Example of a bounded set with no boundary

Question

I read that a "circle in isolation" is bounded but does not have a boundary, but I don't know what "circle in isolation" means. I know both an open circle and a closed circle is bounded in $\mathbb{R^2}$ since you can create a larger open circle around either one, but both also seem to have a boundary as well, since the circumference is the boundary in both cases.

Answer 1

The empty set is an example of such a set. In fact, in Euclidean spaces (or any connected metric space with an unbounded metric), it is the only example, since a set without boundary is necessarily both closed and open.

I'm not sure what "a circle in isolation" may mean.

Answer 2

In topology, the boundary of a set is defined relative to another space, which is either the same space or a larger space. Specifically, the closure of $A$ in $B$ is the intersection of all sets which are closed in $B$ and contain $A$. The interior of $A$ in $B$ is the union of all sets which are open in $B$ and are contained in $A$. The boundary of $A$ in $B$ is the closure minus the interior.

Meanwhile, a subset of a given topological space can be made into a topological space in its own right. Specifically, if $A$ is a subset of $B$ and $B$ is a topological space then we can put a topology on $A$ which consists of all sets $U \cap A$ where $U$ is open in $B$. This is called the subspace topology. It is primarily useful because it makes the inclusion map $f : A \to B$ defined by $f(x)=x$ continuous.

A "circle in isolation" is being assigned the subspace topology from $\mathbb{R}^2$, and is regarded as a subset of itself. It has no boundary because it is both open and closed in itself, which means its interior and closure are the same.

A circle as a subset of $\mathbb{R}^2$ has a boundary, however: it contains no open sets, so it has no interior, and it is a closed subset of $\mathbb{R}^2$, so it is its own closure. Hence the boundary of the circle (as a subset of $\mathbb{R}^2$) is the circle itself.

Answer 3

An example would be $\{1,2,3\}$, viewed as a subset of $\mathbb{R}$, since you could take any neighborhood of radius $r < 1$, then the neighborhood doesn't intersect the set, but intersects its complement, and it's clearly bounded.