# $[0,1]$ in $\mathbb{R}$ with the finite complement topology

Introduction to Topology by Adams states:

On the real line, $\mathbb{R}$, define a topology whose open sets are the empty set and every set in $\mathbb{R}$ with a finite complement.

and

In the finite complement topology on $\mathbb{R}$, every infinite set is dense. Why? In this topology, the closed sets are either finite sets or $\mathbb{R}$ itself. Therefore $\mathbb{R}$ is the only closed set containing an infinite set. Thus, if $A$ is an infinite subset of $\mathbb{R}$, then $Cl(A) = \mathbb{R}$, implying that $A$ is dense in $\mathbb{R}$.

If $A=[0,1]$ which is finite, then $Cl(A)=A$.

Also, $int(A)=\emptyset$ because there are no nonempty open sets contained in $[0, 1]$. Finally, $Bd(A)=\mathbb{R}$. Are my understanding correct?

• $[0,1]$ is not finite - there are infinitely many points in $[0,1]$. It is bounded, but it is not finite. – Thomas Andrews Dec 29 '16 at 14:02
• Should have been $A$ not $V$ - careless typo... Then $Cl(A)=\mathbb{R}$, and the other two - $int(A)=\emptyset$ and $Bd(A)=\mathbb{R}$, are correct? – P Haggerty Dec 29 '16 at 14:09
• I don't quite recall how $\mathrm{Bd}$ is defined, but yes, $\mathrm{Cl}(A)=\mathbb R$ and $\mathrm{int}(A)=\emptyset$. – Thomas Andrews Dec 29 '16 at 14:14
• I think you are confusing the interval $[0,1],$ which is $\{x: 0\leq x\leq 1\}$, with the 2-member set $\{0,1\}.$ – DanielWainfleet Dec 30 '16 at 2:48
• @ThomasAndrews If $Bd$ means boundary, then $Bd(A)= \overline A \cap \overline {A^c},$ where $A^c$ is the complement of $A$ in the space. Equivalently $Bd(A)=\overline A$ \ $int (A).$ More commonly denoted by $\partial A$.... I have also seen it called $Fr(A).$ (For "fringe") – DanielWainfleet Dec 30 '16 at 2:55