ itself; if your topological space is disconnected, there will be others.

There is a standard way to show this using the definition of boundary as difference between closure and interior. Since $A$ is both open and closed, it is equal to both its interior $\mathrm{it}(A)$ and its closure $\mathrm{cl}(A)$. But then $$\partial A = \mathrm{cl}(A) \setminus \mathrm{it}(A) = A \setminus A = \varnothing.$$

In a less direct way, you have the following two facts:

• A set is open iff it is disjoint from its boundary;
• A set is closed iff it contains its boundary.

If $A$ is both open and closed, then both $A \cap \partial A = \varnothing$ and $A \cap \partial A = \partial A$ hold. Hence you may equate $\partial A = \varnothing$.