Is $X:=\{a\}$ Hausdorff space?

Question

I wonder if $X:=\{a\}$ is a Hausdorff space.

Definition;

$A$ is a Hausdorff space $\iff \forall x,y \in A(x\neq y); \exists \text{ open set } U,V s.t. x\in U,y\in V, U\cap V=\phi.$

But we cannot pick up two elements from $X=\{a\}.$

How should I judge whether $X$ is a Hausdorff space or not?

Answer 1

It is a Hausdorff space since that condition is trivially true, by the reason that you have mentioned: there are no two distinct elements on $X$. By the same argument, every function whose domain is a singleton is injective.

Answer 2

It becomes easier when you understand what $\forall x,y\in A(x\neq y)$ really means here. This is an example where the shorthand notation may cause confusion (whereas it usually comes to clarify over-formality).

When we write $(\forall x\in A)\varphi(x)$, what we really mean to say is $\forall x(x\in A\to\varphi(x))$, so that if $x\in A$, then $\varphi(x)$, but if $x\notin A$, then the statement is vacuously true. This extends to the $(x\neq y)$ part in this case. Since we're quantifying over elements of $A$ anyway let me omit it from the notation at the moment, and just write: $$\forall x,y(x\neq y\to\exists U,V(U,V\text{ are open}\land x\in U\land y\in V\land U\cap V=\varnothing)).$$

In other words, if $x\neq y$, then there are disjoint open sets, etc. Now it's easy to see what happens with a singleton: the implication is vacuously true, there are never any $x\neq y$.