Let $\mathbb{R}^2$ be our set and let the topology $\tau$ be given by the smallest topology such that the intersection of two lines is open.
So I take this to mean singletons are open so for any $\vec{x} \in \mathbb{R}^2$, $\{\vec{x}\} \in \tau$.
So the first question I have is that, do we still have for any $\epsilon>0$ and $x \in \mathbb{R}^2$ that $B_\epsilon(x) \in \tau$? (i.e., is it standard TOGETHER with one point sets being open or?)
Then I need to tell if $(\mathbb{R}^2,\tau)$ is 1st countable? And 2nd countable?
Does the first one mean having countable basis at each point? And the second one just says we have a countable basis?
Any hints would help.
Best Answer
You're right, it implies that all sets $\{x\}$, where $x \in \Bbb R^2$, is open in $\tau$.
But recall that all unions of opens sets are open too, so if $A \subseteq \Bbb R^2$ is any subset, $A = \bigcup\{\{x\}\mid x \in A\}$ is also open in $\tau$ so that $\tau$ is the powerset of $\Bbb R^2$; the topology is discrete.
So $\Bbb R^2$ is first countable trivially, every point $x$ has a local base $\{\{x\}\}$, which is finite, so countable.
For every base for $\tau$ we must have $\{x\}$ in the base for any $X$, so no countable base exists for $\tau$ and the space is not second countable.