Show that R is Hausdorff
I have a proof for it
I want to know is there anyway to do it
using set theory alone
For instance the distance between two sets , say U and V the formula is inf{||x-y||,x$\in $U,y$\in$V}
I don’t want to rely on the above too
Any hints would be nice? Thanks
To show that $\Bbb R$ is Hausdorff, you can use different approaches: it is a metric space $(X,d)$ and any metric space is Hausdorff:
Let $x \neq y \in X$. Then $r:=d(x,y)>0$ by properties of the metric. Define $s=\frac{r}{2}>0$. Then $U=B_s(x)$ is an open neighbourhood of $x$ and $V = B_s(y)$ is an open neighbourhood of $y$ and $U \cap V = \emptyset$, for otherwise there would be some $z \in B_s(x) \cap B_s(y)$, so $d(z,x) < s$ and $d(z,y) < s$ and then
$$d(x,y) \le d(x,z) +d(z,y) < s + s =2s = r= d(x,y)$$
which is a contradiction. Hence $(X,d)$ is Hausdorff in the metric topology.
Aother approach is to use that the topology on $\Bbb R$ is induced by its linear order: and $(X,<)$ in the order topology is Hausdorff:
Let $x \neq y$ in $X$. WLOG we can assume $x < y$ (or we interchange the roles of $x$ and $y$).
Case 1:if there is some $z \in X$ such that $x < z < y$ (as will be the case in $\Bbb R$ or $\Bbb Q$, e.g.), then $x \in U:=(\leftarrow, z)$ and $y \in V:=(z,\rightarrow)$ are subbasic open in the order topology and $U \cap V = \emptyset$.
Case 2: if no such $z$ exists, we can define $U:=(\leftarrow, y), V := (x, \rightarrow)$ to be difjoint open (sub)basic open neighbourhoods of $x$ resp. $y$.
So we can prove the fact that $\Bbb R$ is Hausdorff on two possible ways to generate the topology on this set, either by metric (which is Morris' preferred way) or by order.