# Is totally disconnected space, Hausdorff? [duplicate]

Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Is a totally disconnected space, Hausdorff?

I think it is true since if $a$ and $b$ are two distinct points, they can be separated two disjoint open sets, since the main space is totally disconnected (see Theorem at < http://www.emathzone.com/tutorials/general-topology/totally-disconnected-space.html >).

Is this argument true?

## marked as duplicate by user228113, Shailesh, Namaste, zz20s, carmichael561May 9 '17 at 3:15

Let $X=\mathbb{N}\cup\{a,b\}$, where a subset of $X$ is open iff it either is contained in $\mathbb{N}$ or it contains all but finitely many points of $\mathbb{N}$. Then $X$ is not Hausdorff (you can't separate $a$ and $b$). However, it is totally disconnected: suppose $S\subseteq X$ contains more than one point. If $S$ contains a point of $\mathbb{N}$, then that point is clopen and thus $S$ is disconnected. The only other case is where $S=\{a,b\}$, but then $S$ is again disconnected because $\{a\}=S\cap (\mathbb{N}\cup\{a\})$ and $\{b\}=S\cap (\mathbb{N}\cup\{b\})$ are both open in $S$.
• In special case, for a commutative ring with identity, when $Max (R)$ is totally disconnected, can we deduce that it is Hausdorff? – arena Nov 1 '15 at 18:22
• The example $X$ in my answer is compact. I don't know the answer to your second question off the top of my head; I suggest you ask it as a new question. – Eric Wofsey Nov 1 '15 at 18:47