Is $\mathbb{R^2}$ Hausdorff? Give an example of a non-Hausdorff topology on $\mathbb{R}$ these are two questions on Hausdorff topological spaces. The bit I am having particular difficulty with is finding an 'example of a non-Hausdorff topology on $\mathbb{R}$'

A Hausdorff topological space $(X, \tau)$ is such that any distinct points $a, v \in X$ have disjoint open neighbourhoods. i.e. there are open neighbourhoods $U_a, V_b \in \tau$ such that $ a \in U_a  $ and $b \in V_b $ and $U_a \cap V_b = \emptyset$

Is $\mathbb{R^2}$ Hausdorff? 

I believe so. Take $a, b \in \mathbb{R^2}$. Take open neighbourhoods:
$U_a=B_{r_a}(a)=\{(x, y) : |(x, y)-a|<r_a\}$
$V_b=B_{r_b}(b)=\{(x, y) : |(x, y)-b|<r_b\}$
Let $r=d(a,b)$. Take $r_a=r_b=\frac{r}{2}$
So $\mathbb{R^2}$ is Hausdorff.
Is this correct?

Give an example of a non-Hausdorff topology on the set of real numbers

$\mathbb{R}$ is clearly Hausdorff. What is another example of a topology on $\mathbb{R}$ ?
Please could you help me with this one?
 A: There are a few "natural" examples, e.g.:


*

*the trivial topology already mentioned in Mandrathax answer;

*the cofinite topology, where the closed sets are precisely $\Bbb R$, $\emptyset$ and all finite subsets;

*the topology whose non-trivial open sets are the right halflines $(a,\infty)$;

*as the latter but with the left halflines $(-\infty,b)$.

A: Recall that a topology on a set E (here $\mathbb R^2$) is defined by a subset of $\mathcal P(E)$ with special properties (contains $\emptyset$ and E, stable by union and finite intersection) called the set of open sets of E.
Now it appears the simplest possible topology on $\mathbb R^2$, namely $\{\mathbb R^2,\emptyset\}$ (called the trivial topology) perfectly fits your needs, the only neighborhood of any point is $\mathbb R^2$ itself!
To say it with your words, in this topology, given $a,b\in\mathbb R^2$, the only neighborhood $U_a$ of $a$ and $U_b$ of $b$ you can find is $U_a=U_b=\mathbb R^2$ and in this case you definitely don't have $U_a\cap U_b=\emptyset$ ! So it's impossible to separate any two points given this topology.
A: The cofinite topology on a set $X$, which is defined by $$\tau = \{ A \subseteq X \, : \, A = \varnothing \text{ or } X \setminus A \text{ is finite} \}$$ is non-Hausdorff (if $X$ is infinite). 
