There is a theorem which says that every order topology is Hausdorff. Also every Hausdorff follows $T_1$ Axiom.

So suppose $X = \{1,2\}$. Now $1 < 2$. A basis for the topological space $X$ is $B_1=[a,b)$ and $B_2=(a,b]$. We have an open neighborhoods $B_1$ and $B_2$ around $1$ and $2$ respectively, such that $B_1 \cap B_2 = \emptyset$. So $X$ is a Hausdorff space.

But since it also satisfies the $T_1$ Axiom it means $B_1$ and $B_2$, which are singleton sets, are closed. That means $B_1$ and $B_2$ are not open neighborhoods.

If anyone can please tell where am I getting confused?

  • 1
    $\begingroup$ $\{1\}$ and $\{2\}$ are both open and closed at the same time. $\endgroup$ – zozoens Feb 4 '14 at 8:31
  • $\begingroup$ oh.. yes! Thanx a lot :) $\endgroup$ – Richard J Feb 4 '14 at 8:45

As mentioned by zozoens in the comments, the sets $\{1\}$ and $\{2\}$ are both open and closed.

To see this, note that $B_1 = [1, 2) = \{x \in X \mid 1 \leq x < 2\} = \{1\}$ and $B_2 = (1, 2] = \{x \in X \mid 1 < x \leq 2\} = \{2\}$. By the definition of the order topology, $B_1$ and $B_2$ (and hence $\{1\}$ and $\{2\}$) are open. But then $\{1\} = X\setminus B_2$ and $\{2\} = X\setminus B_1$, so they are both closed as well.

Note that $X = \{1\}\cup\{2\}$, with $\{1\}$ and $\{2\}$ both open, so $X$ is disconnected; in fact, as the connected components of $X$ are the singleton sets, $X$ is totally disconnected. In general, the order topology on a finite totally ordered set is totally disconnected.


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