Boundary of $\Bbb{R}$ Is the empty set boundary of $\Bbb{R}$ ? If it is, is it the only boundary of $\Bbb{R}$ ?
I think the empty set is the boundary of $\Bbb{R}$ since any neighborhood set in $\Bbb{R}$ includes the empty set. However, I'm not sure. Besides, I have no idea about is there any other boundary or not.
P.S : It is about my Introduction to Real Analysis course. I haven't taken Topology course yet.
 A: By definition, the boundary of a set $X$ is the complement of its interior in its closure, i.e. $\overline{X} \setminus X_0$. But $\mathbb{R}$ is closed and open, so its interior and closure are both just $\mathbb{R}$. Therefore the boundary is indeed the empty set as you said.
A: The boundary of $\mathbb R$ within $\mathbb R$ is empty.  If $\mathbb R$ is embedded in some larger space, such as $\mathbb C$ or $\mathbb R\cup\{\pm\infty\}$, then that changes.
A boundary point is of a set $A$ is a point whose every open neighborhood intersects both $A$ and the complement of $A$.  Complements are relative: one finds the complement of a set $A$ within a set that includes $A$.  If that set is only $A$ and nothing more, then the complement is empty, and no set intersects the empty set. The complement of $\mathbb R$ within $\mathbb R$ is empty; the complement of $\mathbb R$ within $\mathbb C$ is the union of the upper and lower open half-planes.  The boundary of $\mathbb R$ within $\mathbb C$ is $\mathbb R$; the boundary of $\mathbb R$ within $\mathbb R\cup\{\pm\infty\}$ is $\{\pm\infty\}$.
A: The boundary any set $A \subseteq \Bbb R$ can be thought of as the set of points for which every neighborhood around them intersects both $A$ and $\Bbb R - A$.
Specifically, we should have for every $\epsilon >0$ that $B(x,\epsilon) \cap A \neq \emptyset$ and $B(x, \epsilon) \cap (\Bbb R - A) \neq \emptyset$.  If $x$ satisfies both of these, $x$ is said to be in the boundary of $A$.
So, let's look at the set of $x$ in $\Bbb R$ that satisfy for every $\epsilon > 0$, $B(x, \epsilon) \cap \Bbb R \neq \emptyset$ and $B(x, \epsilon) \cap (\Bbb R - \Bbb R) \neq \emptyset$.
Notice that for the second piece, we are asking that $B(x, \epsilon) \cap \emptyset \neq \emptyset$.  No $x \in \Bbb R$ can satisfy this, so that's why the boundary of $\Bbb R$ is $\emptyset$, the empty set.
