Is the boundary of singleton set in $T_1$ space is empty? The singleton set in $T_1 $ topological space is closed so, I wonder how i use this information to prove that boundary of this set is empty.
 A: $(\Bbb{R}, \tau_{usual}) $ is a $T_1$ space.
$bdd(\{x\}) =cl\{x\} \setminus int\{x\}=\{x\}$
$(X, \tau) \space  T_1 $ space  implies $\{x\}$ is closed.
But $\{x\}$ need not be open.
If $\forall x\in X $ , $\{x\}$ open implies $(X, \tau) $ is a discrete space.
Hence, discrete space is the only example of a $T_1$ space in which every singleton set has empty boundary.
A: Boundary is defined as $\partial A=\overline{A}\cap\overline{X\backslash A}$. Now for any $x\in X$ we have $\overline{\{x\}}=\{x\}$ since the space is $T_1$.
Now if $\{x\}$ is not open, then $X\backslash\{x\}$ is not closed and thus $\overline{X\backslash\{x\}}=X$. Which means that if $\{x\}$ is not open then $\partial\{x\}=\{x\}$.
On the other hand if $\{x\}$ is open, then its complement is closed, i.e. $\overline{X\backslash\{x\}}=X\backslash\{x\}$ and thus $\partial\{x\}=\emptyset$.
All in all, in $T_1$ space, for any $x\in X$ we have:
$$\partial\{x\}=\begin{cases}
\{x\}&\text{if }\{x\}\text{ is not open}\\
\emptyset &\text{otherwise}
\end{cases}$$
