Is it appropriate to ask the community to check my proof? I am rereading Munkres Topology and trying to do the HW. This is my attempt for #10 on page 101.

Show that every order topology is Hausdorff.


Spse that $x_1, x_2$ are elements of $X$, and $x_1 < x_2$.

Case 1:
Spse that $x_2$ is the next element after $x_1$, and suppose that $$x_0 < x_1 < x_2 < x_3.$$ Then $x_1$ is an element of $(x_0, x_2) = U_1$. Then $x_2$ is an element of $(x_1, x_3) = U_2$. $U_1$ and $U_2$ are disjoint.

Case 2: Spse that there might not be "next elements" in $X$ but $x_1 < x_2$. Then there exists $$a < x_1 < b < x_2 < c.$$ Then $x_1$ is an element of $(a, b) = U_1$. Then $x_2$ is an element of $(b, c) = U_2$. $U_1$ and $U_2$ are disjoint.

Thus $X$ is Hausdorff.

Is this on the right track?

  • 3
    $\begingroup$ You really can't be bothered to type the other three letters in "suppose"? $\endgroup$ – MJD Oct 15 '13 at 2:16
  • $\begingroup$ Sorry about that. I guess it is just a habit....... $\endgroup$ – Kara Oct 15 '13 at 2:20
  • $\begingroup$ Note that actually it holds that order topology is hereditarily normal. $\endgroup$ – user87690 Oct 15 '13 at 9:17

It’s almost right: in Case $1$ you forgot to cover the cases in which $x_1$ or $x_2$ is an endpoint of the space. However, there’s no need to treat these separately if you modify your proof a little. I’d arrange it like this:

Let $x,y\in X$, and assume without loss of generality that $x<y$. If there is a $z\in X$ such that $x<z<y$, let $U=(\leftarrow,z)$ and $V=(z,\to)$; then $x\in U$, $y\in V$, and $U\cap V=\varnothing$. If there is no such $z$, let $U=(\leftarrow,y)=(\leftarrow,x]$ and $V=(x,\to)=[y,\to)$; once again $x\in U$, $y\in V$, and $U\cap V=\varnothing$, and $X$ is therefore Hausdorff.

Here $(z,\to)=\{u\in X:u>z\}$, $(\leftarrow,z)=\{u\in X:u<z\}$, $[z,\rightarrow)=\{u\in X:u\ge z\}$, and so on.

| cite | improve this answer | |

Looks almost correct to me.

It might be a bit tighter to phrase you first case not in terms of "next elements," but in terms of whether a there exists an element $b$ such that $x_1 < b < x_2$.

Also, you don't cover cases where $x_1$ is minimal or $x_2$ is maximal. Those should be easy to do from the definition of the order topology, though.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.