# Is a connected space possible if it is Hausdorff? (T2)

Getting into topology this semester and have been studying and doing some exercises and I was wondering if a Hausdorff space i.e: $${\displaystyle \forall \,x,y\in X\,|\,x\neq y\,\exists \,U,V{\text{ open sets of }}x,y\,|\,U\cap V=\varnothing }$$ Could eventually be a connected space ie: X is connected, that is, it cannot be divided into two disjoint non-empty open sets. ,even if at first to me they seem to counter each other there are claims they exists, but I can't find any examples and I just found this very poor wiki link. Of course if anyone has a proof I would more than appreciate it or any literature that could give me an answer

• Every metric space is Hausdorff (for $U$ and $V$ you can take open balls of sufficiently small radius) and many are connected, for example $\mathbb{R}$. Commented Jun 14 at 7:28
• Thank you guys, much more clear now Commented Jun 14 at 7:31
• An example which is not metric space is $\Bbb R$ with $K$-topology. Commented Jun 14 at 7:56

1. Connected + Hausdorff: the real line $$\mathbb{R}$$ with the standard Euclidean topology. Or a single point $$\{x\}$$ with discrete topology $$\tau=\{\emptyset, \{x\}\}$$.
2. Disconnected + Hausdorff: two points $$\{x,y\}$$ with discrete topology $$\tau=\{\emptyset, \{x\}, \{y\}, \{x,y\}\}$$.
3. Connected + non-Hausdorff: two points $$\{x,y\}$$ with Sierpiński topology $$\tau=\{\emptyset, \{x\}, \{x,y\}\}$$. It is similar to discrete, except $$\{y\}$$ is not open.
4. Disconnected + non-Hausdorff: two copies (disjoint union) of Sierpiński space: $$\{x,y\}\sqcup\{x', y'\}$$. Or disjoint union of Sierpiński space with any other non-empty space.