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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.

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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
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    $\begingroup$ 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
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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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