I know this is a basic question, but I am having trouble proving that particular topological spaces are/are not Hausdorff and was wondering if I could get some guidance. For example, I have to decide whether or not the half-open topology is Hausdorff and then prove my decision. I think that it is, and my proof is as follows:

Let $U, V \subset H'$ such that $x \in U$ and $y \in U$ with $x \not= y$. THen for $a,b,c \in \mathbb{R}$, we can construct $U = [a,b)$ and $V=[b,c)$, which are disjoint sets. Thus for any $x,y \in \mathbb{R}$, there exist two open sets $U$ and $V$ that contain $x$ and $y$ such that $U \cap V = \emptyset$. It follows that $H'$ is Hausdorff.

Basically, I feel like this proof is wrong or incomplete or something. Could anyone give me any pointers about this proof and how to go about proofs like this in general? Thanks.


migrated from mathoverflow.net Feb 13 '14 at 21:31

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  • $\begingroup$ Perhaps you want a<x<b<y<c? $\endgroup$ – David Steinberg Feb 13 '14 at 21:23
  • 2
    $\begingroup$ It's hard to tell what you're doing. Instead, find $U$ and $V$ explicitly. For example, if $x<y$ are given, you can set $U=[x,y)$ and $V=[y,y+1)$. $\endgroup$ – David Mitra Feb 13 '14 at 21:38
  • $\begingroup$ So let's take $x=3$ and $y=4$. How does it help me to know that I can construct the sets $[10,11)$ and $[11,12)$ ? $\endgroup$ – WillO Feb 14 '14 at 13:21

It's hard to tell what you are doing. You have to find explicit open sets $U$ and $V$ depending on $x$ and $y$ which are disjoint and contain $x$, $y$, respectively. You can assume that $x<y$, then one of the sets could be $[y,\infty)$ or $[y,y+1)$, the other set could be $[x,y)$. Note also that the complement of a basic open set $[a,b)$ is again open as it's the union of the open basic open sets $$\bigcup_{c<a} [c,a)\cup\bigcup_{b<d} [b,d)$$ Since for any two points $x<y$, the basic set $[x,y)$ contains $x$ but not $y$, there is always a separation of the space between $x$ and $y$. A space with this property (which is stronger than the Hausdorff property) is called totally separated.

If you want to practice some problem solving like this, I suggest that you try and show that $X$ is also normal, i.e. for disjoint closed sets $A$ and $B$ there are disjoint open neighbourhoods. The proof is not very difficult.


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