Interior versus Boundary Points Clarification

Question

Let a,b,c be real numbers such that a<c<c and suppose that U = (a,c)u(c,b) is an open interval punctured at c. Does U have interior points? Does U have boundary points? I understand that all points in U are interior points. However, I am struggling with the boundary points. Is it logical to state that there U has no boundary points since all points in U are interior points? Can someone please tell me if my logic is accurate?

Answer 1

Given any set $A$ in a topological space $X$ (which in your example is $\mathbb{R}$), the boundary of $A$ or the frontier of $A$ is defined by:

$\overline{A}\cap \overline{A^{com}}$, where $A^{com}$ denote the complement of $A$.

In your case $A= (a,c)\cup (c,b)$. Thus, $\overline{A}=[a,b]$ and $A^{com}=(-\infty, a]\cup [b, \infty)\cup \{c\}.$ Note that $A^{com}$ is already closed, since $A$ is open. Therefore, $\overline{A^{com}}=(-\infty, a]\cup [b, \infty)\cup \{c\}.$ Now, it is trivial to see that $\overline{A}\cap \overline{A^{com}}=\{a,b,c\}.$ Therefore, the boundary or the frontier of the set $A$ is given by $\{a,b,c\}$.

You can describe the boundary of a set in a metric space in another way too. Given any set $D$ in a metric space $X$, a point $x_0\in X$ is said to be a boundary point of $D$ if each ball containing $x_0$ contains at least one point of $D$ and at least one point of $D^{com}$. Observe that the boundary $\{a,b,c\}$ (of the set $A$) obtained in case of your example, exactly satisfies this property.