Are singletons open sets? I've recently been learning about metric spaces and one of the very important definitions are of open sets and open balls.
I have a question that was raised when reading on a website that a singleton in a metric space $(X,d)$ such as $[p]$ is considered to be an open set for example let the  set be $[0,1]$ and the metric be the standard metric is it true to say that the singleton $[1]$ is an open set? Since the $B_r$(1)=$[1]$ and this open ball is contained in the set.
However is this true for all singletons in the metric space $(\mathbb{R},d)$.
I'm new to metric spaces and I'm sure this is a very trivial question for most of you but its really confusing me.
Thanks in advance.
 A: You define a metric space by $(X,d)$ where $X$ is a non-empty set and $d$ is the distance function.  In the metric $(X,d),\,\, X$ is the universal set. So $X$ is always an open set. Now if you take $X$ as a singleton set then $X$ is always open.
Consider the Discrete metric space(trivial metric space) with $X=\mathbb Z$ or any subset of $\mathbb  Z$. If you take any $0<r<1$ then every singleton set consisting a single integer is open in $X$.
In the usual metric, (Euclidean metric of degree 1) $(\mathbb R,d)$ no singleton set is open.
A: No, the singleton set $\{1\}$, as a subset of the real line, is not an open set. Formally, this is because any open ball centered at $1$ must contain some number greater than and less than $1$.
When considering subsets of the real line, we can think of open sets as "having no borders;" they're the "opposite" of closed sets, which have a clearly defined boundary. It's the difference between the open interval $(a, b)$ and the closed interval $[a, b]$. One has endpoints, one doesn't.
A singleton set $\{x\}$ has boundaries, namely itself. It's the same as the closed interval $[x, x]$.
However, just note that this is not true in general metric spaces, namely the discrete metric space. In fact, a topology on any set $X$ is equal to the discrete topology on $X$ if and only if every singleton set of $X$ is open.
A: Singleton sets on the real line are not open with the standard metric. To see this let $\{a\}$ be your singleton set, and construct a ball such that $B(a,r)\subseteq\{a\}$ or in other words with the metric we are working with the set of points $x$ such that $a-r<x<a+r\subseteq\{a\}$ but this is a contradiction since $r>0$
